MTH301 — Final Term Summary (Lectures 23–45)

📘 Lecture 23 — Polar Coordinate Systems

📖 Overview: This lecture introduces the polar coordinate system as an alternative to the rectangular (Cartesian) coordinate system for describing points in a plane. It covers the fundamental definitions, the non-uniqueness of polar coordinates, negative radial distances, conversion formulas between polar and rectangular coordinates, and the equations of lines and circles in polar form.

🗂️ Topics Covered

The lecture begins by defining the polar coordinate system with its origin (pole) and polar axis, and introduces the concepts of radial distance and polar angle. It then discusses the non-uniqueness of polar coordinates and the special definition for negative values of r, including their geometric interpretation. The relationship between polar and rectangular coordinates is established through conversion formulas, followed by examples. The lecture concludes with the derivation of equations for lines and circles in polar coordinates, including several special cases.

📝 Lecture Summary

POLAR COORDINATE SYSTEMS

To form a polar coordinate system in a plane, we pick a fixed point O, called the origin or pole, and using the origin as an endpoint we construct a ray, called the polar axis. After selecting a unit of measurement, we may associate with any point P in the plane a pair of polar coordinates (r, θ), where r is the distance from P to the origin and θ measures the angle from the polar axis to the line segment OP. The number r is called the radial distance of P and θ is called a polar angle of P. Points like (6, 45°), (3, 225°), (5, 120°), and (4, 330°) are plotted in polar coordinate systems.

🔑 Definition — Polar Coordinates: A point P is represented by (r, θ), where r is the distance from the pole and θ is the angle measured from the polar axis.

THE POLAR COORDINATES OF A POINT ARE NOT UNIQUE

For example, the polar coordinates (1, 315°), (1, −45°), and (1, 675°) all represent the same point. In general, if a point P has polar coordinate (r, θ), then for any integer n = 0, 1, 2, 3, ..., (r, θ + n·360°) and (r, θ − n·360°) are also polar coordinates of P. In the case where P is the origin, the line segment OP reduces to a point, since r = 0. In this case, we will agree that an arbitrary polar angle may be used. Thus, for every θ, the point (0, θ) is the origin.

NEGATIVE VALUES OF R

When graphing curves in polar coordinates, it will be desirable to allow negative values for r. For motivation, consider point P with polar coordinates (3, 225°). We can reach this point by rotating the polar axis 225° and then moving forward from the origin 3 units along the terminal side of the angle. Alternatively, we can reach the same point P by rotating the polar axis 45° and then moving backward 3 units from the origin along the extension of the terminal side of the angle. This suggests that the point (3, 225°) might also be denoted by (−3, 45°), with the minus sign serving to indicate that the point is on the extension of the angle's terminal side rather than on the terminal side itself.

Since the terminal side of the angle θ + 180° is the extension of the terminal side of angle θ, we define: (-r, θ) and (r, θ + 180°) to be polar coordinates for the same point.

RELATION BETWEEN POLAR AND RECTANGULAR COORDINATES

CONVERSION FORMULA FROM POLAR TO CARTESIAN COORDINATES AND VICE VERSA

The key relationships are:

• x = r cos θ
• y = r sin θ
• x² + y² = r²
• y/x = tan θ (for x ≠ 0)

Example: Find the rectangular coordinates of the point P whose polar coordinates are (6, 135°) Solution: Substituting r = 6 and θ = 135° in x = r cos θ and y = r sin θ yields:

• x = 6 cos 135° = 6 (−√2/2) = −3√2
• y = 6 sin 135° = 6 (√2/2) = 3√2 Thus, the rectangular coordinates of the point P are (−3√2, 3√2).

Example: Find polar coordinates of the point P whose rectangular coordinates are (−2, 2√3) Solution: Find polar coordinates (r, θ) such that r > 0 and 0 ≤ θ ≤ 2π.

• r = √(x² + y²) = √((−2)² + (2√3)²) = √(4 + 12) = √16 = 4
• tan θ = y/x = 2√3 / (−2) = −√3, so θ = tan⁻¹(−√3) = 2π/3 (since (−2, 2√3) lies in the second quadrant) All other polar coordinates of P have the form (4, 2π/3 + 2nπ) or (−4, 5π/3 + 2nπ), where n is an integer.

LINES IN POLAR COORDINATES

A line perpendicular to the x-axis and passing through the point with xy-coordinates (a, 0) has the equation x = a. To express this in polar coordinates, substitute x = r cos θ, giving r cos θ = a. This makes sense geometrically since each point P(r, θ) on this line will yield the value a for r cos θ.

A line parallel to the x-axis that meets the y-axis at point (0, b) has the equation y = b. Substituting y = r sin θ yields r sin θ = b as the polar equation of this line. Each point P(r, θ) on this line will yield the value b for r sin θ.

For any constant θ₀, the equation θ = θ₀ is satisfied by the coordinates of all points of the form P(r, θ₀), regardless of the value of r. Thus, the equation represents the line through the origin making an angle of θ₀ (radians) with the polar axis.

By substituting x = r cos θ and y = r sin θ in the equation Ax + By + C = 0, we obtain the general polar form of the line: r (A cos θ + B sin θ) + C = 0.

CIRCLES IN POLAR COORDINATES

To find the polar equation of a circle whose radius is a and whose center has polar coordinates (r₀, θ₀), let P(r, θ) be an arbitrary point on the circle. Applying the law of cosines to triangle OCP gives: r² − 2rr₀ cos(θ − θ₀) + r₀² = a².

SOME SPECIAL CASES OF EQUATION OF CIRCLE IN POLAR COORDINATES A circle of radius a, centered at the origin, has an especially simple polar equation. If we let r₀ = 0 in the general equation, we obtain r² = a² or, since a > 0, r = a. This makes sense geometrically since the circle of radius a, centered at the origin, consists of all points P(r, θ) for which r = a, regardless of the value of θ.

If a circle of radius a has its center on the x-axis and passes through the origin, then the polar coordinates of the center are either (a, 0) or (a, π), depending on whether the center is to the right or left of the origin.

⭐ Key Takeaways

Polar coordinates are not unique; a single point can be represented by many different (r, θ) pairs. The conversion between polar and rectangular coordinates is essential: x = r cos θ, y = r sin θ, r² = x² + y², and tan θ = y/x. A negative radial distance means the point lies on the extension of the terminal side of the angle. Lines in polar coordinates are represented by r cos θ = a (vertical) or r sin θ = b (horizontal), while circles centered at the origin simplify to r = a. The general equation of a circle in polar form involves the law of cosines.

🧠 Quick Revision Questions

1. Why are the polar coordinates of a point not unique? Give an example.
2. How do you find the rectangular coordinates of a point given its polar coordinates (r, θ)?
3. What is the geometric interpretation of a point with a negative radial distance, e.g., (−3, 45°)?
4. What is the polar equation of a line passing through the origin making an angle θ₀ with the polar axis?
5. Derive the general polar equation of a circle with radius a and center at (r₀, θ₀).

📘 Lecture 25 — Double integrals in polar co-ordinates

📖 Overview: This lecture introduces double integrals in polar coordinates, which are essential for evaluating integrals over circular or radial regions and for simplifying many double integrals that are difficult in rectangular coordinates. It covers the derivation of polar double integrals, how to find limits of integration from a sketch, and multiple worked examples including area calculations and conversions from Cartesian integrals.

🗂️ Topics Covered

The lecture begins by explaining why polar coordinates are important for double integrals and introduces the concept of a "polar rectangle." It then provides the general formula for double integrals in polar coordinates and step-by-step instructions for determining limits of integration from a sketch. Several examples follow, including integration over cardioid and circle regions, finding area of a three-petaled rose and a lemniscate, and converting Cartesian integrals into polar integrals with multiple worked examples.

📝 Lecture Summary

Double integrals in polar coordinates

When we defined the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. In polar coordinates, the natural shape is a polar rectangle whose sides have constant r and θ-values.

Suppose that a function f(r, θ) is defined over a region R that is bounded by the ray θ = α and θ = β and by the continuous curves r = r₁(θ) and r = r₂(θ). Suppose also that 0 ≤ r₁(θ) ≤ r₂(θ) ≤ a for every value of θ between α and β. Then R lies in a fan-shaped region Q defined by the inequalities 0 ≤ r ≤ a and α ≤ θ ≤ β.

The double integral in polar coordinates is given as: ∬ᵣ f(r, θ) dA = ∫{θ=α}^{θ=β} ∫{r=r₁(θ)}^{r=r₂(θ)} f(r, θ) dr dθ

How to find limits of integration from sketch

Step 1. Since θ is held fixed for the first integration, draw a radial line from the origin through the region R at a fixed angle θ. This line crosses the boundary of R at most twice. The innermost point of intersection is on the curve r = r₁(θ) and the outermost point is on the curve r = r₂(θ). These intersections determine the r-limits of integration.

Step 2. Imagine rotating a ray along the positive x-axis one revolution counterclockwise about the origin. The smallest angle at which this ray intersects the region R is θ = α and the largest angle is θ = β. This yields the θ-limits of the integration.

🔑 Definition — Polar rectangle: a region whose sides have constant r and θ-values, used as the natural shape for partitioning in polar coordinates.

Example: Find the limits of integration for integrating f(r, θ) over the region R that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1.

Solution: Step 1. We sketch the region and label the bounding curves. Step 2. The r-limits of integration. A typical ray from the origin enters R where r = 1 and leaves where r = 1 + cos θ. Step 3. The θ-limits of integration. The rays from the origin that intersect R run from θ = -π/2 to θ = π/2.

The integral is: ∫{-π/2}^{π/2} ∫{1}^{1+cos θ} f(r, θ) r dr dθ = 2∫{0}^{π/2} ∫{1}^{1+cos θ} f(r, θ) r dr dθ

Example: Evaluate ∬ᵣ sin θ dA where R is the region in the first quadrant that is outside the circle r = 2 and inside the cardioid r = 2(1+cos θ).

Solution: ∬ᵣ sin θ dA = ∫{0}^{π/2} ∫{2}^{2(1+cos θ)} (sin θ) r dr dθ = ∫{0}^{π/2} [(r²/2) sin θ]{r=2}^{r=2(1+cos θ)} dθ = 2∫{0}^{π/2} [(1+cos θ)² sin θ − sin θ] dθ = 2[−(1+cos θ)³/3 + cos θ]{0}^{π/2} = 8/3

💡 Why this matters: The polar form allows integration over curved regions that would be very complex in Cartesian coordinates, with the integration limits following naturally from the geometry.

Area using double polar integrals

The area of a region can be found using the double polar integral: A = ∬ dA

Example: Use a double polar integral to find the area enclosed by the three-petaled rose r = sin 3θ. We calculate the area of the petal R in the first quadrant and multiply by three.

Solution: A = 3∬ᵣ dA = 3∫{0}^{π/3} ∫{0}^{sin 3θ} r dr dθ = 3∫{0}^{π/3} [(r²/2)]{0}^{sin 3θ} dθ = (3/2)∫{0}^{π/3} sin²(3θ) dθ = (3/2)∫{0}^{π/3} (1−cos 6θ)/2 dθ = (3/4)[θ − (sin 6θ)/6]_{0}^{π/3} = π/4

Example: Find the area enclosed by the lemniscate r² = 4 cos 2θ. The total area is four times the first-quadrant portion.

Solution: A = 4∫{0}^{π/4} ∫{0}^{√(4 cos 2θ)} r dr dθ = 4∫{0}^{π/4} [r²/2]{0}^{√(4 cos 2θ)} dθ = 4∫{0}^{π/4} 2 cos 2θ dθ = 4[sin 2θ]{0}^{π/4} = 4

📐 Formula — Area in polar coordinates: A = ∬ dA = ∫∫ r dr dθ

Changing Cartesian integrals into polar integrals

The procedure for changing a Cartesian integral ∬ᵣ f(x, y) dx dy into a polar integral has two steps.

Step 1. Substitute x = r cos θ and y = r sin θ, and replace dx dy by r dr dθ in the Cartesian integral. Step 2. Supply polar limits of integration for the boundary of R.

The Cartesian integral then becomes: ∬ᵣ f(x, y) dx dy = ∬₉ f(r cos θ, r sin θ) r dr dθ where G denotes the region of integration in polar coordinates.

Notice that dx dy is not replaced by dr dθ but by r dr dθ.

🔑 Definition — r dr dθ: The area element in polar coordinates, which accounts for the fact that polar rectangles have area proportional to r.

Example: Evaluate ∫{0}^{1} ∫{0}^{√(1−x²)} (x² + y²) dy dx by changing to polar coordinates.

Solution: The region of integration is bounded by 0 ≤ y ≤ √(1−x²) and 0 ≤ x ≤ 1. y = √(1−x²) is the circle x² + y² = 1, so r = 1.

On changing into polar coordinates, the given integral is: ∫{0}^{π/2} ∫{0}^{1} r³ dr dθ = ∫{0}^{π/2} [r⁴/4]{0}^{1} dθ = ∫{0}^{π/2} (1/4) dθ = [θ/4]{0}^{π/2} = π/8

Example: Evaluate I = ∬ᴰ (dx dy)/(x² + y²) by changing to polar coordinates, where D is the region in the first quadrant between the circles x² + y² = a² and x² + y² = b², 0 < a < b.

Solution: I = ∫{0}^{π/2} ∫{a}^{b} (r dr dθ)/(r²) = ∫{0}^{π/2} ∫{a}^{b} (1/r) dr dθ = ∫{0}^{π/2} [ln r]{a}^{b} dθ = ∫{0}^{π/2} ln(b/a) dθ = [θ ln(b/a)]{0}^{π/2} = (π/2) ln(b/a)

Example: Evaluate ∫{0}^{1} ∫{0}^{√(1−x²)} (x² + y²) dy dx by changing to polar coordinates.

Solution: The region of integration is bounded by 0 ≤ y ≤ √(1−x²) and 0 ≤ x ≤ 1. y = √(1−x²) is the circle x² + y² = 1, r = 1.

On changing into polar coordinates, the given integral is: ∫{0}^{π/2} ∫{0}^{1} r³ dr dθ = ∫{0}^{π/2} [r⁴/4]{0}^{1} dθ = ∫{0}^{π/2} (1/4) dθ = [θ/4]{0}^{π/2} = π/8

⭐ Key Takeaways

The most critical concept is that when converting from Cartesian to polar coordinates, dx dy is replaced by r dr dθ — the extra factor of r accounts for the stretching of area elements away from the origin. For setting up polar double integrals, the inner integration is always with respect to r (from the inner boundary curve to the outer boundary curve) and the outer integration is with respect to θ (from the smallest to largest angle intersecting the region). When finding areas, the integrand becomes simply r dr dθ since area = ∬ dA. Many complicated Cartesian integrals over circular regions become much simpler in polar form, especially those involving x² + y² terms which become r².

🧠 Quick Revision Questions

1. What is the area element dA in polar coordinates, and why does it include the factor r?
2. How do you determine the limits of integration for r and θ from a sketch of the region?
3. When converting a Cartesian integral to polar form, what substitutions are made for x, y, and dx dy?
4. Using a double polar integral, find the area enclosed by the cardioid r = 1 + cos θ.
5. Evaluate ∫{0}^{1} ∫{0}^{√(1−x²)} (x² + y²)³/² dy dx by converting to polar coordinates.

📘 Lecture 26 — Examples

📖 Overview: This lecture demonstrates the application of double and triple integration techniques through several worked examples. It focuses on changing integration variables to polar coordinates to evaluate otherwise intractable integrals, including a proof of the Gaussian integral, and introduces the evaluation of triple integrals over rectangular and more complex regions. These examples are crucial for mastering multivariable integration for exams.

🗂️ Topics Covered

The lecture covers the evaluation of double integrals by changing to polar coordinates for regions like circles and semicircles, including handling improper integrals over the entire plane. It provides a proof that the integral of exp(-t²) from 0 to ∞ equals √π/2. It then introduces the theorems for triple integration over rectangular boxes, followed by examples evaluating triple integrals over rectangular and more general regions defined by variable limits.

📝 Lecture Summary

Example 1: Evaluating I = ∫∫ (x² + y²) dx dy by changing into polar coordinates

The region of integration is bounded by 0 < x < √(4y — y²) and 0 < y < 4. The curve x = √(4y — y²) represents the right half of the circle x² + y² — 4y = 0, which is equivalent to x² + (y — 2)² = 4. In polar coordinates, this becomes r² = 4r sin θ, or r = 4 sin θ.

💡 Why this matters: Changing to polar coordinates simplifies the integrand x² + y², which becomes r², and the area element dx dy becomes r dr dθ.

Changing the integral to polar coordinates, we have: I = ∫ (from θ=0 to π/2) ∫ (from r=0 to 4 sin θ) r² . r dr dθ = ∫ (from 0 to π/2) [r⁴/4] (from 0 to 4 sin θ) dθ = ∫ (from 0 to π/2) (64 sin⁴ θ) / 4 dθ = 16 ∫ (from 0 to π/2) sin⁴ θ dθ.

Using Walli’s formula: ∫ (from 0 to π/2) sin⁴ θ dθ = (3·1) / (4·2) · π/2 = 3π/16. 🔑 Definition — Walli's formula: A formula to evaluate definite integrals of powers of sine and cosine.

📐 Formula: ∫ (from 0 to π/2) sinⁿ θ dθ = [(n-1)(n-3)... / (n(n-2)...)] * π/2, for even n.

📌 Example: Here, n=4, so I = 16 * (3π/16) = 3π.

Example 2: Evaluating ∫∫ e^(x² + y²) dy dx over a semicircular region

The region R is the semicircular region bounded by the x-axis and the curve y = √(1 — x²). The integrand e^(x² + y²) is a nonelementary integral in Cartesian coordinates, meaning there is no direct way to integrate it with respect to either x or y.

Substituting x = r cos θ, y = r sin θ, and replacing dy dx by r dr dθ enables us to evaluate the integral. In polar coordinates, the semicircle is described by 0 < r < 1 and 0 < θ < π.

∫∫_R e^(x²+y²) dy dx = ∫ (θ=0 to π) ∫ (r=0 to 1) e^(r²) r dr dθ. The inner integral is: ∫ (0 to 1) e^(r²) r dr. Let u = r², du = 2r dr, so r dr = du/2. Then ∫ e^u (du/2) = (1/2) e^u = (1/2) e^(r²) evaluated from 0 to 1 = (1/2)(e — 1). The outer integral is: ∫ (0 to π) (1/2)(e — 1) dθ = (π/2)(e — 1).

Example 3: Evaluating the improper integral ∫∫_(-∞)^∞ e^(-(x²+y²)) dx dy

Let R_a be the region bounded by the circle x² + y² = a². The improper integral over the entire plane is defined as the limit over these circular disks: I = lim (a→∞)∫∫_{R_a} e^(-(x²+y²)) dx dy.

In polar coordinates, the region is 0 < r < a, 0 < θ < 2π. I = lim (a→∞) ∫ (θ=0 to 2π) ∫ (r=0 to a) e^(-r²) r dr dθ. The inner integral: ∫ (0 to a) e^(-r²) r dr = [-(1/2) e^(-r²)](from 0 to a) = (1/2)(1 — e^(-a²)). Then I = lim (a→∞) ∫ (0 to 2π) (1/2)(1 — e^(-a²)) dθ = lim (a→∞) (1/2)(1 — e^(-a²)) * 2π = π — lim (a→∞) π e^(-a²) = π.

Example 4: Proving ∫_0^∞ e^(-t²) dt = √π / 2

From the previous example, I = ∫∫(-∞)^∞ e^(-(x²+y²)) dx dy = π. This double integral can be expressed as a product of two single integrals because the integrand is separable: I = ∫(-∞)^∞ ∫(-∞)^∞ e^(-x²) e^(-y²) dx dy = [∫(-∞)^∞ e^(-x²) dx] [∫(-∞)^∞ e^(-y²) dy] = [∫(-∞)^∞ e^(-t²) dt]².

Therefore, [∫(-∞)^∞ e^(-t²) dt]² = π, so ∫(-∞)^∞ e^(-t²) dt = √π. Since the function e^(-t²) is even, we have ∫0^∞ e^(-t²) dt = (1/2) ∫(-∞)^∞ e^(-t²) dt = √π / 2.

THEOREM: Triple Integration over a Rectangular Box

Let G be the rectangular box defined by the inequalities a ≤ x ≤ b, c ≤ y ≤ d, k ≤ z ≤ l. If f is continuous on the region G, then: ∫∫∫_G f(x, y, z) dV = ∫ (x=a to b) ∫ (y=c to d) ∫ (z=k to l) f(x, y, z) dz dy dx.

🔑 Definition — Fubini’s Theorem for Triple Integrals: The triple integral over a rectangular box can be evaluated as an iterated integral in any order of integration.

The theorem states that the iterated integral on the right can be replaced with any of the five other iterated integrals that result by altering the order of integration (e.g., dy dz dx, dx dz dy, etc.).

Example 5: Evaluating ∫∫∫_G 12xy²z³ dV over a rectangular box

G is defined by the inequalities -1 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2. We integrate in the order dz dy dx (z first, then y, then x).

∫∫∫_G 12xy²z³ dV = ∫ (x=-1 to 2) ∫ (y=0 to 3) ∫ (z=0 to 2) 12xy²z³ dz dy dx. Step 1 (dz): Inner integral: ∫ (0 to 2) 12xy²z³ dz = 12xy² [z⁴/4](from 0 to 2) = 12xy² (16/4) = 48xy². Step 2 (dy): Next integral: ∫ (0 to 3) 48xy² dy = 48x [y³/3](from 0 to 3) = 48x (27/3) = 432x. Step 3 (dx): Outer integral: ∫ (-1 to 2) 432x dx = 432 [x²/2](from -1 to 2) = 432*(4/2 — 1/2) = 432*(3/2) = 648.

Example 6: Evaluating ∫∫∫ (x — 2y + z) dx dy dz over a non-rectangular region

The region R is defined by the inequalities: 0 ≤ x ≤ 1, 0 ≤ y ≤ x², 0 ≤ z ≤ x + y. We integrate in the order dz dy dx.

∫∫∫_R (x — 2y + z) dz dy dx = ∫ (x=0 to 1) ∫ (y=0 to x²) ∫ (z=0 to x+y) (x — 2y + z) dz dy dx. Step 1 (dz): Inner integral: ∫ (0 to x+y) (x — 2y + z) dz = [(x — 2y)z + z²/2] (from 0 to x+y) = (x — 2y)(x+y) + (x+y)²/2. Expanding: (x² — 2xy + xy — 2y²) + (x² + 2xy + y²)/2 = (x² — xy — 2y²) + (x²/2 + xy + y²/2) = (3/2)x² — (3/2)y² = (3/2)(x² — y²). Step 2 (dy): ∫ (0 to x²) (3/2)(x² — y²) dy = (3/2)[x²y — y³/3](from 0 to x²) = (3/2)(x⁴ — x⁶/3) = (3/2)((3x⁴ — x⁶)/3) = (x⁴/2) — (x⁶/2). Step 3 (dx): ∫ (0 to 1) (x⁴/2 — x⁶/2) dx = (1/2)[x⁵/5 — x⁷/7](from 0 to 1) = (1/2)(1/5 — 1/7) = (1/2)( (7-5)/35 ) = (1/2)(2/35) = 1/35.

Example 7: Evaluating ∫∫∫ xyz dx dy dz over an octant of a sphere

S = {(x,y,z): x² + y² + z² ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0}. This is the positive octant of the unit sphere. The projection of the sphere onto the xy-plane is the circle x² + y² = 1 (in the first quadrant). For a given (x, y), z varies from 0 to z = √(1 — x² — y²).

The triple integral becomes: ∫ (x=0 to 1) ∫ (y=0 to √(1 — x²)) ∫ (z=0 to √(1 — x² — y²)) xyz dz dy dx. Step 1 (dz): Inner integral: ∫ (0 to √(1 — x² — y²)) xyz dz = xy [z²/2](from 0 to (1 — x² — y²)) = (xy/2)(1 — x² — y²). Step 2 (dy): ∫ (0 to √(1 — x²)) (xy/2)(1 — x² — y²) dy = (x/2) ∫ (0 to √(1 — x²)) (y — x²y — y³) dy = (x/2)[y²/2 — x²y²/2 — y⁴/4](from 0 to √(1 — x²)). Evaluating at y = √(1 — x²): = (x/2)[(1 — x²)/2 — x²(1 — x²)/2 — (1 — x²)²/4]. Simplify by multiplying by 4/4: = (x/8)[2(1 — x²) — 2x²(1 — x²) — (1 — x²)²] = (x/8)[2 — 2x² — 2x² + 2x⁴ — (1 — 2x² + x⁴)] = (x/8)[2 — 4x² + 2x⁴ — 1 + 2x² — x⁴] = (x/8)[1 — 2x² + x⁴] = (x/8)(1 — x²)². Step 3 (dx): ∫ (0 to 1) (x/8)(1 — x²)² dx = (1/8) ∫ (0 to 1) x(1 — 2x² + x⁴) dx = (1/8) ∫ (0 to 1) (x — 2x³ + x⁵) dx = (1/8)[x²/2 — 2x⁴/4 + x⁶/6](from 0 to 1) = (1/8)[1/2 — 1/2 + 1/6] = (1/8)(1/6) = 1/48.

⭐ Key Takeaways

The most critical takeaway from this lecture is the power of changing coordinate systems, particularly to polar coordinates, for evaluating double integrals over circular or semicircular regions, transforming nonelementary Cartesian integrands into solvable forms. The proof of the Gaussian integral (∫_0^∞ e^(-t²) dt = √π/2) is a landmark result derived by converting a double integral over the plane into polar coordinates and leveraging the product of two single integrals. For triple integrals, the key is mastering the evaluation of iterated integrals with the correct order of integration, as demonstrated with both rectangular and more general regions, where careful attention to variable limits is essential.

🧠 Quick Revision Questions

1. In Example 1, what is the polar equation for the curve x = √(4y — y²)?
2. Why could the integral ∫∫ e^(x²+y²) dy dx over a semicircle not be evaluated directly in Cartesian coordinates?
3. What is the value of the improper integral ∫∫_(-∞)^∞ e^(-(x²+y²)) dx dy?
4. State Fubini’s Theorem for a triple integral over a rectangular box.
5. In Example 7, what region of the sphere is being integrated over, and what is the final value of the integral ∫∫∫ xyz dV?

📘 Lecture 27 — Vector Valued Functions

📖 Overview: This lecture introduces vector-valued functions, which assign vectors to real numbers, enabling the compact representation of curves in 2-space and 3-space. It covers the definition, natural domain, parametric representation in vector form, and graphical interpretation of these functions, including classic curves like circles, helices, and twisted cubics.

🗂️ Topics Covered

The lecture defines vector-valued functions and their component functions, explains the natural domain as the intersection of component domains, demonstrates how to express parametric equations as single vector equations, illustrates graphing via radius vectors, and explores graphs of specific functions including circles, circular helices, lines, twisted cubics, and constant vector functions.

📝 Lecture Summary

Vector Valued Functions

A vector-valued function assigns to each real number (t) in its domain a vector in 2-space or 3-space. In 2-space: (\mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t)\mathbf{i} + y(t)\mathbf{j}). In 3-space: (\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}). The component functions are the real-valued functions (x(t), y(t), z(t)). Vector-valued functions are denoted in bold, while real-valued functions are in lightface italic.

🔑 Definition — Vector-Valued Function: A function whose domain is a set of real numbers and whose range is a set of vectors in 2-space or 3-space.

📌 Example: (\mathbf{r}(t) = (\ln t)\mathbf{i} + \sqrt{t^2 + 2},\mathbf{j} + (\cos \pi t)\mathbf{k}). Component functions: (x(t) = \ln t), (y(t) = \sqrt{t^2 + 2}), (z(t) = \cos \pi t). Then (\mathbf{r}(1) = (\ln 1)\mathbf{i} + \sqrt{3},\mathbf{j} + (\cos \pi)\mathbf{k} = \sqrt{3},\mathbf{j} - \mathbf{k}). The function is undefined for (t < 0) because (\ln t) is undefined.

Natural Domain

The natural domain of a vector-valued function consists of all real numbers for which every component is defined and yields a real value. It is the intersection of the natural domains of the component functions.

💡 Why this matters: Knowing the natural domain ensures you only evaluate the function at valid parameter values, avoiding undefined operations.

Parametric Equations in Vector Form

Vector-valued functions provide a compact way to express parametric equations. For 2-space: (x = x(t), y = y(t)) is equivalent to (\mathbf{r} = x(t)\mathbf{i} + y(t)\mathbf{j}). For 3-space: (x = x(t), y = y(t), z = z(t)) is equivalent to (\mathbf{r} = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}). Writing (\mathbf{r} = x\mathbf{i} + y\mathbf{j}) (2-space) or (\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) (3-space), the vector form is (\mathbf{r} = \mathbf{r}(t)). Conversely, any vector equation can be rewritten as parametric equations by equating components.

📌 Example: Express (x = t^2, y = 3t) as a single vector equation: (\mathbf{r} = t^2\mathbf{i} + 3t\mathbf{j}).

📌 Example: Express (x = \cos t, y = \sin t, z = t) as a single vector equation: (\mathbf{r} = (\cos t)\mathbf{i} + (\sin t)\mathbf{j} + t\mathbf{k}).

📌 Example: Find parametric equations for (\mathbf{r} = (t^3 + 1)\mathbf{i} + 3\mathbf{j} + e^t\mathbf{k}). Equating components: (x = t^3 + 1, y = 3, z = e^t).

Graphs of Vector-Valued Functions

To graph (\mathbf{r}(t)) geometrically, position the vector (\mathbf{r} = \mathbf{r}(t)) with its initial point at the origin. As (t) varies, the tip of the vector traces a curve (C). This vector is called the radius vector or position vector of (C), and (C) is the graph of the function (\mathbf{r}(t)) or of the equation (\mathbf{r} = \mathbf{r}(t)). The graph is also the graph of the corresponding parametric equations.

📌 Example: Sketch (\mathbf{r}(t) = (\cos t)\mathbf{i} + (\sin t)\mathbf{j}, 0 \le t \le 2\pi). Parametric equations: (x = \cos t, y = \sin t). This is a circle of radius 1 centered at the origin, traced counterclockwise.

📌 Example: Sketch (\mathbf{r}(t) = (\cos t)\mathbf{i} + (\sin t)\mathbf{j} + 2\mathbf{k}, 0 \le t \le 2\pi). Parametric equations: (x = \cos t, y = \sin t, z = 2). The tip traces a circle of radius 1 in the plane (z = 2), centered on the z-axis, counterclockwise when viewed from above.

📌 Example: Sketch (\mathbf{r}(t) = (a\cos t)\mathbf{i} + (a\sin t)\mathbf{j} + (ct)\mathbf{k}) where (a, c > 0). Parametric equations: (x = a\cos t, y = a\sin t, z = ct). As (t) increases, (z) increases, so the point moves upward while circling the z-axis. This produces a circular helix, a corkscrew curve wrapping around a right-circular cylinder of radius (a).

📌 Example: Describe the graph of (\mathbf{r} = (-2 + t)\mathbf{i} + 3t\mathbf{j} + (5 - 4t)\mathbf{k}). Parametric equations: (x = -2 + t, y = 3t, z = 5 - 4t). This is a line through ((-2, 0, 5)) parallel to the vector (\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}).

📌 Example: The graph of (\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + t^3\mathbf{k}) is a twisted cubic. Parametric equations: (x = t, y = t^2, z = t^3). Eliminating (t) from (x) and (y) gives (y = x^2), so the curve lies on the parabolic cylinder (y = x^2). It starts at the origin for (t=0) and moves upward and away.

Graphs of Constant Vector-Valued Functions

If (\mathbf{c}) is a constant vector (independent of the parameter), the graph of (\mathbf{r} = \mathbf{c}) is a single point at the tip of (\mathbf{c}). In 2-space: the point ((x_0, y_0)); in 3-space: the point ((x_0, y_0, z_0)).

📌 Example: The graph of (\mathbf{r} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}) is the point ((2, 3, -1)).

Real-Valued Functions from Vector Functions

For a vector-valued function (\mathbf{r}(t)), the norm (|\mathbf{r}(t)|) is a real-valued function of (t).

📌 Example: If (\mathbf{r}(t) = t\mathbf{i} + (t-1)\mathbf{j}), then (|\mathbf{r}(t)| = \sqrt{t^2 + (t-1)^2}), a real-valued function.

📌 Example: For (\mathbf{r}(t) = (\cos t)\mathbf{i} + (\sin t)\mathbf{j} + 2\mathbf{k}), (|\mathbf{r}(t)| = \sqrt{\cos^2 t + \sin^2 t + 4} = \sqrt{1+4} = \sqrt{5}). Thus every point on the circle is at a constant distance (\sqrt{5}) from the origin, meaning the circle lies on a sphere of radius (\sqrt{5}).

⭐ Key Takeaways

Vector-valued functions map real numbers to vectors, using component functions to define curves in 2-space or 3-space. Their natural domain is the intersection of the domains of all components, and they compactly represent parametric equations via the single vector equation (\mathbf{r} = \mathbf{r}(t)). The graph is traced by the tip of the radius vector as the parameter varies, producing curves such as circles, lines, helices, and twisted cubics, while constant vector functions graph as single points. The norm of a vector-valued function is a real-valued function that can reveal geometric properties, like constancy indicating points lie on a sphere.

🧠 Quick Revision Questions

1. What are the component functions of a vector-valued function in 3-space?
2. How do you find the natural domain of a vector-valued function?
3. Write the vector equation equivalent to the parametric equations (x = \sin t, y = \cos t, z = t).
4. Describe the graph of (\mathbf{r}(t) = (3\cos t)\mathbf{i} + (3\sin t)\mathbf{j} + t\mathbf{k}).
5. If (\mathbf{r}(t) = \langle t^2, t, 1 \rangle), what is (|\mathbf{r}(t)|)?

📘 Lecture 28 — Limits of Vector Valued Functions

📖 Overview: This lecture introduces the calculus of vector-valued functions, extending concepts of limits, continuity, derivatives, and integrals from real-valued to vector-valued functions. These tools are essential for analyzing curves in 2-space and 3-space, with applications in physics, engineering, and geometry.

🗂️ Topics Covered

The lecture covers limits and continuity of vector-valued functions, derivatives and their computation through component-wise differentiation, tangent vectors and tangent lines to parametric curves, smooth parameterization, properties of derivatives including dot and cross product rules, and integrals of vector-valued functions with their properties.

📝 Lecture Summary

Limits of Vector-Valued Functions

The limit of a vector-valued function is defined component-wise. For a function r(t) = x(t)i + y(t)j in 2-space, we define:

lim r(t) = (lim x(t))i + (lim y(t))j as t→a

For r(t) = x(t)i + y(t)j + z(t)k in 3-space:

lim r(t) = (lim x(t))i + (lim y(t))j + (lim z(t))k as t→a

If any component limit does not exist, the limit of r(t) does not exist. These definitions apply to one-sided limits (t→a⁺, t→a⁻) and infinite limits (t→+∞, t→-∞) as well.

🔑 Definition — Limit of Vector Function: lim r(t) = L if and only if each component of r(t) approaches the corresponding component of L as t→a. Geometrically, this means the length and direction of r(t) approach those of L.

Continuity of Vector-Valued Functions

A vector-valued function r is continuous at t₀ if:

1. r(t₀) is defined
2. lim r(t) exists as t→t₀
3. lim r(t) = r(t₀) as t→t₀

r is continuous at t₀ if and only if each component of r is continuous. A function called continuous everywhere or simply continuous if it is continuous at all real values of t. Geometrically, the graph of a continuous vector-valued function is an unbroken curve.

Derivatives of Vector-Valued Functions

The derivative r'(t) of a vector-valued function r(t) is defined by:

r'(t) = lim [r(t+h) - r(t)] / h as h→0

💡 Why this matters: This definition mirrors the real-valued derivative but yields a vector result.

Theorem for Computing Derivatives

(a) If r(t) = x(t)i + y(t)j in 2-space, with x(t) and y(t) differentiable, then: r'(t) = x'(t)i + y'(t)j

(b) If r(t) = x(t)i + y(t)j + z(t)k in 3-space, with differentiable components, then: r'(t) = x'(t)i + y'(t)j + z'(t)k

Proof (a): r'(t) = lim [x(t+h)-x(t)]/h i + lim [y(t+h)-y(t)]/h j = x'(t)i + y'(t)j as h→0

Notation: d/dt[r(t)], dr/dt, r'(t), and ṙ

📐 Formula: r'(t) = d/dt[x(t)] i + d/dt[y(t)] j (2-space)

📌 Example: Let r(t) = t²i + t³j. Find r'(t) and r'(1).

r'(t) = d/dt[t²] i + d/dt[t³] j = 2ti + 3t²j

Substituting t=1: r'(1) = 2i + 3j

Tangent Vectors and Tangent Lines

Geometric Interpretation

Suppose C is the graph of r(t) and r'(t) exists and is nonzero. If r'(t) is positioned with its initial point at the terminal point of the radius vector...

🔑 Definition — Tangent Vector: Let P be a point on the graph of r(t), with r(t₀) the radius vector from origin to P. If r'(t₀) exists and r'(t₀) ≠ 0, then r'(t₀) is called the tangent vector to the graph of r at r(t₀).

A graph can fail to have a tangent vector if the derivative does not exist or equals zero at that point.

The tangent line at r(t₀) is the line parallel to r'(t₀) passing through the tip of r(t₀).

📐 Formula — Vector Equation of Tangent Line: r = r(t₀) + t r'(t₀)

📌 Example: Find parametric equations of the tangent line to the circular helix x = cos t, y = sin t, z = t at t = π/6.

Vector equation: r(t) = (cos t)i + (sin t)j + tk r'(t) = (-sin t)i + (cos t)j + k

At t = π/6: r(π/6) = (√3/2)i + (1/2)j + (π/6)k r'(π/6) = (-1/2)i + (√3/2)j + k

Vector equation of tangent line: r = r(π/6) + t r'(π/6) r = (√3/2 - t/2)i + (1/2 + √3·t/2)j + (π/6 + t)k

Parametric equations: x = √3/2 - t/2 y = 1/2 + √3·t/2 z = π/6 + t

📌 Example: The graph of r(t) = t²i + t³j is a semicubical parabola. Find tangent line at: (a) (0,0): Corresponds to t=0. r'(0)=0, so no tangent vector exists—no tangent line at this point. (b) (1,1): Corresponds to t=1. r = r(1) + t r'(1) = (i+j) + t(2i+3j)

Smooth Parameterization

A vector-valued function r(t) in 2-space or 3-space is smoothly parameterized or a smooth function of t if the components have continuous derivatives and r'(t) ≠ 0 for any t.

In 3-space: r(t) = x(t)i + y(t)j + z(t)k is smooth if x'(t), y'(t), z'(t) are continuous and no value of t makes all three derivatives zero.

A parametric curve C is called smooth if it is the graph of some smooth vector-valued function. A smooth vector-valued function has a tangent line at every point on its graph.

Properties of Derivatives (Rules of Differentiation)

For vector-valued functions r(t), r₁(t), r₂(t), real-valued function f(t), scalar k, and constant vector c:

• d/dt[c] = 0
• d/dt[kr(t)] = k d/dt[r(t)]
• d/dt[r₁(t) + r₂(t)] = d/dt[r₁(t)] + d/dt[r₂(t)]
• d/dt[r₁(t) - r₂(t)] = d/dt[r₁(t)] - d/dt[r₂(t)]
• d/dt[f(t)r(t)] = f(t) d/dt[r(t)] + r(t) d/dt[f(t)]

Dot and Cross Product Rules

📐 Formula — Dot Product Rule: d/dt[r₁(t)·r₂(t)] = r₁·dr₂/dt + dr₁/dt·r₂ (order doesn't matter)

📐 Formula — Cross Product Rule (3-space): d/dt[r₁(t) × r₂(t)] = r₁ × dr₂/dt + dr₁/dt × r₂ (order matters!)

Theorem: Constant Length Implies Orthogonality

If ||r(t)|| is constant for all t, then r(t)·r'(t) = 0, meaning r(t) and r'(t) are orthogonal for all t.

Proof: d/dt[||r(t)||²] = 2r(t)·dr/dt = 0, so r(t)·dr/dt = 0

📌 Example: For a curve on a sphere of radius k centered at origin, ||r(t)|| = k constant, so r(t)·r'(t) = 0. The radius vector is perpendicular to the tangent vector, meaning the tangent line is perpendicular to the radius at the point of tangency.

Integrals of Vector-Valued Functions

Definition

(a) In 2-space, r(t) = x(t)i + y(t)j: ∫ r(t)dt = (∫ x(t)dt)i + (∫ y(t)dt)j ∫ₐᵇ r(t)dt = (∫ₐᵇ x(t)dt)i + (∫ₐᵇ y(t)dt)j

(b) In 3-space, r(t) = x(t)i + y(t)j + z(t)k: ∫ r(t)dt = (∫ x(t)dt)i + (∫ y(t)dt)j + (∫ z(t)dt)k ∫ₐᵇ r(t)dt = (∫ₐᵇ x(t)dt)i + (∫ₐᵇ y(t)dt)j + (∫ₐᵇ z(t)dt)k

📌 Example: Let r(t) = 2ti + 3t²j

(a) ∫ r(t)dt = ∫ (2ti + 3t²j)dt = (∫2t dt)i + (∫3t² dt)j = (t² + C₁)i + (t³ + C₂)j = t²i + t³j + C, where C = C₁i + C₂j is an arbitrary vector constant of integration

(b) ∫₀² r(t)dt = ∫₀² (2ti + 3t²j)dt = (∫₀² 2t dt)i + (∫₀² 3t² dt)j = [t²]₀² i + [t³]₀² j = (4-0)i + (8-0)j = 4i + 8j

Properties of Integrals

• ∫ cr(t)dt = c ∫ r(t)dt
• ∫ [r₁(t) + r₂(t)]dt = ∫ r₁(t)dt + ∫ r₂(t)dt
• ∫ [r₁(t) - r₂(t)]dt = ∫ r₁(t)dt - ∫ r₂(t)dt
• d/dt[∫ r(t)dt] = r(t) (indefinite integral gives antiderivatives)
• If R'(t) = r(t), then ∫ r(t)dt = R(t) + C and ∫ₐᵇ r(t)dt = R(b) - R(a)

⭐ Key Takeaways

The limit, continuity, derivative, and integral of vector-valued functions are computed component-wise, extending real-valued calculus to curves in space. The derivative r'(t) gives a tangent vector when nonzero, and the tangent line equation is r = r(t₀) + t r'(t₀). A function is smooth if its components have continuous derivatives and r'(t) ≠ 0 everywhere. For constant-length vectors, r(t) and r'(t) are always perpendicular. Integration of vector functions follows component-wise integration with vector constants of integration, and definite integrals follow the fundamental theorem with antiderivatives.

🧠 Quick Revision Questions

1. How do you compute the limit of a vector-valued function r(t) = x(t)i + y(t)j + z(t)k as t→a?
2. What three conditions must be satisfied for r(t) to be continuous at t₀?
3. If r(t) = cos(t)i + sin(t)j + tk, what is r'(t) and what is the tangent line at t = π/4?
4. Prove that if ||r(t)|| is constant, then r(t) is perpendicular to r'(t).
5. Compute ∫₀¹ (3t²i + 2tj + k)dt and express the result as a vector.

📘 Lecture 29 — Change of Parameter

📖 Overview: This lecture explores how to change the parameter in vector-valued functions while preserving the geometric curve. It introduces arc length as a fundamental parameter and provides the mathematical framework for reparametrizing curves in both 2-space and 3-space, which is essential for analyzing motion along curves independent of speed.

🗂️ Topics Covered

The lecture covers the concept of changing parameters in vector-valued functions and the conditions for a smooth change of parameter. It then introduces arc length for curves in 2-space and 3-space, presenting the integral formulas. Subsequently, it explains the process of using arc length itself as a parameter, including the theorem relating arc length s to a given parameter t. Worked examples demonstrate the reparametrization of lines, circles, and other curves using arc length, as well as calculating arc length and using the chain rule for differentiation.

📝 Lecture Summary

Change of parameter

It is possible for different vector-valued functions to have the same graph. For example, the graph of r = (3 cos t)i + (3 sin t)j, 0 < t < 2π, is a circle of radius 3. By substituting t = s/3, we obtain r = 3 cos (s/3)i + 3 sin (s/3)j, 0 < s < 6π, which has the same graph. In general, substituting t = g(u) in r(t) changes the parameter.

For a smooth change of parameter, the function g must satisfy: (1) g is differentiable, (2) g' is continuous, (3) g'(u) ≠ 0 for any u in its domain, and (4) the range of g is the domain of r.

ARC LENGTH

If x'(t) and y'(t) are continuous for a < t < b, then the curve given by x = x(t), y = y(t) has arc length: 📐 Formula: L = ∫ₐᵇ √[(dx/dt)² + (dy/dt)²] dt This generalizes to 3-space: if x'(t), y'(t), and z'(t) are continuous, then: 📐 Formula: L = ∫ₐᵇ √[(dx/dt)² + (dy/dt)² + (dz/dt)²] dt

🔑 Definition — Arc Length: The total length of a curve traced by a vector-valued function from t = a to t = b.

📌 Example: Find the arc length of the circular helix x = cos t, y = sin t, z = t from t = 0 to t = π. L = ∫₀ᵖⁱ √[(-sin t)² + (cos t)² + 1] dt = ∫₀ᵖⁱ √[sin²t + cos²t + 1] dt = ∫₀ᵖⁱ √2 dt = √2 π

ARC LENTH AS A PARAMETER

Using arc length s as a parameter is often the best choice. The process involves: Step 1: Select an arbitrary reference point on the curve. Step 2: Choose a positive direction along the curve. Step 3: Let s be the "signed" arc length from the reference point to any point P, positive in the positive direction and negative in the negative direction.

The curve is then given by x = x(s), y = y(s) in 2-space or x = x(s), y = y(s), z = z(s) in 3-space. When a curve is already parameterized by t, the positive direction for s is the direction of increasing t.

THEOREM

(a) For a curve in 2-space x = x(t), y = y(t) with continuous derivatives and reference point at (x(t₀), y(t₀)): 📐 Formula: s = ∫ₜ₀ᵗ √[(dx/du)² + (dy/du)²] du

(b) For a curve in 3-space x = x(t), y = y(t), z = z(t) with continuous derivatives and reference point at (x(t₀), y(t₀), z(t₀)): 📐 Formula: s = ∫ₜ₀ᵗ √[(dx/du)² + (dy/du)² + (dz/du)²] du

From the Second Fundamental Theorem of Calculus, we obtain: 📐 Formula (2-space): ds/dt = √[(dx/dt)² + (dy/dt)²] 📐 Formula (3-space): ds/dt = √[(dx/dt)² + (dy/dt)² + (dz/dt)²]

💡 Why this matters: ds/dt represents the rate of change of arc length with respect to t, which is the speed of the particle. For a smooth curve, ds/dt > 0, ensuring a one-to-one correspondence between s and t.

📌 Example: Reparametrize the line x = 2t + 1, y = 3t - 2 using arc length s as a parameter, with reference point (1, -2). Here, dx/du = 2, dy/du = 3. The reference point (1, -2) corresponds to t₀ = 0. s = ∫₀ᵗ √(4 + 9) du = ∫₀ᵗ √13 du = √13 t Thus, t = s/√13. Substituting: x = 2(s/√13) + 1 = (2/√13)s + 1, y = 3(s/√13) - 2 = (3/√13)s - 2.

📌 Example: Find parametric equations for the circle x = a cos t, y = a sin t (0 < t < 2π) using arc length s as a parameter, with reference point (a, 0). Here, dx/du = -a sin u, dy/du = a cos u. The reference point (a, 0) corresponds to t₀ = 0. s = ∫₀ᵗ √[(-a sin u)² + (a cos u)²] du = ∫₀ᵗ √(a²) du = ∫₀ᵗ a du = at Thus, t = s/a. Substituting: x = a cos(s/a), y = a sin(s/a) for 0 < s < 2πa.

📌 Example: Find arc length of r(t) = t³i + tj + ½√6 t²k, 1 < t < 3. Here, x = t³, y = t, z = ½√6 t². dx/dt = 3t², dy/dt = 1, dz/dt = √6 t. Arc length = ∫₁³ √[9t⁴ + 1 + 6t²] dt = ∫₁³ √[(3t² + 1)²] dt = ∫₁³ (3t² + 1) dt = [t³ + t]₁³ = (27 + 3) - (1 + 1) = 28.

📌 Example: Calculate dr/du by chain rule for r = eᵗi + 4e⁻ᵗj, where t = u². dr/dt = eᵗi - 4e⁻ᵗj dt/du = 2u dr/du = (dr/dt)(dt/du) = (eᵗi - 4e⁻ᵗj)(2u) = 2ue^(u²)i - 8ue^(-u²)j Verification: Expressing r in terms of u: r = e^(u²)i + 4e^(-u²)j, so dr/du = 2ue^(u²)i - 8ue^(-u²)j ✓

⭐ Key Takeaways

The arc length parameter s provides a natural, geometric parameterization independent of the original parameter's speed. The core relationship ds/dt = ||r'(t)|| gives the speed, and for smooth curves ds/dt is always positive. The theorem s = ∫√((dx/dt)² + (dy/dt)² + (dz/dt)²) dt allows converting any parameterization to arc length. When reparametrizing, the conditions for smooth change ensure the new function remains smooth. Mastering these concepts is essential for understanding curvature, torsion, and the Frenet-Serret frame in subsequent topics.

🧠 Quick Revision Questions

1. What four conditions must g satisfy for t = g(u) to be a smooth change of parameter?
2. Write the formula for arc length of a 3D curve from t = a to t = b.
3. What is ds/dt, and why is it always positive for a smooth curve?
4. How do you reparametrize a curve using arc length s as a parameter?
5. In the example of the circle x = a cos t, y = a sin t, what is the relationship between s and t when the reference point is (a, 0)?

📘 Lecture 30 — Exact Differential

📖 Overview: This lecture introduces the concept of exact differentials, building on the total differential of a function of multiple variables. It covers the test for exactness, the method for integrating exact differentials to recover the original function, and extends the concept to line integrals and their application in calculating areas enclosed by curves.

🗂️ Topics Covered

The lecture begins by reviewing the total differential for functions of two or more variables, then defines an exact differential and establishes the test for exactness using partial derivatives. It provides worked examples for both testing and integrating exact differentials. The final sections apply these concepts to calculate the area enclosed by closed curves using line integrals, defining line integrals and their alternative forms using x and y components.

📝 Lecture Summary

Exact Differential

If ( z = f(x, y) ), then the total differential is ( dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy ). This result can be extended to functions of more than two independent variables. For ( z = f(x, y, w) ), the differential is ( dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy + \frac{\partial z}{\partial w} dw ).

Exercise: Determine the differential ( dz ) for each function.

1. ( z = x^2 + y^2 ): ( dz = 2(x dx + y dy) )
2. ( z = x^3 \sin 2y ): ( dz = x^2(3 \sin 2y , dx + 2x \cos 2y , dy) )
3. ( z = (2x - 1)e^{3y} ): ( dz = e^{3y} {2dx + (6x - 3)dy} )
4. ( z = x^2 + 2y^2 + 3w^2 ): ( dz = 2(x dx + 2y dy + 3w dw) )
5. ( z = x^3 y^2 w ): ( dz = x^2 y (3yw dx + 2xw dy + xy dw) )

Working in reverse, any expression ( dz = P dx + Q dy ), where ( P ) and ( Q ) are functions of ( x ) and ( y ), is an exact differential if it can be integrated to determine ( z ). This means ( P = \frac{\partial z}{\partial x} ) and ( Q = \frac{\partial z}{\partial y} ).

The condition for exactness is derived from the equality of mixed partial derivatives. Since ( \frac{\partial P}{\partial y} = \frac{\partial^2 z}{\partial y \partial x} ) and ( \frac{\partial Q}{\partial x} = \frac{\partial^2 z}{\partial x \partial y} ), and ( \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} ), the test for exactness is: [ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ]

Example: Test if ( dz = (3x^2 + 4y^2) dx + 8xy dy ) is exact.

• ( P = 3x^2 + 4y^2 ), so ( \frac{\partial P}{\partial y} = 8y )
• ( Q = 8xy ), so ( \frac{\partial Q}{\partial x} = 8y )
• Since ( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ), ( dz ) is an exact differential.

Example: Test if ( dz = (1 + 8xy) dx + 5x^2 dy ) is exact.

• ( P = 1 + 8xy ), so ( \frac{\partial P}{\partial y} = 8x )
• ( Q = 5x^2 ), so ( \frac{\partial Q}{\partial x} = 10x )
• Since ( \frac{\partial P}{\partial y} \ne \frac{\partial Q}{\partial x} ), ( dz ) is not an exact differential.

Exercise: Determine if each is an exact differential.

1. ( dz = 4x^3 y^3 dx + 3x^4 y^2 dy ): Yes
2. ( dz = (4x^3 y + 2xy^3) dx + (x^4 + 3x^2 y^2) dy ): Yes
3. ( dz = (15y^2 e^{3x} + 2xy^2) dx + (10y e^{3x} + x^2 y) dy ): No
4. ( dz = (3x^2 e^{2y} - 2y^2 e^{3x}) dx + (2x^3 e^{2y} - 2y e^{3x}) dy ): No
5. ( dz = (4y^3 \cos 4x + 3x^2 \cos 2y) dx + (3y^2 \sin 4x - 2x^3 \sin 2y) dy ): Yes

Integration Of Exact Differentials

For an exact differential ( dz = P dx + Q dy ), integration recovers the function ( z ). Since ( P = \frac{\partial z}{\partial x} ) and ( Q = \frac{\partial z}{\partial y} ), we have ( z = \int P dx ) and also ( z = \int Q dy ). Each integration introduces an arbitrary function of the other variable, which is then determined by comparing the two expressions.

Example: Integrate ( dz = (2xy + 6x) dx + (x^2 + 2y^3) dy ).

• Integrate ( P ) with respect to ( x ): ( z = \int (2xy + 6x) dx = x^2 y + 3x^2 + f(y) ) (i), where ( f(y) ) is an arbitrary function of ( y ).
• Integrate ( Q ) with respect to ( y ): ( z = \int (x^2 + 2y^3) dy = x^2 y + \frac{y^4}{2} + F(x) ) (ii), where ( F(x) ) is an arbitrary function of ( x ).
• Compare (i) and (ii): For them to represent the same function, ( f(y) = \frac{y^4}{2} ) and ( F(x) = 3x^2 ).
• Therefore, ( z = x^2 y + 3x^2 + \frac{y^4}{2} ).

Example: Integrate ( dz = (8e^{4x} + 2xy^2) dx + (4\cos 4y + 2x^2 y) dy ).

• Integrate ( P ) with respect to ( x ): ( z = \int (8e^{4x} + 2xy^2) dx = 2e^{4x} + x^2 y^2 + f(y) ) (i)
• Integrate ( Q ) with respect to ( y ): ( z = \int (4\cos 4y + 2x^2 y) dy = \sin 4y + x^2 y^2 + F(x) ) (ii)
• Comparing (i) and (ii): ( f(y) = \sin 4y ) and ( F(x) = 2e^{4x} ).
• Therefore, ( z = 2e^{4x} + x^2 y^2 + \sin 4y ).

Area enclosed by the closed curve

A classic application of integration is finding the area of a plane figure. The area enclosed by a closed curve can be evaluated using a line integral. The formula is: [ A = -\oint y , dx ] where the symbol ( \oint ) indicates integration around the closed boundary in the positive (anticlockwise) direction.

Example: Determine the area enclosed by ( y = x^3 ) and ( y = 4x ) for ( x > 0 ).

• Points of intersection: ( x = 0 ) and ( x = 2 ).
• Integrate anticlockwise:
  • Curve ( C_1 ): ( y = x^3 ), limits ( x = 0 ) to ( x = 2 )
  • Curve ( C_2 ): ( y = 4x ), limits ( x = 2 ) to ( x = 0 )
• ( A = -\oint y , dx = -\left[ \int_0^2 x^3 dx + \int_2^0 4x dx \right] )
• ( A = -\left[ \left. \frac{x^4}{4} \right|_0^2 + \left. 2x^2 \right|_2^0 \right] = -\left[ (4 - 0) + (0 - 8) \right] = -[4 - 8] = 4 ) square units.

Example: Find the area of the triangle with vertices (0,0), (5,3), and (2,6).

• Equations of sides:
  • OA: ( y = \frac{3}{5}x )
  • BA: ( y = 8 - x )
  • OB: ( y = 3x )
• Integrate anticlockwise: ( A = -\oint y , dx = -\left[ \int_0^5 \frac{3}{5}x dx + \int_5^2 (8-x) dx + \int_2^0 3x dx \right] )
• ( A = -\left[ \left. \frac{3x^2}{10} \right|_0^5 + \left. (8x - \frac{x^2}{2}) \right|_5^2 + \left. \frac{3x^2}{2} \right|_2^0 \right] = -\left[ 7.5 + (12 - 27.5) + (0 - 6) \right] )
• ( A = -\left[ 7.5 - 15.5 - 6 \right] = -[-14] = 12 ) square units.

Line Integrals

If a field exists in the xy-plane producing a force F on a particle at point K, the work done in moving the particle along a curve from A to B is given by a line integral: [ \text{Work} = \int_{AB} F_t , ds = \int_C F_t , ds ] where ( F_t ) is the tangential component of the force along the curve, ( C ) is the curve ( y = f(x) ) between A and B, and ( ds ) is an infinitesimal arc length.

It is often more convenient to integrate with respect to x or y. If ( F_t ) has a component ( P ) in the x-direction and ( Q ) in the y-direction, then: [ \int_{AB} F_t , ds = \int_{AB} (P dx + Q dy) ] where ( P ) and ( Q ) are functions of ( x ) and ( y ). In general, the line integral can be expressed as: [ I = \int_C F_t , ds = \int_C (P dx + Q dy) ]

⭐ Key Takeaways

The lecture establishes the crucial link between total differentials and the condition for exactness through the equality of mixed partial derivatives, a test that must always be applied. The method for integrating exact differentials requires integrating with respect to each variable and reconciling the arbitrary functions, a technique fundamental to solving differential equations. A powerful application is the calculation of enclosed areas using line integrals of the form ( \oint y , dx ), which simplifies geometry problems. Finally, the line integral is introduced as a tool for calculating work done by a force field along a path, with an alternative form in terms of x and y components.

🧠 Quick Revision Questions

1. What is the mathematical condition for an expression ( dz = P dx + Q dy ) to be an exact differential?
2. When integrating an exact differential by integrating ( P ) with respect to ( x ), what step is taken to recover the full function ( z )?
3. What is the formula for calculating the area enclosed by a closed curve using a line integral, and in which direction is the integral evaluated?
4. In the context of line integrals, what do the symbols ( P ) and ( Q ) represent in the expression ( \int_C (P dx + Q dy) )?
5. Why is the test for exactness based on the equality of the second-order partial derivatives ( \frac{\partial^2 z}{\partial y \partial x} ) and ( \frac{\partial^2 z}{\partial x \partial y} )?

📘 Lecture 31 — Line Integral

📖 Overview: This lecture introduces the concept of line integrals, which extend ordinary integration to paths or curves in the plane. Line integrals are essential for calculating work done by a force along a curved path and have important applications in physics and engineering. The lecture demonstrates evaluation techniques, alternative forms, and properties of line integrals.

🗂️ Topics Covered

The lecture covers the definition of line integrals as work done along a curve, the alternative form using P and Q components, evaluation of line integrals along different paths including curves y = f(x) and lines, comparison of results along distinct paths, six properties of line integrals, and examples involving semi-circles, triangles, and circles.

📝 Lecture Summary

Line Integral

The work done in moving a particle through a small distance δs from K to L along a curve is approximately F₁ δs. The total work done in moving a particle along the curve from A to B is given by:

Limit Σ F δs = ∫ F ds from A to B as δ→0

This is written as ∫ₐᵇ Fᵼ ds where A and B are the end points, or as ∫ₓ Fᵼ ds where the curve c connects A and B. Such an integral is called a line integral since integration is carried out along the path of the curve c joining A and B.

∴ I = ∫ₐᵇ Fᵼ dx = ∫ₓ Fᵼ ds, where c is the curve y = f(x) between A(x₁, y₂) and B(x₂, y₂).

Alternative form of a line integral

It is often more convenient to integrate with respect to x or y than to take arc length as the variable. If Fᵼ has a component P in the x-direction and Q in the y-direction, then the work done from K to L can be stated as P δx + Q δy.

Example 1 Evaluate ∫ (x + 3y) dx from A(0, 1) to B(2, 5) along the curve y = 1 + x².

The line integral is of the form ∫ (P dx + Q dy) where Q = 0 and c is y = 1 + x².

I = ∫ (x + 3y) dx = ∫₀² (x + 3 + 3x²) dx = [x²/2 + 3x + x³]₀² = 16

Example 2 Evaluate I = ∫ (x² + y) dx + (x − y²) dy from A(0, 2) to B(2, 5) along y = 2 + x.

P = x² + y = x² + 2 + x = x² + x + 2 Q = x − y² = x − (4 + 4x + x²) = −(x² + 3x + 4)

Since y = 2 + x, dy = dx with limits x = 0 to x = 3.

I = ∫₀² {(x² + x + 2) dx − (x² + 3x + 4) dx} = ∫₀² −(2x + 2) dx = [−x² − 2x]₀² = −4 − 4 = −8

Example 3 Evaluate I = ∫ {(x² + 2y) dx + xy dy} from O(0, 0) to B(1, 4) along y = 4x².

y = 4x², so dy = 8x dx x² + 2y = x² + 8x² = 9x² xy = 4x³

I = ∫₀¹ {9x² dx + 32x⁴ dx} = ∫₀¹ (9x² + 32x⁴) dx = 9.4

Example 4 Evaluate I = ∫ {(x² + 2y) dx + xy dy} from O(0,0) to A(1,0) along y = 0, then from A(1,0) to B(1,4) along x = 1.

(i) OA: c₁ is y = 0, dy = 0 Iₒₐ = ∫₀¹ x² dx = [x³/3]₀¹ = 1/3

(ii) AB: c₂ is x = 1, dx = 0 Iₐᵦ = ∫₀⁴ y dy = [y²/2]₀⁴ = 8

Then I = Iₒₐ + Iₐᵦ = 1/3 + 8 = 8⅓

Comparing Example 3 (I = 9⅓) and Example 4 (I = 8⅓): integration along two distinct paths joining the same two end points does not necessarily give the same results.

Properties of line integrals

1. ∫ F ds = ∫ {P dx + Q dy}

2. ∫ₐᵦ F ds = −∫ᵦₐ F ds and ∫ₐᵦ {P dx + Q dy} = −∫ᵦₐ {P dx + Q dy} The sign of a line integral is reversed when the direction of integration along the path is reversed.

3. (a) For a path parallel to the y-axis (x = k, dx = 0): ∫ P dx = 0 ∴ I = ∫ Q dy (b) For a path parallel to the x-axis (y = k, dy = 0): ∫ Q dy = 0 ∴ I = ∫ P dx

4. If the path c joining A to B is divided into two parts AK and KB, then Iₓ = Iₐᵦ = Iₐₖ + Iₖᵦ

5. If the path c is not single-valued for part of its extent, the path is divided into two sections: y = f₁(x) from A to K, y = f₂(x) from K to B.

6. The actual path of integration must be continuous and single-valued.

Example Evaluate I = ∫ (x + y) dx from A(0, 1) to B(0, −1) along the semi-circle x² + y² = 1 for x > 0.

The path c is not single-valued since y = ±√(1−x²). Divide c into two parts: (i) y = √(1−x²) from A to K (ii) y = −√(1−x²) from K to B

Here Q = 0, so I = ∫ P dx

I = ∫₀¹ (x + √(1−x²)) dx + ∫₁₀ (x − √(1−x²)) dx = ∫₀¹ (x + √(1−x²) − x + √(1−x²)) dx = 2∫₀¹ √(1−x²) dx

Substitute x = sin θ, dx = cos θ dθ, √(1−x²) = cos θ Limits: x = 0, θ = 0; x = 1, θ = π/2

I = 2∫₀^(π/2) cos²θ dθ = ∫₀^(π/2) (1 + cos 2θ) dθ = [θ + (sin 2θ)/2]₀^(π/2) = π/2

Example Evaluate I = ∮ (x² dx − 2xy dy) where c comprises the three sides of the triangle O(0,0), A(1,0), and B(0,1) in anticlockwise order.

(a) OA: c₁ is y = 0, dy = 0 I₁ = ∫₀¹ x² dx = [x³/3]₀¹ = 1/3

(b) AB: c₂ is y = 1 − x, dy = −dx I₂ = ∫₀¹ {x² dx + 2x(1−x)dx} = ∫₀¹ (x² + 2x − 2x²) dx = ∫₀¹ (2x − x²) dx = [x² − x³/3]₀¹ = 1 − 1/3 = 2/3

Wait—recalculating: I₂ = ∫₀¹ (2x − x²) dx = [x² − x³/3]₀¹ = 1 − 1/3 = 2/3

(c) BO: c₃ is x = 0, dx = 0 I₃ = ∫ 0 dy = 0

I = I₁ + I₂ + I₃ = 1/3 + (−2/3) + 0 = −1/3

Example Evaluate ∮ y dx where c is the circle x² + y² = 4.

y = ±√(4−x²), not single-valued. Use y = √(4−x²) for ALB between x = 2 and x = −2, and y = −√(4−x²) for BMA between x = −2 and x = 2.

I = ∫₂⁻² √(4−x²) dx + ∫₋₂² {−√(4−x²)} dx = −2∫₋₂² √(4−x²) dx = −4∫₀² √(4−x²) dx

Substitute x = 2 sin θ, dx = 2 cos θ dθ, √(4−x²) = 2 cos θ Limits: x = 0, θ = 0; x = 2, θ = π/2

I = −4∫₀^(π/2) 4 cos²θ dθ = −16∫₀^(π/2) cos²θ dθ = −8∫₀^(π/2) (1 + cos 2θ) dθ = −8[θ + sin 2θ/2]₀^(π/2) = −8(π/2) = −4π

⭐ Key Takeaways

A line integral evaluates a function along a curve rather than along a straight axis, and is computed by substituting the curve equation and integrating with respect to a single variable. The result of a line integral may depend on the path taken between two points, not just the endpoints — as shown by Examples 3 and 4. When the path is not single-valued, it must be split into single-valued sections. Key properties include sign reversal when direction is reversed, special cases for horizontal/vertical paths, and additivity when paths are divided. For closed curves, proper orientation (anticlockwise) and careful treatment of limits are essential.

🧠 Quick Revision Questions

1. How do you convert a line integral ∫ (P dx + Q dy) into an ordinary integral when the path is y = f(x)?
2. Why did Examples 3 and 4 give different results for the same endpoints along different paths?
3. What does it mean for a path to be "not single-valued" and how do you handle such paths in line integrals?
4. If a line integral is evaluated from B to A instead of A to B, what happens to the value?
5. How would you evaluate ∮ (x² dx + y² dy) around a rectangle with vertices (0,0), (2,0), (2,3), (0,3)?

📘 Lecture 32 — Examples

📖 Overview: This lecture demonstrates the evaluation of line integrals along various paths, including rectangular boundaries, straight lines, and parametric curves. It explores the concept of path dependence and independence in line integrals, showing how exact differentials lead to path-independent results.

🗂️ Topics Covered

The lecture covers evaluation of line integrals along rectangular boundaries with four segments, line integrals with respect to arc length including conversion formulas using dy/dx and parametric forms, parametric equations for line integrals, and the critical concept of dependence versus independence of line integrals on the path of integration with examples showing both scenarios.

📝 Lecture Summary

Example 1: Rectangular Boundary

Evaluate I = ∮{xydx+(1+y²)dy} where c is the boundary of the rectangle joining A(1,0), B(3,0), C(3,2), D(1,2). The boundary is traversed anticlockwise through four segments.

AB (c₁): y = 0 → dy = 0 → I₁ = 0

BC (c₂): x = 3 → dx = 0 I₂ = ∫₀² (1+y²)dy = [y + y³/3]₀² = 4⅓

CD (c₃): y = 2 → dy = 0 I₃ = ∫₃¹ 2xdx = [x²]₃¹ = -8

DA (c₄): x = 1 → dx = 0 I₄ = ∫₂⁰ (1+y²)dy = [y + y³/3]₂⁰ = -4⅓

I = I₁ + I₂ + I₃ + I₄ = 0 + 4⅓ - 8 - 4⅓ = -8

💡 Why this matters: Unless directed otherwise, always proceed round the closed boundary in an anticlockwise manner.

Line Integral with respect to Arc Length

The integral I = ∫ f(x,y) ds can be converted into an integral in terms of x:

📐 Formula: I = ∫ f(x,y) dx where ds/dx = √(1 + (dy/dx)²)

I = ∫ₓ₁ˣ² f(x,y)√(1 + (dy/dx)²) dx

Example 2: Straight Line Path

Evaluate I = ∫(4x+3xy)ds where c is the straight line joining O(0,0) to A(1,2).

c is the line y = 2x → dy/dx = 2

ds/dx = √(1 + 2²) = √5

I = ∫₀¹ (4x+3xy)(√5) dx

Since y = 2x: I = ∫₀¹ (4x+6x²)(√5) dx = 2√5 ∫₀¹ (2x+3x²) dx = 4√5

Parametric Equations

When x = φ(t) and y = g(t):

📐 Formula: ds/dt = √((dx/dt)² + (dy/dt)²) → ds = √((dx/dt)² + (dy/dt)²) dt

I = ∫ₜ₁ᵗ² f(x,y) √((dx/dt)² + (dy/dt)²) dt

Example 3: Parametric Curve

Evaluate I = ∮ 4xyds where c is defined as x = sin t, y = cos t between t=0 and t=π/4.

dx/dt = cos t, dy/dt = -sin t

ds/dt = √(cos²t + sin²t) = 1

I = ∫₀^(π/4) 4 sin t cos t (1) dt = 2∫₀^(π/4) sin 2t dt

= -2[cos 2t/2]₀^(π/4) = -[cos(π/2) - cos 0] = -[0 - 1] = 1

Dependence of Line Integral on Path of Integration

Integration along two separate paths joining the same two endpoints does not necessarily give identical results.

Example 4: Path Independence Demonstration

Evaluate I = ∮{3x²y² dx + 2x³y dy} between O(0,0) and A(2,4) along three different paths.

Path c₁: y = x² dy = 2x dx I₁ = ∫₀² (3x²x⁴ + 2x³x²·2x)dx = ∫₀² (3x⁶ + 4x⁶)dx = ∫₀² 7x⁶ dx = [x⁷]₀² = 128

Path c₂: y = 2x dy = 2dx I₂ = ∫₀² (3x²·4x² + 2x³·2x·2)dx = ∫₀² (12x⁴ + 8x⁴)dx = ∫₀² 20x⁴ dx = 4[x⁵]₀² = 128

Path c₃: x=0 from (0,0) to (0,4), then y=4 from (0,4) to (2,4) Segment 1: x=0, dx=0 → I₃ₐ = 0 Segment 2: y=4, dy=0 → I₃ᵦ = ∫₀² 48x² dx = 128 I₃ = 128

All three paths give the same result: 128.

🔑 Definition — Exact Differential: The integrand 3x²y² dx + 2x³y dy is an exact differential of z = x³y², since ∂z/∂x = 3x²y² and ∂z/∂y = 2x³y.

If the integrand is an exact differential, the value of the line integral is independent of the path and depends only on the coordinates of the two endpoints.

⭐ Key Takeaways

Line integrals around closed boundaries are evaluated by breaking them into segments and summing individual contributions, always traversing anticlockwise. Arc length integrals require converting ds using ds/dx = √(1+(dy/dx)²) for Cartesian form or ds/dt = √((dx/dt)²+(dy/dt)²) for parametric form. The most critical concept is that when the integrand Pdx+Qdy is an exact differential of some function, the line integral becomes path-independent, depending only on the endpoints. This was demonstrated when 3x²y²dx+2x³ydy was shown to be the exact differential of x³y², giving identical results (128) for three different paths. Students must be able to recognize exact differentials and understand their implications for path independence.

🧠 Quick Revision Questions

1. What is the value of the line integral around the rectangle A(1,0), B(3,0), C(3,2), D(1,2) for the integrand xydx+(1+y²)dy?

2. How do you convert a line integral with respect to arc length into an integral with respect to x?

3. What is the parametric formula for ds when x = φ(t) and y = g(t)?

4. What condition must the integrand Pdx+Qdy satisfy for the line integral to be independent of path?

5. Why did all three different paths in Example 4 give the same result of 128?

📘 Lecture 33 — Examples

📖 Overview: This lecture provides worked examples demonstrating the integration of exact differentials in both two and three independent variables. It also introduces Green’s theorem, showing how a double integral over a plane region can be transformed into a line integral over its boundary, and vice versa, with practical examples to illustrate its application.

🗂️ Topics Covered

The lecture covers evaluating line integrals of exact differentials in two variables with examples, extending the concept to exact differentials in three independent variables with a verification and evaluation example, and then introduces and applies Green’s theorem to evaluate line integrals around closed boundaries, including an ellipse and a triangle, comparing the method to direct path integration.

📝 Lecture Summary

Example

Evaluate ( I = \int_C {3y dx + (3x+2y) dy} ) from A(1, 2) to B(3, 5).

No path is given, so the integrand is doubtless an exact differential of some function ( z = f(x, y) ). In fact ( \frac{\partial P}{\partial y} = 3 = \frac{\partial Q}{\partial x} ). We have already dealt with the integration of exact differentials, so there is no difficulty. Compare with ( I = \int_C {P dx + Q dy} ).

( P = \frac{\partial z}{\partial x} = 3y \Rightarrow z = \int 3y dx = 3xy + f(y) ) (i) ( Q = \frac{\partial z}{\partial y} = 3x + 2y \Rightarrow z = \int (3x+2y) dy = 3xy + y^2 + F(x) ) (ii)

For (i) and (ii) to agree, ( f(y) = y^2 ) and ( F(x) = 0 ). Hence ( z = 3xy + y^2 ).

( \Rightarrow I = \int_{(1,2)}^{(3,5)} {3y dx + (3x+2y) dy} = \int_{(1,2)}^{(3,5)} d(3xy+y^2) = [3xy+y^2]_{(1,2)}^{(3,5)} = (45+25) - (6+4) = 60 )

Example

Evaluate ( I = \int_C {(x^2+ye^x) dx + (e^x+y) dy} ) between A(0, 1) and B(1, 2).

As before, compare with ( \int_C {P dx+Q dy} ).

( P = \frac{\partial z}{\partial x} = x^2 + ye^x \Rightarrow z = \frac{x^3}{3} + ye^x + f(y) ) ( Q = \frac{\partial z}{\partial y} = e^x + y \Rightarrow z = ye^x + \frac{y^2}{2} + F(x) )

For these expressions to agree, ( f(y) = \frac{y^2}{2} ) and ( F(x) = \frac{x^3}{3} ).

Then ( I = \left[ \frac{x^3}{3} + ye^x + \frac{y^2}{2} \right]_{(0,1)}^{(1,2)} = \frac{5}{6} + 2e )

So the main points are that, if ( (Pdx+Qdy) ) is an exact differential: (a) ( I = \int_C (Pdx + Qdy) ) is independent of the path of integration. (b) ( \oint_C (P dx + Q dy) ) is zero.

If ( I = \int_C {P dx + Q dy} ) and ( (Pdx+Qdy) ) is an exact differential, then ( I_{c_1} = I_{c_2} ) and ( I_{c_1} + I_{c_2} = 0 ). Hence, the integration taken round a closed curve is zero, provided ( (Pdx+Q dy) ) is an exact differential. Therefore, if ( (P dx + Q dy) ) is an exact differential, ( \oint_C (P dx + Q dy) = 0 ).

Exact differentials in three independent variables

A line integral in space naturally involves three independent variables, but the method is very much like that for two independent variables.

( dz = Pdx + Q dy + R dw ) is an exact differential of ( z = f(x, y, w) ) if: ( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ; \frac{\partial P}{\partial w} = \frac{\partial R}{\partial x} ; \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial w} )

If the test is successful, then: (a) ( \int_C (P dx + Q dy + R dw) ) is independent of the path of integration. (b) ( \oint_C (P dx + Q dy + R dw) ) is zero.

Example Verify that ( dz = (3x^2 y w + 6x) dx + (x^3 w - 8y) dy + (x^3 y + 1) dw ) is an exact differential and hence evaluate ( \int_C dz ) from A(1, 2, 4) to B(2, 1, 3).

First check that ( dz ) is an exact differential by finding the partial derivatives, where ( P = 3x^2 y w + 6x ), ( Q = x^3 w - 8y ), and ( R = x^3 y + 1 ).

( \frac{\partial P}{\partial y} = 3x^2 w ; \frac{\partial Q}{\partial x} = 3x^2 w \Rightarrow \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ) ( \frac{\partial P}{\partial w} = 3x^2 y ; \frac{\partial R}{\partial x} = 3x^2 y \Rightarrow \frac{\partial P}{\partial w} = \frac{\partial R}{\partial x} ) ( \frac{\partial R}{\partial y} = x^3 ; \frac{\partial Q}{\partial w} = x^3 \Rightarrow \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial w} ) Therefore, ( dz ) is an exact differential.

Now to find ( z ). ( P = \frac{\partial z}{\partial x} ; Q = \frac{\partial z}{\partial y} ; R = \frac{\partial z}{\partial w} ).

( \frac{\partial z}{\partial x} = 3x^2 y w + 6x \Rightarrow z = \int (3x^2 y w + 6x) dx = x^3 y w + 3x^2 + f(y) + F(w) ) ( \frac{\partial z}{\partial y} = x^3 w - 8x \Rightarrow z = \int (x^3 w - 8y) dy = x^3 y w - 4y^2 + g(x) + F(w) ) ( \frac{\partial z}{\partial w} = x^3 y + 1 \Rightarrow z = \int (x^3 y + 1) dw = x^3 y w + w + h(y) + g(x) )

For these three expressions for ( z ) to agree: ( f(y) = -4y^2 ; F(w) = w ; g(x) = 3x^2 ). Therefore, ( z = x^3 y w + 3x^2 - 4y^2 + w ).

( \Rightarrow I = \int_{(1,2,4)}^{(2,1,3)} dz = [x^3 y w + 3x^2 - 4y^2 + w]_{(1,2,4)}^{(2,1,3)} = (24+12-4+3) - (8+3-16+4) = 36 )

The extension to line integrals in space is thus quite straightforward.

Finally, we have a theorem that can be very helpful on occasions and which links up with the work we have been doing.

Green’s Theorem

Let ( P ) and ( Q ) be two functions of ( x ) and ( y ) that are finite and continuous inside and on the boundary ( c ) of a region ( R ) in the xy-plane. If the first partial derivatives are continuous within the region and on the boundary, then Green’s theorem states that: ( \iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx dy = - \oint_C (P dx + Q dy) )

That is, a double integral over the plane region ( R ) can be transformed into a line integral over the boundary ( c ) of the region – and the action is reversible.

EXAMPLE Evaluate ( I = \oint_C {(2x - y) dx + (2y + x) dy} ) around the boundary ( c ) of the ellipse ( x^2 + 9y^2 = 16 ).

The integral is of the form ( I = \oint_C {P dx + Q dy} ) where ( P = 2x - y \Rightarrow \frac{\partial P}{\partial y} = -1 ) and ( Q = 2y + x \Rightarrow \frac{\partial Q}{\partial x} = 1 ).

( \Rightarrow I = - \iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx dy = - \iint_R (-1 - 1) dx dy = 2 \iint_R dx dy )

But ( \iint_R dx dy ) over any closed region gives the area of the figure. In this case, then, ( I = 2A ) where ( A ) is the area of the ellipse ( x^2 + 9y^2 = 16 ) i.e. ( \frac{x^2}{16} + \frac{9y^2}{16} = 1 ). Therefore, ( a = 4 ) and ( b = \frac{4}{3} ), so ( A = \frac{16\pi}{3} ). Hence, ( I = 2A = \frac{32\pi}{3} ).

To demonstrate the advantage of Green’s theorem, let us work through the next example (a) by the previous method, and (b) by applying Green’s theorem.

Example Evaluate ( I = \oint_C {(2x+y) dx + (3x-2y) dy} ) taken in anticlockwise manner round the triangle with vertices at O(0,0), A(1,0), B(1,2).

(a) By the previous method There are clearly three stages with ( c_1, c_2, c_3 ).

(i) ( c_1 ) is ( y = 0 \Rightarrow dy = 0 ). ( I_1 = \int_0^1 2x dx = [x^2]_0^1 = 1 ).

(ii) ( c_2 ) is ( x = 1 \Rightarrow dx = 0 ). ( I_2 = \int_0^2 (3-2y) dy = [3y - y^2]_0^2 = 2 ).

(iii) ( c_3 ) is ( y = 2x \Rightarrow dy = 2 dx ). ( I_3 = \int_1^0 {4x dx + (3x - 4x)2 dx} = \int_1^0 2x dx = [x^2]_1^0 = -1 ).

( I = I_1 + I_2 + I_3 = 1 + 2 + (-1) = 2 ).

(b) By Green’s theorem ( I = \oint_C {(2x+y) dx + (3x-2y) dy} ) ( P = 2x + y \Rightarrow \frac{\partial P}{\partial y} = 1 ; Q = 3x - 2y \Rightarrow \frac{\partial Q}{\partial x} = 3 ).

( I = - \iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx dy = - \iint_R (1 - 3) dx dy = 2 \iint_R dx dy = 2A )

( I = 2 \times ) the area of the triangle ( = 2 \times 1 = 2 ).

Application of Green’s theorem is not always the quickest method. It is useful, however, to have both methods available.

📐 Formula: ( \iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx dy = - \oint_C (P dx + Q dy) ) → The double integral of the difference of partial derivatives over a plane region equals the negative of the line integral around the closed boundary.

📌 Example (Ellipse): The line integral around ellipse ( x^2 + 9y^2 = 16 ) is ( 32\pi/3 ), found by using Green's theorem to convert it to ( 2 \times ) the area of the ellipse.

📌 Example (Triangle): The line integral around the triangle (0,0), (1,0), (1,2) is 2, found both by direct path integration and by Green's theorem as ( 2 \times ) the triangle area (area=1).

💡 Why this matters: Green's theorem provides a powerful alternative for evaluating line integrals, especially for closed curves, by converting them into double integrals over the enclosed area, which can often be simpler to compute.

⭐ Key Takeaways

The key points from this lecture are understanding that for an exact differential ( Pdx+Qdy ), its line integral is path-independent and zero around any closed loop, with this property extending to three independent variables. Green's theorem provides a crucial connection between a line integral around a closed curve and a double integral over the enclosed region, allowing conversion between the two forms. The condition for a differential to be exact involves equality of mixed partial derivatives, which must be verified before applying path independence. Worked examples demonstrate that using integration by recognizing exact differentials can be more efficient than direct path integration, and using Green's theorem is often simpler than evaluating a line integral directly along multiple segments. Finally, both direct and theorem-based methods serve as valuable tools, with the choice depending on the specific problem.

🧠 Quick Revision Questions

1. State the condition for which ( \oint_C (P dx + Q dy) = 0 ).
2. What are the three conditions that must hold for ( Pdx+Qdy+Rdw ) to be an exact differential in three variables?
3. Write down Green’s theorem relating a line integral around a closed curve C to a double integral over the region R bounded by C.
4. For the integral ( I = \oint_C {(2x+y) dx + (3x-2y) dy} ) around a triangle, what does Green’s theorem reduce the calculation to, and what is its value?
5. If ( dz = Pdx + Qdy ) is an exact differential, what can be said about the value of the line integral ( \int_C dz ) between two points, regardless of the path?

📘 Lecture 34 — Examples

📖 Overview: This lecture applies Green’s theorem to evaluate line integrals and compute areas bounded by curves. It also introduces the fundamental vector calculus operators — gradient, divergence, and curl — along with their properties and multiple operations, concluding with general results like curl grad φ = 0 and div curl A = 0.

🗂️ Topics Covered

The lecture begins with examples using Green’s theorem to compute line integrals and areas enclosed by curves, including regions bounded by y = x² and x = y², y = 3x² and y = 6x, and y = 2x³ with y = x³ + 1. It then presents a revision summary of line integral properties and Green’s theorem. The second half introduces the gradient of a scalar function, divergence of a vector function, and curl of a vector function, with worked examples for each. Multiple operations such as grad div A, div grad φ, and curl curl F are demonstrated, followed by proofs that curl grad φ = 0 and div curl A = 0.

📝 Lecture Summary

Example 1: Line Integral Using Green’s Theorem

Evaluate the line integral ( I = \oint_C {xy , dx + (2x - y) , dy} ) around the region bounded by ( y = x^2 ) and ( x = y^2 ). Points of intersection are O(0,0) and A(1,1).

Given ( P = xy ), so ( \frac{\partial P}{\partial y} = x ); ( Q = 2x - y ), so ( \frac{\partial Q}{\partial x} = 2 ).

By Green’s theorem: ( I = -\iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx , dy = -\iint_R (x - 2) dx , dy ) [ I = -\int_{0}^{1} \int_{y=x^2}^{y=\sqrt{x}} (x - 2) dy , dx = -\int_{0}^{1} (x - 2)(\sqrt{x} - x^2) dx ] [ = -\int_{0}^{1} (x^{3/2} - x^3 - 2x^{1/2} + 2x^2) dx = -\left[ \frac{2}{5}x^{5/2} - \frac{1}{4}x^4 - \frac{4}{3}x^{3/2} + \frac{2}{3}x^3 \right]_0^1 = \frac{31}{60} ]

Area Using Green’s Theorem (Special Case)

When ( P = y ) and ( Q = -x ), we have ( \frac{\partial P}{\partial y} = 1 ) and ( \frac{\partial Q}{\partial x} = -1 ). Green’s theorem gives: [ \iint_R {1 - (-1)} dx , dy = -\oint_C (y , dx - x , dy) ] [ 2 \iint_R dx , dy = \oint_C (x , dy - y , dx) ] Therefore, the area of the closed region ( A = \iint_R dx , dy = \frac{1}{2} \oint_C (x , dy - y , dx) )

🔑 Definition — Area formula using line integral: ( A = \frac{1}{2} \oint_C (x , dy - y , dx) )

Example 2: Area Enclosed by y = 3x² and y = 6x

Points of intersection: ( 3x^2 = 6x \Rightarrow x = 0 ) or ( x = 2 ). Evaluate along two paths: OA along c₁ (y = 3x²) and AO along c₂ (y = 6x).

• I₁: c₁ is y = 3x² ⇒ dy = 6x dx. ( I_1 = \int_0^2 (6x^2 dx - 3x^2 dx) = \int_0^2 3x^2 dx = [x^3]_0^2 = 8 )
• I₂: c₂ is y = 6x ⇒ dy = 6 dx. ( I_2 = \int_2^0 (6x dx - 6x dx) = 0 ) So ( 2A = I_1 + I_2 = 8 \Rightarrow A = 4 ) square units.

📌 Example: The area bounded by y = 3x² and y = 6x is 4 square units.

Example 3: Area Bounded by y = 2x³, y = x³ + 1, and x = 0 (for x > 0)

Point of intersection: ( 2x^3 = x^3 + 1 \Rightarrow x = 1 ). Evaluate along three paths:

• OA (c₁): y = 2x³, dy = 6x² dx. ( I_1 = \int_0^1 (6x^3 - 2x^3) dx = \int_0^1 4x^3 dx = [x^4]_0^1 = 1 )
• AB (c₂): y = x³ + 1, dy = 3x² dx. ( I_2 = \int_1^0 (3x^3 - (x^3 + 1)) dx = \int_1^0 (2x^3 - 1) dx = \left[ \frac{1}{2}x^4 - x \right]_1^0 = \frac{1}{2} )
• BO (c₃): x = 0, dx = 0. ( I_3 = 0 ) So ( 2A = 1 + \frac{1}{2} + 0 = \frac{3}{2} \Rightarrow A = \frac{3}{4} ) square units.

Revision Summary — Properties of Line Integrals

• Sign reversal: Reversing the direction of integration reverses the sign of the line integral.
• Path parallel to y-axis: dx = 0 ⇒ ( I_c = \int_c Q , dy )
• Path parallel to x-axis: dy = 0 ⇒ ( I_c = \int_c P , dx )
• Path of integration must be continuous and single-valued.
• In general, the value of the line integral depends on the particular path.
• Exact differential: If P dx + Q dy is an exact differential, then: (a) ( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ) (b) ( I = \int_c (P dx + Q dy) ) is independent of path (c) ( \oint_C (P dx + Q dy) = 0 )
• Exact differential in three variables: For P dx + Q dy + R dw: (a) ( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \frac{\partial P}{\partial w} = \frac{\partial R}{\partial x}, \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial w} ) (b) Path independent; (c) ( \oint = 0 )
• Green’s theorem: ( \oint_C (P dx + Q dy) = -\iint_R \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) dx , dy ) and ( \oint_C (x dy - y dx) = 2 \iint_R dx , dy = 2A )

Gradient of a Scalar Function

The del operator is given by ( \nabla = \left( i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z} \right) ) [ \nabla \phi = \text{grad } \phi = \left( i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z} \right) \phi = i \frac{\partial \phi}{\partial x} + j \frac{\partial \phi}{\partial y} + k \frac{\partial \phi}{\partial z} ]

🔑 Definition — Gradient: ( \text{grad } \phi = \nabla \phi = \frac{\partial \phi}{\partial x} i + \frac{\partial \phi}{\partial y} j + \frac{\partial \phi}{\partial z} k )

Divergence of a Vector Function

If ( A = a_1 i + a_2 j + a_3 k ), then: [ \text{div } A = \nabla \cdot A = \left( i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z} \right) \cdot (a_1 i + a_2 j + a_3 k) = \frac{\partial a_1}{\partial x} + \frac{\partial a_2}{\partial y} + \frac{\partial a_3}{\partial z} ]

💡 Why this matters: The grad operator acts on a scalar and gives a vector; the div operator acts on a vector and gives a scalar.

📌 Example: If ( A = x^2 y i - xyz j + yz^2 k ), then div A = ( \frac{\partial}{\partial x}(x^2 y) - \frac{\partial}{\partial y}(xyz) + \frac{\partial}{\partial z}(yz^4) = 2xy - xz + 2yz )

📌 Example: If ( A = 2x^2 y i - 2(xy^2 + y^3)j + 3y^2 z^2 k ), then div A = 4xy - 2(2xy + 3y²z) + 6y²z = 0. Such a vector with ( \nabla \cdot A = 0 ) at all points is called a solenoid vector.

Curl of a Vector Function

The curl operator ( \nabla \times ) acts on a vector and gives another vector.

If ( A = a_1 i + a_2 j + a_3 k ), then: [ \text{curl } A = \nabla \times A = \begin{vmatrix} i & j & k \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ a_1 & a_2 & a_3 \end{vmatrix} ] [ = i \left( \frac{\partial a_3}{\partial y} - \frac{\partial a_2}{\partial z} \right) + j \left( \frac{\partial a_1}{\partial z} - \frac{\partial a_3}{\partial x} \right) + k \left( \frac{\partial a_2}{\partial x} - \frac{\partial a_1}{\partial y} \right) ]

🔑 Definition — Curl: curl A = ( \nabla \times A ) is a vector function.

📌 Example: If ( A = (y^4 - x^2 z^2)i + (x^2 + y^2)j - x^2 yz k ), find curl A at (1, 3, -2). ( \nabla \times A = i{ -x^2 z } - j{ -2xyz + 2x^2 z } + k{ 2x - 4y^3 } ) At (1, 3, -2): ( \nabla \times A = i(2) - j(12 - 4) + k(2 - 108) = 2i - 8j - 106k )

📌 Example: Determine curl F at (2, 0, 3) given ( F = ze^{2xy} i + 2xy \cos y j + (x + 2y)k ). Expanding the determinant: ( \nabla \times F = i{2 - 2x \cos y} - j{1 - e^{2xy}} + k{2z \cos y - 2xz e^{2xy}} ) At (2, 0, 3): ( \nabla \times F = i(2-4) - j(1-1) + k(6-12) = -2i - 6k = -2(i + 3k) )

Summary of Grad, Div, and Curl

(a) Grad operator ( \nabla ) acts on a scalar field → gives a vector field. (b) Div operator ( \nabla \cdot ) acts on a vector field → gives a scalar field. (c) Curl operator ( \nabla \times ) acts on a vector field → gives a vector field. (d) For scalar function ( \phi(x,y,z) ): ( \text{grad } \phi = \nabla \phi = \frac{\partial \phi}{\partial x} i + \frac{\partial \phi}{\partial y} j + \frac{\partial \phi}{\partial z} k ) (e) For vector function ( A = a_x i + a_y j + a_z k ): (i) ( \text{div } A = \nabla \cdot A = \frac{\partial a_x}{\partial x} + \frac{\partial a_y}{\partial y} + \frac{\partial a_z}{\partial z} ) (ii) ( \text{curl } A = \nabla \times A = \begin{vmatrix} i & j & k \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ a_x & a_y & a_z \end{vmatrix} )

Multiple Operations — Grad div A

If ( A = x^2 y i + y z^3 j - z x^3 k ), then div A = ( 2xy + z^3 + x^3 = \phi ) (say). Then grad (div A) = ( \nabla(\nabla \cdot A) = \frac{\partial \phi}{\partial x} i + \frac{\partial \phi}{\partial y} j + \frac{\partial \phi}{\partial z} k = (2y + 3x^2)i + (2x)j + (3z^2)k )

📌 Example: If ( \phi = xyz - 2y^2 z + x^2 z^2 ), find div grad ( \phi ) at (2, 4, 1). grad ( \phi = (yz + 2xz^2)i + (xz - 4yz)j + (xy - 2y^2 + 2x^2 z)k ) div grad ( \phi = \nabla \cdot (\nabla \phi) = 2z^2 - 4z + 2x^2 ) At (2, 4, 1): ( \nabla \cdot (\nabla \phi) = 2 - 4 + 8 = 6 )

📐 Formula: div grad ( \phi = \nabla \cdot (\nabla \phi) = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} )

Multiple Operations — Curl curl F

If ( F = x^2 yz i + xyz^2 j + y^2 z k ), find curl curl F at (2, 1, 1). First: curl F = ( \nabla \times F = (2yz - 2xyz)i + x^2 y j + (yz^2 - x^2 z)k ) Then: curl curl F = ( \nabla \times (\nabla \times F) = z^2 i - (-2xz - 2y + 2xy)j + (2xy - 2z + 2xz)k ) At (2, 1, 1): curl curl F = i + 2j + 6k

Two Interesting General Results

(a) curl grad ( \phi = 0 ): Let ( \phi ) be a scalar. Then grad ( \phi = \frac{\partial \phi}{\partial x} i + \frac{\partial \phi}{\partial y} j + \frac{\partial \phi}{\partial z} k ). Computing the curl: [ \nabla \times (\nabla \phi) = i\left( \frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y} \right) - j\left( \frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z} \right) + k\left( \frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x} \right) = 0 ] ✅ Result: curl grad ( \phi = \nabla \times (\nabla \phi) = 0 )

(b) div curl A = 0: Let ( A = a_x i + a_y j + a_z k ). curl A = ( i\left( \frac{\partial a_z}{\partial y} - \frac{\partial a_y}{\partial z} \right) - j\left( \frac{\partial a_z}{\partial x} - \frac{\partial a_x}{\partial z} \right) + k\left( \frac{\partial a_y}{\partial x} - \frac{\partial a_x}{\partial y} \right) ) Then div curl A = ( \frac{\partial}{\partial x}\left( \frac{\partial a_z}{\partial y} - \frac{\partial a_y}{\partial z} \right) - \frac{\partial}{\partial y}\left( \frac{\partial a_z}{\partial x} - \frac{\partial a_x}{\partial z} \right) + \frac{\partial}{\partial z}\left( \frac{\partial a_y}{\partial x} - \frac{\partial a_x}{\partial y} \right) = 0 ) ✅ Result: div curl A = ( \nabla \cdot (\nabla \times A) = 0 )

(c) div grad ( \phi ): [ \nabla \cdot (\nabla \phi) = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} ] This is sometimes denoted by ( \nabla^2 \phi ).

⭐ Key Takeaways

Green’s theorem converts a line integral over a closed curve into a double integral over the enclosed area, with the area formula ( A = \frac{1}{2} \oint (x dy - y dx) ) as a special case. The three fundamental vector operators are: gradient (( \nabla \phi ), vector result from scalar), divergence (( \nabla \cdot A ), scalar result from vector), and curl (( \nabla \times A ), vector result from vector). Two essential identities are curl grad φ = 0 and div curl A = 0, which are crucial for verifying conservative fields and vector field properties. Multiple operations like div grad φ produce the Laplacian ( \nabla^2 \phi ), combining second partial derivatives. Mastering these operators and identities is foundational for advanced topics in electromagnetism and fluid dynamics.

🧠 Quick Revision Questions

1. What is the formula for area using Green’s theorem, and how is it derived from the special case P = y, Q = -x?
2. Compute the area bounded by the curves y = 3x² and y = 6x using the line integral method.
3. If A = 2x²y i - 2(xy² + y³)j + 3y²z²k, show that div A = 0 and state what type of vector this is called.
4. Prove that curl grad φ = 0 for any scalar function φ.
5. Given F = x²yz i + xyz² j + y²z k, compute curl curl F at the point (2, 1, 1).

📘 Lecture 35 — Definite Integrals

📖 Overview: This lecture focuses on evaluating definite integrals of powers of sine and cosine functions over the interval [0, π/2], leading to the Wallis Sine Formula. It also introduces Integration by Parts as a technique for integrating products of functions, and begins the topic of Line Integrals for scalar fields along curves.

🗂️ Topics Covered

The lecture covers definite integrals for sinⁿx and cosⁿx from 0 to π/2 for n = 2 through 12, establishing patterns for even and odd exponents. It then formalizes these patterns into the Wallis Sine Formula. The second half introduces Integration by Parts with worked examples, and concludes with an introduction to Line Integrals of scalar fields along curves, including parametric evaluation.

📝 Lecture Summary

Definite integral for sinⁿx and cosⁿx , 0 ≤ x ≤ π/2

The lecture begins by evaluating specific definite integrals of sinⁿx and cosⁿx from 0 to π/2. For n=2, using the identity sin²x = (1 - cos 2x)/2, the result is π/4. Similarly, cos²x integrates to π/4. For n=3, using sin³x = sin²x sin x = (1 - cos²x) sin x, the integral evaluates to 2/3. For n=4, using sin⁴x = (sin²x)² and the double-angle formula again, the result is 3π/16. The same process is repeated for cos⁴x, yielding the same value.

🔑 Definition — Definite Integral of sinⁿx: The integral of sinⁿx from 0 to π/2 depends on whether n is even or odd. 📐 Formula: For n=2, ∫₀^{π/2} sin²x dx = π/4 📌 Example: ∫₀^{π/2} sin⁴x dx = 1/4 [ (3/2)(π/2) - sin π + (sin 2π)/8 ] = 3π/16

The pattern continues: ∫₀^{π/2} sin⁵x dx = (4·2)/(5·3) = 8/15. ∫₀^{π/2} sin⁶x dx = (5·3·1)/(6·4·2) · π/2 = 5π/32. ∫₀^{π/2} sin⁷x dx = (6·4·2)/(7·5·3) = 48/105. ∫₀^{π/2} sin⁸x dx = (7·5·3·1)/(8·6·4·2) · π/2 = 35π/256. The same patterns hold for cosⁿx: ∫₀^{π/2} cos⁵x dx = 8/15, ∫₀^{π/2} cos⁶x dx = 5π/32, etc.

💡 Why this matters: These integrals appear frequently in physics and engineering, especially in Fourier series and wave analysis.

Wallis Sine Formula

The Wallis Sine Formula provides a compact way to evaluate ∫₀^{π/2} sinⁿx dx and ∫₀^{π/2} cosⁿx dx for any positive integer n.

🔑 Definition — Wallis Sine Formula: A formula for the definite integral of sinⁿx (or cosⁿx) from 0 to π/2. 📐 Formula: When n is even: ∫₀^{π/2} sinⁿx dx = (n-1)/(n) · (n-3)/(n-2) · (n-5)/(n-4) · ... · (5·3·1)/(6·4·2) · π/2. When n is odd: ∫₀^{π/2} sinⁿx dx = (n-1)/(n) · (n-3)/(n-2) · ... · (6·4·2)/(7·5·3). 📌 Example: For n=11 (odd): ∫₀^{π/2} sin¹¹x dx = (10·8·6·4·2)/(11·9·7·5·3). For n=12 (even): ∫₀^{π/2} sin¹²x dx = (11·9·7·5·3·1)/(10·8·6·4·2) · π/2.

The same formula applies for cosⁿx.

Integration By Parts

Integration by Parts is a technique derived from the product rule for differentiation.

🔑 Definition — Integration by Parts: A method to integrate the product of two functions. 📐 Formula: ∫UV dx = U∫V dx - ∫[∫V dx · (dU/dx)] dx 📌 Example: Evaluate ∫x ln x dx. Let U = ln x, V = x. Then ∫x ln x dx = ln x (x²/2) - ∫[(x²/2)(1/x)] dx = (x²/2) ln x - ∫(x/2) dx = (x²/2) ln x - x²/4 + C. 📌 Example: Evaluate ∫x sin x dx. Let U = x, V = sin x. Then ∫x sin x dx = x(-cos x) - ∫(-cos x)(1) dx = -x cos x + ∫cos x dx = -x cos x + sin x + C.

Line Integrals

A line integral extends integration along a curve in space.

🔑 Definition — Line Integral: ∫_c V(r) dr, where V is a scalar field and the integral is taken along curve c from point A to point B. 🔑 Definition — Scalar Field: A function V(r) that assigns a scalar value to every point in space. 🔑 Definition — Position Vector r: A vector from the origin to a point on the curve.

The curve c is divided into small vector elements dr_p. As dr → 0, the sum of V(r)dr_p becomes the line integral. To evaluate, express dr in components: dr = dx i + dy j + dz k, and parameterize x, y, z in terms of a parameter u.

📌 Example: If x = x(u), y = y(u), z = z(u), then dx = (dx/du) du, etc., and the integral becomes ∫_c V(x,y,z) √[(dx/du)² + (dy/du)² + (dz/du)²] du.

⭐ Key Takeaways

For definite integrals of sinⁿx and cosⁿx from 0 to π/2, a clear pattern emerges: when n is even, the result includes a factor of π/2; when n is odd, the result is a rational number. The Wallis Sine Formula captures this pattern and allows quick computation without repeated integration. Integration by Parts requires careful selection of U (to be differentiated) and V (to be integrated), and is essential for products like x ln x or x sin x. Line integrals extend integration to curves in space and require parametric representation of the curve for evaluation. The scalar field V(r) is integrated along the curve by summing contributions from infinitesimal vector arc elements dr.

🧠 Quick Revision Questions

1. What is ∫₀^{π/2} sin⁴x dx equal to?
2. State the Wallis Sine Formula for even n.
3. What is ∫₀^{π/2} cos⁹x dx using the Wallis pattern?
4. Using Integration by Parts, evaluate ∫x cos x dx.
5. How do you evaluate a line integral of a scalar field V along a curve c?

📘 Lecture 36 — Scalar Field

📖 Overview: This lecture introduces line integrals of scalar and vector fields, explaining how to evaluate them along specified curves. It demonstrates the path-dependent nature of line integrals for vector fields, which is a fundamental concept in vector calculus with applications in physics and engineering.

🗂️ Topics Covered

The lecture covers the definition and evaluation of line integrals for scalar fields, with examples using parametric equations to express the curve. It then extends the concept to vector fields, showing how to compute the scalar product line integral along a curve. The lecture includes detailed examples demonstrating the integration process for both scalar and vector fields over various paths, including piecewise linear paths and straight lines.

📝 Lecture Summary

Scalar Field

If a scalar field V(r) exists for all points on a curve c, the line integral of V is defined as ∑ V(r) dr over n segments, and as dr → 0, this becomes ∫c V(r) dr. This integral is represented by the area of a curved surface erected above the curve between points A and B. To evaluate, the integrand is expressed in terms of x, y, z with dr = dx i + dy j + dz k. In practice, x, y, and z are often expressed in parametric form x = x(u), y = y(u), z = z(u), allowing dx, dy, dz to be written in terms of u.

🔑 Definition — Line integral of a scalar field: ∫c V(r) dr is the integral of the scalar field V along curve c, evaluated by expressing V and dr in terms of a parameter.

📐 Formula: ∫c V(r) dr = ∫ V(u) [x'(u) i + y'(u) j + z'(u) k] du

📌 Example: If V = xy²z, evaluate ∫c V(r) dr along curve c: x = 3u, y = 2u², z = u³ between A(0,0,0) and B(3,2,1).

• V = (3u)(4u⁴)(u³) = 12u⁸
• dr = 3du i + 4u du j + 3u² du k (since dx=3du, dy=4u du, dz=3u² du)
• Limits: u=0 at A, u=1 at B
• ∫c V dr = ∫₀¹ 12u⁸ (3 i + 4u j + 3u² k) du = ∫₀¹ (36u⁸ i + 48u⁹ j + 36u¹⁰ k) du = [36u⁹/9 i + 48u¹⁰/10 j + 36u¹¹/11 k]₀¹ = 4i + 24/5 j + 36/11 k

📌 Example: If V = xy + y²z, evaluate ∫c V(r) dr along curve c: x = t², y = 2t, z = t+5 between A(0,0,5) and B(4,4,7).

• V = (t²)(2t) + (4t²)(t+5) = 6t³ + 20t²
• dr = 2t dt i + 2 dt j + dt k
• Limits: t=0 at A, t=2 at B
• ∫c V dr = ∫₀² (6t³+20t²)(2t i + 2 j + k) dt = 2∫₀² [(6t⁴+20t³)i + (6t³+20t²)j + (3t³+10t²)k] dt = (8/15)(444i + 290j + 145k)

Vector Field

If a vector field F(r) exists for all points on curve c, the line integral from A to B is ∫c F·dr. Since F·dr is a scalar product, the line integral is a scalar. To evaluate, F and dr are expressed in terms of x, y, z: F = F₁ i + F₂ j + F₃ k and dr = dx i + dy j + dz k, so F·dr = F₁dx + F₂dy + F₃dz.

🔑 Definition — Line integral of a vector field: ∫c F·dr = ∫c F₁dx + F₂dy + F₃dz

📐 Formula: ∫c F·dr = ∫ [F₁(x(u),y(u),z(u)) x'(u) + F₂(u) y'(u) + F₃(u) z'(u)] du

📌 Example: If F(r) = x²y i + xz j + 2yz k, evaluate ∫c F·dr between A(0,0,0) and B(4,2,1) along curve c: x=4t, y=2t², z=t³.

• dx=4dt, dy=4t dt, dz=3t² dt
• x²y = 32t⁴, xz = 4t⁴, 2yz = 4t⁵
• F = 32t⁴ i + 4t⁴ j - 4t⁵ k (note: the text shows -4t⁵ k but the calculation uses +12t⁷)
• dr = 4dt i + 4t dt j + 3t² dt k
• F·dr = (32t⁴·4) + (4t⁴·4t) + (-4t⁵·3t²) = 128t⁴ + 16t⁵ - 12t⁷ (the example in text shows +12t⁷)
• Limits: t=0 at A, t=1 at B
• ∫c F·dr = ∫₀¹ (128t⁴+16t⁵+12t⁷) dt = [128t⁵/5 + 16t⁶/6 + 12t⁸/8]₀¹ = 128/5 + 8/3 + 3/2 = 29.76

💡 Why this matters: The value of a line integral for vector fields can depend on the path taken between endpoints, as the next example shows.

📌 Example: Evaluate ∫c F·dr where F = x²y i + 2yz j + 3z²x k between A(0,0,0) and B(1,2,3) via two different paths.

Path (a): Along c₁, c₂, c₃ piecewise

• c₁: from (0,0,0) to (1,0,0): y=0, z=0, dy=0, dz=0 → ∫c₁ = 0
• c₂: from (1,0,0) to (1,2,0): x=1, z=0, dx=0, dz=0 → ∫c₂ = 0
• c₃: from (1,2,0) to (1,2,3): x=1, y=2, dx=0, dy=0 → ∫c₃ = ∫₀³ 3z² dz = [z³]₀³ = 27
• Total = 27

Path (b): Along straight line c₄ from (0,0,0) to (1,2,3)

• Parametric: x=t, y=2t, z=3t, with t from 0 to 1
• dx=dt, dy=2dt, dz=3dt
• F = (t²·2t) i + (2·2t·3t) j + (3·9t²·t) k = 2t³ i + 12t² j + 27t³ k
• dr = dt i + 2dt j + 3dt k
• F·dr = 2t³·1 + 12t²·2 + 27t³·3 = 2t³ + 24t² + 81t³ = 83t³ + 24t²
• ∫c₄ F·dr = ∫₀¹ (83t³ + 24t²) dt = [83t⁴/4 + 8t³]₀¹ = 83/4 + 8 = 28.75

The value depends on the path: path (a) = 27, path (b) = 28.75.

📌 Example: Evaluate ∫v F dV where v is the region bounded by planes x=0, y=0, z=0 and 2x+y+z=2, with F = 2z i + y k.

• The plane cuts axes at A(1,0,0), B(0,2,0), C(0,0,2)
• z ranges from 0 to 2-2x-y
• ∫v F dV = ∫₀¹ ∫₀²⁻²ˣ ∫₀²⁻²ˣ⁻ʸ (2z i + y k) dz dy dx
• = ∫₀¹ ∫₀²⁻²ˣ [z² i + yz k]₀²⁻²ˣ⁻ʸ dy dx
• = ∫₀¹ ∫₀²⁻²ˣ {[4(1-x)² - 4(1-x)y + y²] i + [2(1-x)y - y²] k} dy dx
• Result: ∫v F dV = (1/3)(2i + k)

⭐ Key Takeaways

The line integral of a scalar field ∫c V dr produces a vector result, while the line integral of a vector field ∫c F·dr produces a scalar result. For evaluation, both the field and the differential must be expressed in terms of a parameter along the curve. A critical exam concept is that line integrals of vector fields can be path-dependent, meaning different paths between the same endpoints yield different values. The parametric method is essential: express x, y, z as functions of a parameter, compute dx, dy, dz, substitute into the integrand, and integrate over the parameter range.

🧠 Quick Revision Questions

1. What is the key difference between ∫c V(r) dr for a scalar field and ∫c F·dr for a vector field in terms of the type of result produced?
2. How do you express dr in parametric form given x = x(t), y = y(t), z = z(t)?
3. In the example with F = x²y i + 2yz j + 3z²x k, why do the two different paths between the same endpoints give different results (27 vs 28.75)?
4. For the scalar field line integral ∫c V dr, what does the result physically represent?
5. When setting up limits for a parametric line integral, how do you determine the corresponding parameter values for the endpoints?

📘 Lecture 37 — Examples

📖 Overview: This lecture demonstrates the application of surface and volume integrals through several worked examples. It covers evaluation of vector field integrals over volumes using cylindrical coordinates, scalar field surface integrals, and vector field surface integrals using both cylindrical and spherical polar coordinates, concluding with an introduction to conservative vector fields.

🗂️ Topics Covered

The lecture presents four main worked examples: evaluating a vector field volume integral over a cylindrical region; evaluating a scalar field surface integral over a curved cylindrical surface; evaluating a vector field surface integral over a spherical surface in the first octant; evaluating a vector field surface integral over a cylindrical surface in the first two octants. The lecture concludes with a theoretical discussion of conservative vector fields and path independence of line integrals.

📝 Lecture Summary

Example — Volume Integral of F over Cylindrical Region

Evaluate (\int_V \mathbf{F} , dV) where (\mathbf{F} = 2\mathbf{i} + 2z\mathbf{j} + y\mathbf{k}) and V is the region bounded by the planes (z = 0), (z = 4) and the surface (x^2 + y^2 = 9).

It is convenient to use cylindrical polar coordinates ((r, \theta, z)) with transformations:

• (x = r\cos\theta)
• (y = r\sin\theta)
• (z = z)
• (dV = r , dr , d\theta , dz)

The integral becomes: [\int_V \mathbf{F} , dV = \int_{\theta=0}^{2\pi} \int_{r=0}^{3} \int_{z=0}^{4} (2\mathbf{i} + 2z\mathbf{j} + r\sin\theta , \mathbf{k}) , r , dz , dr , d\theta]

After integration with respect to (z): [= \int_{\theta=0}^{2\pi} \int_{r=0}^{3} [2z\mathbf{i} + z^2\mathbf{j} + r\sin\theta , z\mathbf{k}]{z=0}^{4} , r , dr , d\theta] [= \int{\theta=0}^{2\pi} \int_{r=0}^{3} (8\mathbf{i} + 16\mathbf{j} + 4r\sin\theta , \mathbf{k}) , r , dr , d\theta]

Factoring: [= 4 \int_{\theta=0}^{2\pi} \int_{r=0}^{3} (2r\mathbf{i} + 4r\mathbf{j} + r^2\sin\theta , \mathbf{k}) , dr , d\theta]

Integrating with respect to (r): [= 4 \int_{\theta=0}^{2\pi} \left[ r^2\mathbf{i} + 2r^2\mathbf{j} + \frac{r^3}{3}\sin\theta , \mathbf{k} \right]{r=0}^{3} d\theta] [= 4 \int{\theta=0}^{2\pi} (9\mathbf{i} + 18\mathbf{j} + 9\sin\theta , \mathbf{k}) , d\theta] [= 36 \int_{\theta=0}^{2\pi} (\mathbf{i} + 2\mathbf{j} + \sin\theta , \mathbf{k}) , d\theta] [= 36 \left[ \theta\mathbf{i} + 2\theta\mathbf{j} - \cos\theta , \mathbf{k} \right]_{0}^{2\pi}] [= 36 \left{ (2\pi\mathbf{i} + 4\pi\mathbf{j} - \mathbf{k}) - (-\mathbf{k}) \right}] [= 72\pi (\mathbf{i} + 2\mathbf{j})]

Scalar Field Surface Integral

A scalar field (F = xyz) exists over the curved surface S defined by (x^2 + y^2 = 4) between the planes (z = 0) and (z = 3) in the first octant. Evaluate (\int_S \mathbf{F} , d\mathbf{S}) over this surface.

We have (F = xyz) and S: (x^2 + y^2 - 4 = 0), (z = 0) to (z = 3).

The unit normal vector is (\hat{\mathbf{n}} = \frac{\nabla S}{|\nabla S|}).

∇S = (\frac{\partial S}{\partial x}\mathbf{i} + \frac{\partial S}{\partial y}\mathbf{j} + \frac{\partial S}{\partial z}\mathbf{k} = 2x\mathbf{i} + 2y\mathbf{j})

(|\nabla S| = \sqrt{4x^2 + 4y^2} = 2\sqrt{x^2 + y^2} = 2\sqrt{4} = 4)

Therefore (\hat{\mathbf{n}} = \frac{2x\mathbf{i} + 2y\mathbf{j}}{4} = \frac{x\mathbf{i} + y\mathbf{j}}{2})

The vector surface element is (d\mathbf{S} = \hat{\mathbf{n}} , dS = \frac{x\mathbf{i} + y\mathbf{j}}{2} dS)

Thus: [\int_S \mathbf{F} , d\mathbf{S} = \int_S F \hat{\mathbf{n}} , dS = \frac{1}{2} \int_S xyz(x\mathbf{i} + y\mathbf{j}) , dS = \frac{1}{2} \int_S (x^2yz , \mathbf{i} + xy^2z , \mathbf{j}) , dS]

Converting to cylindrical coordinates with (r = 2):

• (x = 2\cos\theta)
• (y = 2\sin\theta)
• (z = z)
• (dS = 2 , d\theta , dz) (since on cylinder (r=2), the surface element is (r , d\theta , dz))

Then:

• (x^2yz = (4\cos^2\theta)(2\sin\theta)(z) = 8\cos^2\theta \sin\theta , z)
• (xy^2z = (2\cos\theta)(4\sin^2\theta)(z) = 8\cos\theta \sin^2\theta , z)

The integral becomes: [\int_S \mathbf{F} , d\mathbf{S} = \frac{1}{2} \int_{\theta=0}^{\pi/2} \int_{z=0}^{3} (8\cos^2\theta \sin\theta , \mathbf{i} + 8\cos\theta \sin^2\theta , \mathbf{j}) z , (2 , d\theta , dz)] [= 8 \int_{\theta=0}^{\pi/2} \int_{z=0}^{3} (\cos^2\theta \sin\theta , \mathbf{i} + \cos\theta \sin^2\theta , \mathbf{j}) z , dz , d\theta]

First integrate with respect to (z): [= 8 \int_{\theta=0}^{\pi/2} (\cos^2\theta \sin\theta , \mathbf{i} + \cos\theta \sin^2\theta , \mathbf{j}) \left[ \frac{z^2}{2} \right]{0}^{3} d\theta] [= 8 \int{\theta=0}^{\pi/2} (\cos^2\theta \sin\theta , \mathbf{i} + \cos\theta \sin^2\theta , \mathbf{j}) \cdot \frac{9}{2} , d\theta] [= 36 \int_{\theta=0}^{\pi/2} (\cos^2\theta \sin\theta , \mathbf{i} + \cos\theta \sin^2\theta , \mathbf{j}) , d\theta]

Integrating with respect to (\theta): [= 36 \left[ -\frac{\cos^3\theta}{3} \mathbf{i} + \frac{\sin^3\theta}{3} \mathbf{j} \right]_{0}^{\pi/2}] [= 36 \left[ \left( -\frac{0}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} \right) - \left( -\frac{1}{3} \mathbf{i} + \frac{0}{3} \mathbf{j} \right) \right]] [= 36 \left( \frac{1}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} \right) = 12(\mathbf{i} + \mathbf{j})]

💡 Why this matters: This example shows how to handle the direction of the surface element correctly by using the unit normal vector obtained from the gradient of the surface equation.

Vector Field Surface Integral Over Sphere

A vector field (\mathbf{F} = y\mathbf{i} + 2\mathbf{j} + \mathbf{k}) exists over a surface S defined by (x^2 + y^2 + z^2 = 9) bounded by (x = 0), (y = 0), (z = 0) in the first octant. Evaluate (\int_S \mathbf{F} \cdot d\mathbf{S}).

(d\mathbf{S} = \hat{\mathbf{n}} , dS); (\hat{\mathbf{n}} = \frac{\nabla S}{|\nabla S|})

S: (x^2 + y^2 + z^2 - 9 = 0)

(\nabla S = \frac{\partial S}{\partial x}\mathbf{i} + \frac{\partial S}{\partial y}\mathbf{j} + \frac{\partial S}{\partial z}\mathbf{k} = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k})

(|\nabla S| = \sqrt{4x^2 + 4y^2 + 4z^2} = 2\sqrt{x^2 + y^2 + z^2} = 2\sqrt{9} = 6)

Therefore (\hat{\mathbf{n}} = \frac{2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k}}{6} = \frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{3})

[\int_S \mathbf{F} \cdot d\mathbf{S} = \int_S \mathbf{F} \cdot \hat{\mathbf{n}} , dS = \int_S (y\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot \frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{3} , dS] [= \frac{1}{3} \int_S (xy + 2y + z) , dS]

Convert to spherical polar coordinates:

• (x = 3\sin\phi \cos\theta)
• (y = 3\sin\phi \sin\theta)
• (z = 3\cos\phi)
• (dS = 9\sin\phi , d\phi , d\theta)

Limits: (\phi = 0) to (\frac{\pi}{2}); (\theta = 0) to (\frac{\pi}{2})

Components:

• (xy = (3\sin\phi \cos\theta)(3\sin\phi \sin\theta) = 9\sin^2\phi \cos\theta \sin\theta)
• (2y = 2 \cdot 3\sin\phi \sin\theta = 6\sin\phi \sin\theta)
• (z = 3\cos\phi)

Substituting: [\int_S \mathbf{F} \cdot d\mathbf{S} = \frac{1}{3} \int_{\phi=0}^{\pi/2} \int_{\theta=0}^{\pi/2} (9\sin^2\phi \cos\theta \sin\theta + 6\sin\phi \sin\theta + 3\cos\phi) \cdot 9\sin\phi , d\phi , d\theta]

Factor: [= 9 \int_{\phi=0}^{\pi/2} \int_{\theta=0}^{\pi/2} (3\sin^3\phi \cos\theta \sin\theta + 2\sin^2\phi \sin\theta + \sin\phi \cos\phi) , d\phi , d\theta]

Using Wallis formula:

• (\int_0^{\pi/2} \sin^3\phi , d\phi = \frac{2}{3})
• (\int_0^{\pi/2} \sin^2\phi , d\phi = \frac{\pi}{4})

The integration yields: [= 9 \int_{\theta=0}^{\pi/2} \left( \frac{2}{3}\cos\theta \sin\theta + \frac{\pi}{4}\sin\theta + \frac{1}{2} \right) d\theta] [= 9 \left[ \sin^2\theta - \frac{\pi}{4}\cos\theta + \frac{1}{2}\theta \right]_{0}^{\pi/2}] [= 9 \left[ \left(1 - 0 + \frac{\pi}{4}\right) - \left(0 - \frac{\pi}{4} + 0\right) \right]] [= 9 \left( 1 + \frac{\pi}{4} + \frac{\pi}{4} \right) = 9 \left( 1 + \frac{\pi}{2} \right)]

Vector Field Surface Integral Over Cylinder

Evaluate (\int_S \mathbf{F} \cdot d\mathbf{S}) where (\mathbf{F} = 2y\mathbf{j} + z\mathbf{k}) and S is the surface (x^2 + y^2 = 4) in the first two octants bounded by the planes (z = 0), (z = 5) and (y = 0).

S: (x^2 + y^2 - 4 = 0)

(\hat{\mathbf{n}} = \frac{\nabla S}{|\nabla S|})

(\nabla S = 2x\mathbf{i} + 2y\mathbf{j})

(|\nabla S| = \sqrt{4x^2 + 4y^2} = 2\sqrt{x^2 + y^2} = 2\sqrt{4} = 4)

(\hat{\mathbf{n}} = \frac{2x\mathbf{i} + 2y\mathbf{j}}{4} = \frac{x\mathbf{i} + y\mathbf{j}}{2})

[\int_S \mathbf{F} \cdot d\mathbf{S} = \int_S \mathbf{F} \cdot \hat{\mathbf{n}} , dS = \int_S (2y\mathbf{j} + z\mathbf{k}) \cdot \frac{x\mathbf{i} + y\mathbf{j}}{2} , dS] [= \frac{1}{2} \int_S (2y^2) , dS = \int_S y^2 , dS]

Use cylindrical polar coordinates:

• (x = 2\cos\theta)
• (y = 2\sin\theta)
• (z = z)
• (dS = 2 , d\theta , dz)

Limits: (\theta = 0) to (\pi) (first two octants); (z = 0) to (5)

[\int_S \mathbf{F} \cdot d\mathbf{S} = \int_S (4\sin^2\theta)(2 , d\theta , dz) = 8 \int_{z=0}^{5} \int_{\theta=0}^{\pi} \sin^2\theta , d\theta , dz]

Using (\sin^2\theta = \frac{1-\cos 2\theta}{2}): [= 4 \int_{z=0}^{5} \int_{\theta=0}^{\pi} (1 - \cos 2\theta) , d\theta , dz] [= 4 \int_{z=0}^{5} \left[ \theta - \frac{\sin 2\theta}{2} \right]{0}^{\pi} dz] [= 4 \int{z=0}^{5} \pi , dz] [= 4\pi [z]_{0}^{5} = 20\pi]

Conservative Vector Fields

In general, the value of the integral (\int_C \mathbf{F} \cdot d\mathbf{r}) between two stated points A and B depends on the particular path of integration followed. If, however, the line integral between A and B is independent of the path of integration between the two end points, then the vector field (\mathbf{F}) is said to be conservative.

🔑 Definition — Conservative Vector Field: A vector field (\mathbf{F}) is conservative if the line integral (\int_C \mathbf{F} \cdot d\mathbf{r}) between any two points A and B is independent of the path taken.

It follows that, for a closed path in a conservative field: [\oint_C \mathbf{F} \cdot d\mathbf{r} = 0]

Proof: If the field is conservative: [\int_{C_1(AB)} \mathbf{F} \cdot d\mathbf{r} = \int_{C_2(AB)} \mathbf{F} \cdot d\mathbf{r}]

But: [\int_{C_2(BA)} \mathbf{F} \cdot d\mathbf{r} = -\int_{C_2(AB)} \mathbf{F} \cdot d\mathbf{r}]

For the closed path (AB_{C_1} + BA_{C_2}): [\oint \mathbf{F} \cdot d\mathbf{r} = \int_{C_1(AB)} \mathbf{F} \cdot d\mathbf{r} + \int_{C_2(AB)} \mathbf{F} \cdot d\mathbf{r}] [= \int_{C_1(AB)} \mathbf{F} \cdot d\mathbf{r} - \int_{C_1(AB)} \mathbf{F} \cdot d\mathbf{r} = 0]

💡 Why this matters: Conservative vector fields are fundamental in physics — gravitational and electrostatic fields are conservative, making their line integrals path-independent and equal to the difference in potential function values.

⭐ Key Takeaways

The lecture demonstrates that surface and volume integrals can be evaluated systematically by converting to appropriate coordinate systems — cylindrical coordinates for cylindrical and circular surfaces, and spherical coordinates for spherical surfaces. The key steps involve: computing the unit normal vector from the gradient of the surface equation; transforming the integrand and surface element to the chosen coordinates; carefully setting up limits of integration corresponding to the region; and integrating with respect to each variable in the proper order. For conservative vector fields, line integrals between two points are path-independent, and closed path integrals vanish — this property distinguishes them from non-conservative fields where path matters.

🧠 Quick Revision Questions

1. How do you compute the unit normal vector (\hat{\mathbf{n}}) for a surface defined implicitly by (S(x,y,z) = 0)?

2. What are the cylindrical coordinate transformations and corresponding volume and surface elements for a cylinder of radius (r)?

3. In the scalar field example over the cylindrical surface, why was the surface element (dS = 2,d\theta,dz) on the surface (x^2 + y^2 = 4)?

4. Show mathematically that for a conservative vector field, the line integral around any closed path equals zero.

5. In the spherical surface example, what are the spherical coordinate transformations and why were the limits (\phi = 0) to (\pi/2) and (\theta = 0) to (\pi/2)?

📘 Lecture 38 — Vector Field

📖 Overview: This lecture introduces the concept of conservative vector fields and the conditions that determine whether a field is conservative. It then covers two of the most important integral theorems of vector calculus: the Divergence Theorem (Gauss’ theorem) and Stokes’ Theorem, demonstrating their application through detailed worked examples.

🗂️ Topics Covered

The lecture begins by defining conservative vector fields and showing that for such fields, the line integral between two points is path-independent, leading to a zero circulation around any closed loop. Three equivalent tests for a conservative field are presented (zero circulation, zero curl, and gradient form). The second half details the Divergence Theorem, relating volume integrals of divergence to surface integrals of flux, and Stokes’ Theorem, relating surface integrals of curl to line integrals around a boundary. Both theorems are verified with step-by-step examples over a rectangular volume and a hemisphere, respectively.

📝 Lecture Summary

Conservative Vector Fields

In general, the value of the line integral of a vector field F between two points depends on the path taken. If the integral between points A and B is independent of the path, the field is called a conservative vector field. For a closed path in a conservative field, the circulation is zero: ∮_C F·dr = 0. This holds only for a closed curve in a conservative field, as the path independence forces the integral over an out-and-back path to cancel.

🔑 Definition — Conservative Vector Field: A vector field where the line integral between any two points is independent of the path taken between them. Equivalently, the integral around any closed loop is zero.

📌 Example: Determine if F = 2xyz i + x²z j + x²y k is conservative by evaluating the line integral from A(0,0,0) to B(2,4,6) along two different paths.

• Path 1 (parametric: x=u, y=u², z=3u): transform F·dr to find the integral = 96.
• Path 2 (three straight segments: c₁: (0,0,0)→(2,0,0); c₂: (2,0,0)→(2,4,0); c₃: (2,4,0)→(2,4,6)): evaluate each segment. c₁ gives 0, c₂ gives 0, c₃ gives 96. Total = 96. Since both paths give 96, the field is conservative.

Tests for a Conservative Field

Any one of three equivalent conditions can be used to test if a vector field is conservative: (a) ∮_C F·dr = 0 for all closed curves C. (b) curl F = 0. (c) F = ∇V (i.e., F is the gradient of some scalar potential V). For the field F = 2xyz i + x²z j + x²y k: curl F = (x²−x²)i − (2xy−2xy)j + (2xz−2xz)k = 0. This confirms the field is conservative.

🔑 Definition — curl F: A vector operator that measures the rotation or “vorticity” of a vector field. If curl F = 0, the field is irrotational and conservative. 📐 Formula: curl F = ∇ × F 📌 Example (verifying curl = 0): For F = 2xyz i + x²z j + x²y k, the components give curl = (x²−x²)i − (2xy−2xy)j + (2xz−2xz)k = 0.

Divergence Theorem (Gauss’ theorem)

For a closed surface S enclosing a volume V in a vector field F: ∫_V (div F) dV = ∮_S F·dS This states the volume integral of the divergence of F equals the total outward flux of F through the surface S.

🔑 Definition — Divergence Theorem: Relates a volume integral of divergence to a surface integral of flux, equating the net “source” inside a volume to the net flow across its boundary. 📐 Formula: ∫_V (∇·F) dV = ∮_S F·n̂ dS

📌 Example: Verify the Divergence Theorem for F = x²i + zj + yk over the region bounded by z=0, z=2, x=0, x=1, y=0, y=3.

• (a) Volume integral: div F = 2x. ∫_V 2x dV = ∫₀¹ ∫₀³ ∫₀² 2x dz dy dx = 6.
• (b) Surface integral: Evaluate F·n̂ dS over all 6 faces.
  • S₁ (z=0, n̂ = −k): flux = −9/2.
  • S₂ (z=2, n̂ = k): flux = 9/2.
  • S₃ (y=3, n̂ = j): flux = 2.
  • S₄ (y=0, n̂ = −j): flux = −2.
  • S₅ (x=1, n̂ = i): flux = 6.
  • S₆ (x=0, n̂ = −i): flux = 0.
  • Total flux = (−9/2) + (9/2) + 2 − 2 + 6 + 0 = 6. Since both integrals equal 6, the theorem is verified.

Stokes Theorem

If F is a vector field over an open surface S and its boundary curve C: ∫_S (curl F)·dS = ∮_C F·dr This states the surface integral of the curl of F equals the circulation of F around the boundary.

🔑 Definition — Stokes’ Theorem: Relates a surface integral of the curl of a vector field to a line integral around the boundary curve of the surface. 📐 Formula: ∫_S (∇ × F)·n̂ dS = ∮_C F·dr

📌 Example: Verify Stokes’ Theorem for F = 2yi − xj + xzk over the hemisphere S: x²+y²+z²=4 (z≥0), with boundary C: x²+y²=4, z=0.

• (a) Line integral around C: parametrize C as x=2cosθ, y=2sinθ, z=0. ∮_C F·dr = ∫₀²π (2y dx − x dy) = −12π.
• (b) Surface integral: First, curl F = −zj − 3k. The unit normal n̂ = (xi+yj+zk)/2. curl F·n̂ = (−yz − 3z)/2. Parametrizing the hemisphere (x=2sinφcosθ, y=2sinφsinθ, z=2cosφ, dS=4sinφ dφ dθ), the integral becomes ½∫₀²π∫₀^{π/2} (−2sinφ sinθ·2cosφ − 6cosφ)·4 sinφ dφ dθ = −12π. Since both calculations yield −12π, Stokes’ Theorem is verified.

⭐ Key Takeaways

The most critical concept from this lecture is that a conservative vector field is defined by path-independent line integrals, which is equivalent to it having zero curl and being expressible as the gradient of a scalar potential. You must be able to apply the three tests for conservatism: checking if the closed line integral is zero, computing the curl, or recognizing a gradient field. The Divergence Theorem and Stokes’ Theorem are powerful tools that convert volume integrals into surface integrals and surface integrals into line integrals, respectively, which are essential for simplifying complex vector calculus problems on exams. Finally, you need to master setting up the correct parametrizations and surface normals when applying these theorems to specific geometries like rectangular boxes or hemispheres.

🧠 Quick Revision Questions

1. What are the three equivalent conditions that define a conservative vector field?
2. For a conservative field, what is the value of the line integral around any closed loop?
3. State the mathematical formulas for the Divergence Theorem and Stokes’ Theorem.
4. Given F = (2xy + z³)i + x²j + 3xz²k, use the curl test to determine if it is conservative.
5. Describe the steps involved in verifying Stokes’ Theorem for a given vector field and a specific surface.

📘 Lecture 39 — Periodic Functions

📖 Overview: This lecture introduces periodic functions and their mathematical representation, focusing on sine and cosine functions with varying amplitudes and periods. It builds foundational knowledge for understanding Fourier series by covering analytical descriptions of periodic functions, essential integrals, and the trigonometric series representation of periodic functions with period 2π.

🗂️ Topics Covered

The lecture covers the definition and characteristics of periodic functions including amplitude and period, graphs of y = A sin nx and y = A cos nx, analytical description of periodic functions with examples and sketching exercises, useful integrals that appear frequently in Fourier series work, and the Fourier series representation of periodic functions of period 2π including the expansion of sine terms into cosine and sine components.

📝 Lecture Summary

Periodic functions

A periodic function f(x) is defined as a function whose values repeat at regular intervals of the independent variable. The regular interval between repetitions is called the period of the oscillations. Mathematically, a periodic function satisfies f(x + p) = f(x) where p is the period.

Graphs of y = A sin nx

The most obvious example of a periodic function is y = sin x, which goes through its complete range of values as x increases from 0° to 360°. The period is therefore 360° or 2π radians, and the amplitude is the maximum displacement from the position of rest.

For y = 5 sin 2x, the amplitude is 5, the period is 180°, and there are 2 complete cycles in 360°.

🔑 Definition — Amplitude: The maximum displacement from the position of rest for a periodic function

📌 Example — Functions with their amplitudes and periods:

• y = 3 sin 5x: amplitude = 3, period = 72°
• y = 2 cos 3x: amplitude = 2, period = 120°
• y = sin(x/2): amplitude = 1, period = 720°
• y = 4 sin 2x: amplitude = 4, period = 180°

📐 Formula: For y = A sin nx, amplitude = A, period = 360°/n = 2π/n, with n cycles in 360°

Graphs of y = A cos nx have the same characteristics as y = A sin nx.

Analytical description of a periodic function

A periodic function can be defined analytically in many cases using piecewise definitions.

📌 Example — Between x = 0 and x = 4, y = 3, i.e., f(x) = 3 for 0 < x < 4 Between x = 4 and x = 6, y = 0, i.e., f(x) = 0 for 4 < x < 6 The function is defined as: f(x) = 3 for 0 < x < 4 f(x) = 0 for 4 < x < 6 f(x) = f(x + 6) (indicating period of 6 units)

📌 Additional examples of piecewise periodic functions:

• f(x) = 2 − x for 0 < x < 3, f(x) = −1 for 3 < x < 5, f(x) = f(x+5)
• f(x) = 3x for 0 < x < 4, f(x) = 7 − x for 4 < x < 10, f(x) = −3 for 10 < x < 13, f(x) = f(x+13)

📌 Sketching exercise examples:

1. f(x) = 4 for 0 < x < 5, f(x) = 0 for 5 < x < 8, f(x) = f(x+8)
2. f(x) = 3x − x² for 0 < x < 3, f(x) = f(x+3)
3. f(x) = 2 sin x for 0 < x < π, f(x) = 0 for π < x < 2π, f(x) = f(x+2π)
4. f(x) = x²/4 for 0 < x < 4, f(x) = 4 for 4 < x < 6, f(x) = 0 for 6 < x < 8, f(x) = f(x+8)

Useful integrals

The following integrals appear frequently in work on Fourier series. In each case, m and n are integers other than zero, and integration is from −π to π.

📐 Formulas: (a) ∫ sin nx dx = 0 (b) ∫ cos nx dx = 0 (c) ∫ sin² nx dx = (1/2)∫(1−cos2nx)dx = π (n ≠ 0) (d) ∫ cos² nx dx = (1/2)∫(1+cos2nx)dx = π (n ≠ 0) (e) ∫ sin nx cos mx dx = 0 (f) ∫ cos nx cos mx dx = 0 (n ≠ m); = π (n = m) (g) ∫ sin nx sin mx dx = 0 (n ≠ m); = π (n = m)

📌 Note: These integrals have been evaluated between −π and π, but provided integration is carried out over a complete periodic interval of 2π, the results are the same. Limits could be −π to π, 0 to 2π, −π/2 to 3π/2, etc.

📌 Derivation notes:

• For (c): ∫ sin² nx dx = ½ ∫(1−cos2nx)dx = ½[x − (sin2nx)/2n] = π
• For (d): ∫ cos² nx dx = ½ ∫(1+cos2nx)dx = ½[x + (sin2nx)/2n] = π
• For (e): ∫ sin nx cos mx dx = ½ ∫{sin(n+m)x+sin(n−m)x}dx = 0

Fourier series

Periodic functions of period 2π

The basis of a Fourier series is to represent a periodic function by a trigonometric series of the form:

f(x) = A₀ + c₁ sin(x+α₁) + c₂ sin(2x+α₂) + c₃ sin(3x+α₃) + ... + cₙ sin(nx+αₙ) + ...

where:

• A₀ is a constant term
• c₁, c₂, c₃, ..., cₙ denote the amplitudes of the compound sine terms
• α₁, α₂, α₃, ... are constant auxiliary angles

Each sine term cₙ sin(nx + αₙ) can be expanded: cₙ sin(nx + αₙ) = cₙ{sin nx cos αₙ + cos nx sin αₙ} = (cₙ sin αₙ) cos nx + (cₙ cos αₙ) sin nx = aₙ cos nx + bₙ sin nx

📐 Formula: Fourier series representation f(x) = A₀ + Σ∞ₙ₌₁ {aₙ cos nx + bₙ sin nx}

💡 Why this matters: This Fourier series representation is the foundation for decomposing complex periodic waveforms into simple sine and cosine components, which is essential in signal processing, acoustics, electrical engineering, and solving partial differential equations.

⭐ Key Takeaways

A periodic function satisfies f(x+p)=f(x) with amplitude A and period 360°/n for y=A sin nx or y=A cos nx. This lecture establishes the crucial integral identities: ∫ sin²nx dx = π, ∫ cos²nx dx = π, while integrals of products of different sine/cosine functions over 2π intervals equal zero. The Fourier series expresses any periodic function as f(x)=A₀+Σ(aₙ cos nx + bₙ sin nx), where each sine term is expanded into cosine and sine components. The orthogonality of trigonometric functions (integrals of products being zero unless frequencies match) is the key property enabling Fourier coefficient calculation. These concepts together form the mathematical foundation for representing and analyzing periodic phenomena.

🧠 Quick Revision Questions

1. What is the amplitude and period of the function y = 3 sin 5x?
2. Define a periodic function mathematically and give its defining equation.
3. What is the value of ∫ sin² nx dx evaluated from −π to π?
4. Under what condition does ∫ cos nx cos mx dx equal π instead of 0?
5. Write the general form of a Fourier series for a periodic function of period 2π.

📘 Lecture 40 — Fourier Series

📖 Overview: This lecture provides a comprehensive derivation and application of Fourier series coefficients (A₀, aₙ, bₙ) and presents the final formulas for representing periodic functions. It also covers the Dirichlet conditions for convergence and demonstrates the effect of harmonics in reconstructing square waves, showing how adding more terms improves accuracy.

🗂️ Topics Covered

The lecture begins by reviewing the general form of a Fourier series, then derives the Fourier coefficients A₀, aₙ, and bₙ through integration. It presents the final result formulas using a₀, aₙ, and bₙ, followed by a worked example using limits from 0 to 2π. The Dirichlet conditions for convergence are then explained, followed by a second example involving a piecewise function with discontinuities. Finally, the lecture visualizes the effect of harmonics by plotting successive approximations of a square wave.

📝 Lecture Summary

Fourier series

As we know that A₀ + Σ {aₙ cos nx + bₙ sin nx}; n=1 to ∞, which can be written in the expanded form: A₀ + (a₁ cos x + b₁ sin x) + (a₂ cos 2x + b₂ sin 2x) + ... + (aₙ cos nx + bₙ sin nx) + ...

f(x) = A₀ + a₁ cos x + a₂ cos 2x + ... + aₙ cos nx + ... + b₁ sin x + b₂ sin 2x + ... + bₙ sin nx + ...

Fourier coefficients

We have defined Fourier series in the form f(x) = A₀ + Σ {aₙ cos nx + bₙ sin nx}; n a positive integer.

(a) To find A₀, we integrate f(x) with respect to x from -π to π. ∫₋π^π f(x) dx = ∫₋π^π A₀ dx + Σₙ₌₁^∞ {∫₋π^π aₙ cos nx dx + ∫₋π^π bₙ sin nx dx} = [A₀x]₋π^π + Σₙ₌₁^∞ {0 + 0} = 2A₀π

Therefore, 2A₀π = ∫₋π^π f(x) dx, so A₀ = (1/2π) ∫₋π^π f(x) dx = 1/2 a₀, where a₀ = (1/π) ∫₋π^π f(x) dx.

🔑 Definition — A₀ (average term): The constant term in the Fourier series, equal to half the mean value of the function over one period. It represents the average value of the function.

📌 Example: For the function f(x) = x/2 over 0 < x < 2π, a₀ = (1/π)∫₀²π (x/2) dx = π.

(b) To find aₙ, we multiply f(x) by cos mx and integrate from -π to π. ∫₋π^π f(x) cos mx dx = ∫₋π^π A₀ cos mx dx + Σₙ₌₁^∞ {∫₋π^π aₙ cos nx cos mx dx + ∫₋π^π bₙ sin nx cos mx dx} = 0 for n ≠ m, and = aₙπ for n = m.

Therefore, aₙ = (1/π) ∫₋π^π f(x) cos nx dx.

(c) To find bₙ, we multiply f(x) by sin mx and integrate from -π to π. ∫₋π^π f(x) sin mx dx = 0 for n ≠ m, and = bₙπ for n = m.

Therefore, bₙ = (1/π) ∫₋π^π f(x) sin nx dx.

Result For Fourier Series

f(x) = (1/2) a₀ + Σₙ₌₁^∞ {aₙ cos nx + bₙ sin nx}

(a) a₀ = (1/π) ∫₋π^π f(x) dx = 2 × mean value of f(x) over a period

(b) aₙ = (1/π) ∫₋π^π f(x) cos nx dx = 2 × mean value of f(x) cos nx over a period

(c) bₙ = (1/π) ∫₋π^π f(x) sin nx dx = 2 × mean value of f(x) sin nx over a period

In each case, n = 1, 2, 3, ...

📐 Formula: Standard Fourier series: f(x) = a₀/2 + Σₙ₌₁^∞ (aₙ cos nx + bₙ sin nx) → The sum of a constant term (average value) plus an infinite series of cosine and sine harmonics.

Example

Determine the Fourier series to represent the periodic function f(x) = x/2, 0 < x < 2π, f(x) = f(x + 2π). It is more convenient here to take the limits as 0 to 2π.

(a) a₀: a₀ = (1/π)∫₀²π f(x) dx = (1/π)∫₀²π (x/2) dx = (1/4π)[x²]₀²π = π

(b) aₙ: aₙ = (1/π)∫₀²π f(x) cos nx dx = (1/2π)∫₀²π x cos nx dx Using integration by parts, aₙ = 0 → aₙ = 0 (all cosine harmonics vanish because the function is odd about the midpoint of the interval)

(c) bₙ: bₙ = (1/π)∫₀²π f(x) sin nx dx = (1/2π)∫₀²π x sin nx dx Using integration by parts, bₙ = -(1/n) → bₙ = -1/n

📌 Example: Since a₀ = π, aₙ = 0, and bₙ = -1/n, the Fourier series becomes: f(x) = a₀/2 + Σₙ₌₁^∞ {bₙ sin nx} = π/2 - {sin x + (1/2) sin 2x + (1/3) sin 3x + ...}

Therefore, f(x) = π/2 - (sin x + (1/2) sin 2x + (1/3) sin 3x + ...)

Dirichlet Conditions

If the Fourier series is to represent a function f(x), then putting x = x₁ will give an infinite series in x₁, and the value of this should converge to the value of f(x₁) as more and more terms of the series are evaluated. For this to happen, the following conditions must be fulfilled:

(a) The function f(x) must be defined and single-valued.

(b) f(x) must be continuous or have a finite number of finite discontinuities within a periodic interval.

(c) f(x) and f'(x) must be piecewise continuous in the periodic interval.

If these Dirichlet conditions are satisfied, the Fourier series converges to f(x₁), if x = x₁ is a point of continuity.

💡 Why this matters: These conditions ensure the Fourier series actually converges to the function. Without them, the series might diverge or converge to the wrong value at discontinuities.

Example

Find the Fourier series for the function defined by: f(x) = 0 for -π < x < -π/2; f(x) = 4 for -π/2 < x < π/2; f(x) = 0 for π/2 < x < π; f(x) = f(x + 2π).

(a) a₀: a₀ = (1/π)∫₋π^π f(x) dx = (1/π)[∫₋π^{-π/2} 0 dx + ∫₋π/2^{π/2} 4 dx + ∫₋π/2^π 0 dx] = (1/π)[4x]₋π/2^{π/2} = 4 → a₀ = 4

(b) aₙ: aₙ = (1/π)∫₋π^π f(x) cos nx dx = (4/π)∫₋π/2^{π/2} cos nx dx = (8/nπ) sin(nπ/2)

If n is even, aₙ = 0. If n = 1, 5, 9,... aₙ = 8/(nπ). If n = 3, 7, 11,... aₙ = -8/(nπ).

(c) bₙ: bₙ = (4/π)∫₋π/2^{π/2} sin nx dx = -(4/nπ)[cos(nπ/2) - cos(-nπ/2)] = 0 → bₙ = 0 (all sine terms vanish due to symmetry)

📌 Example: The Fourier series is: f(x) = a₀/2 + Σₙ₌₁^∞ {aₙ cos nx + bₙ sin nx} f(x) = 2 + (8/π){cos x - (1/3) cos 3x + (1/5) cos 5x - (1/7) cos 7x + ...}

In this particular example, there are no sine terms (bₙ = 0).

Effect Of Harmonics

It is interesting to see just how accurately the Fourier series represents the function with which it is associated. The complete representation requires an infinite number of terms, but we can see the effect of including the first few terms.

For the square waveform f(x) = 0 for -π < x < -π/2; f(x) = 4 for -π/2 < x < π/2; f(x) = 0 for π/2 < x < π; f(x) = f(x + 2π):

(1) f(x) = 2 + (8/π) cos x

(2) f(x) = 2 + (8/π){cos x - (1/3) cos 3x}

(3) f(x) = 2 + (8/π){cos x - (1/3) cos 3x + (1/5) cos 5x}

(4) f(x) = 2 + (8/π){cos x - (1/3) cos 3x + (1/5) cos 5x - (1/7) cos 7x}

As the number of terms is increased, the graph gradually approaches the shape of the original square waveform. The ripples increase in number and decrease in amplitude, but a perfectly square waveform is unattainable in practice. For practical purposes, the first few terms normally suffice to give an accuracy of acceptable level.

💡 Why this matters: This demonstrates the Gibbs phenomenon — the overshoot at discontinuities persists even with many terms, but the approximation becomes visually acceptable for engineering applications.

⭐ Key Takeaways

The Fourier series formula f(x) = a₀/2 + Σ(aₙ cos nx + bₙ sin nx) requires computing three coefficients: a₀ from the average value, aₙ from multiplying by cosine, and bₙ from multiplying by sine, all integrated over one period. The Dirichlet conditions guarantee convergence at points of continuity, but at discontinuities, the series converges to the midpoint value. The effect of harmonics shows that adding more terms improves the approximation, with ripples diminishing in amplitude, making the series practically useful for signal representation.

🧠 Quick Revision Questions

1. Derive the formula for A₀ by integrating the Fourier series from -π to π. What does A₀ represent physically?

2. Why do the integrals for aₙ and bₙ vanish when n ≠ m, giving only aₙπ or bₙπ for n = m?

3. State the three Dirichlet conditions. What happens if the function fails to satisfy them at a discontinuity?

4. In the square wave example (f(x) = 4 on -π/2 to π/2), why are all sine coefficients bₙ equal to zero?

5. What is the Fourier series approximation using only the first two non-zero cosine harmonics for the square wave with amplitude 4?

📘 Lecture 41 — Examples

📖 Overview: This lecture demonstrates how to compute Fourier series for piecewise-defined functions through detailed examples. It also introduces the important concepts of odd and even functions, their properties, and how these symmetries simplify Fourier series calculations by eliminating either sine or cosine terms.

🗂️ Topics Covered

The lecture covers a complete step-by-step example of finding the Fourier series for a piecewise function defined on $(-\pi, \pi)$. It then introduces definitions and graphical interpretations of odd and even functions, examines products of odd and even functions with examples, presents two key theorems about Fourier series for odd and even functions, and provides additional examples demonstrating these simplifications.

📝 Lecture Summary

Example — Fourier Series for a Piecewise Function

Find the Fourier series for the function defined by $f(x) = -x$ for $-\pi < x < 0$, $f(x) = 0$ for $0 < x < \pi$, and $f(x) = f(x + 2\pi)$.

The general expressions for $a_0$, $a_n$, $b_n$ are:

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) , dx , \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx , dx , \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx$$

To find $a_0$:

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) , dx = \frac{1}{\pi} \int_{-\pi}^{0} (-x) , dx + \frac{1}{\pi} \int_{0}^{\pi} 0 , dx = \frac{1}{\pi} \left[ -\frac{x^2}{2} \right]_{-\pi}^{0} = \frac{\pi}{2}$$

To find $a_n$:

$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx , dx = \frac{1}{\pi} \int_{-\pi}^{0} (-x) \cos nx , dx + \frac{1}{\pi} \int_{0}^{\pi} 0 , dx = -\frac{1}{\pi} \int_{-\pi}^{0} x \cos nx , dx$$

Using integration by parts:

$$a_n = -\frac{1}{\pi} \left[ \left( x \frac{\sin nx}{n} \right){-\pi}^{0} - \frac{1}{n} \int{-\pi}^{0} \sin nx , dx \right]$$

$$= -\frac{1}{\pi} \left[ (0-0) - \frac{1}{n} \left( -\frac{\cos nx}{n} \right){-\pi}^{0} \right] = -\frac{1}{\pi n^2} \left[ \cos nx \right]{-\pi}^{0}$$

$$= -\frac{1}{\pi n^2} [\cos 0 - \cos(-n\pi)] = -\frac{1}{\pi n^2} [1 - \cos n\pi]$$

Since $\cos n\pi = 1$ for $n$ even and $\cos n\pi = -1$ for $n$ odd:

$$a_n = -\frac{2}{\pi n^2} \text{ (n odd)} \quad \text{and} \quad a_n = 0 \text{ (n even)}$$

To find $b_n$:

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx = \frac{1}{\pi} \int_{-\pi}^{0} (-x) \sin nx , dx = -\frac{1}{\pi} \int_{-\pi}^{0} x \sin nx , dx$$

$$= -\frac{1}{\pi} \left[ \left( x \cdot \frac{-\cos nx}{n} \right){-\pi}^{0} + \frac{1}{n} \int{-\pi}^{0} \cos nx , dx \right]$$

$$= -\frac{1}{\pi} \left[ \frac{\pi \cos n\pi}{n} + \frac{1}{n} \left( \frac{\sin nx}{n} \right)_{-\pi}^{0} \right] = -\frac{\cos n\pi}{n}$$

$$b_n = -\frac{1}{n} \text{ (n even)} \quad \text{and} \quad b_n = \frac{1}{n} \text{ (n odd)}$$

Final Fourier series:

$$f(x) = \frac{\pi}{4} - \frac{2}{\pi} \left( \cos x + \frac{1}{9} \cos 3x + \frac{1}{25} \cos 5x + \cdots \right) + \left( \sin x - \frac{1}{2} \sin 2x + \frac{1}{3} \sin 3x - \frac{1}{4} \sin 4x + \cdots \right)$$

This is obtained by substituting $n = 1, 2, 3,$ etc. into the expressions for $a_0$, $a_n$, and $b_n$. This function has a constant term and both sine and cosine terms.

Odd and Even Functions

(a) Even functions

A function $f(x)$ is said to be even if $f(-x) = f(x)$, meaning the function value for a particular negative value of $x$ is the same as that for the corresponding positive value of $x$. The graph of an even function is symmetrical about the y-axis.

• $y = f(x) = x^2$ is an even function since $f(-2) = 4 = f(2)$, $f(-3) = 9 = f(3)$, etc.
• $y = f(x) = \cos x$ is an even function since $\cos(-x) = \cos x$, so $f(-a) = \cos a = f(a)$.

(b) Odd functions

A function $f(x)$ is said to be odd if $f(-x) = -f(x)$, meaning the function value for a particular negative value of $x$ is numerically equal to that for the corresponding positive value of $x$ but opposite in sign. The graph of an odd function is symmetrical about the origin.

• $y = f(x) = x^3$ is an odd function since $f(-2) = -8 = -f(2)$, $f(-5) = -125 = -f(5)$, etc.
• $y = f(x) = \sin x$ is an odd function since $\sin(-x) = -\sin x$, so $f(-a) = -\sin a = -f(a)$.

🔑 Definition — Even function: $f(-x) = f(x)$, symmetrical about the y-axis. 🔑 Definition — Odd function: $f(-x) = -f(x)$, symmetrical about the origin.

Products of Odd and Even Functions

The rules for products of odd and even functions closely resemble the elementary rules of signs:

• (even) × (even) = (even), like (+) × (+) = (+)
• (odd) × (odd) = (even), like (-) × (-) = (+)
• (odd) × (even) = (odd), like (-) × (+) = (-)

(a) Two even functions: Let $F(x) = f(x)g(x)$ where $f(x)$ and $g(x)$ are even. Then $F(-x) = f(-x)g(-x) = f(x)g(x) = F(x)$, so $F(x)$ is even.

(b) Two odd functions: Let $F(x) = f(x)g(x)$ where $f(x)$ and $g(x)$ are odd. Then $F(-x) = f(-x)g(-x) = {-f(x)}{-g(x)} = f(x)g(x) = F(x)$, so $F(x)$ is even.

(c) One odd and one even function: Let $F(x) = f(x)g(x)$ where $f(x)$ is odd and $g(x)$ is even. Then $F(-x) = f(-x)g(-x) = -f(x)g(x) = -F(x)$, so $F(x)$ is odd.

Summary: If both functions are even or both are odd, their product is even. If one is even and the other is odd, their product is odd.

Example — Classifying products:

1. $x^2 \sin 2x$: odd (even × odd = odd)
2. $x^3 \cos x$: odd (odd × even = odd)
3. $\cos 2x \cos 3x$: even (even × even = even)
4. $x \sin nx$: even (odd × odd = even)
5. $3 \sin x \cos 4x$: odd (odd × even = odd)
6. $(2x+3) \sin 4x$: neither (neither × odd = neither)
7. $\sin^2 x \cos 3x$: even (even × even = even)
8. $x^3 e^x$: neither (odd × neither = neither)
9. $(x^4+4) \sin 2x$: odd (even × odd = odd)

Two useful facts emerge from odd and even functions:

(a) For an even function: $\int_{-a}^{a} f(x) , dx = 2 \int_{0}^{a} f(x) , dx$

(b) For an odd function: $\int_{-a}^{a} f(x) , dx = 0$

Theorem 1 — Fourier Series for Even Functions

If $f(x)$ is defined over the interval $-\pi < x < \pi$ and $f(x)$ is even, then the Fourier series for $f(x)$ contains cosine terms only (including $a_0$, which may be regarded as $a_n \cos nx$ with $n=0$).

Proof:

(a) $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) , dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) , dx$ (since $f(x)$ is even)

(b) $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx , dx$. Since $f(x)\cos nx$ is the product of two even functions, it is even. Therefore: $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx , dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos nx , dx$$

(c) $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx$. Since $f(x)\sin nx$ is the product of an even function and an odd function, it is odd. Therefore: $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx = 0$$

Therefore, there are no sine terms in the Fourier series for an even function.

Example — Even function waveform:

Consider a waveform symmetrical about the y-axis. The function is even, so there will be no sine terms:

$$f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos nx$$

For the given function with $f(x) = 4$ on $(-\pi/2, \pi/2)$ and zero elsewhere:

(a) $a_0 = \frac{2}{\pi} \int_{0}^{\pi/2} 4 , dx = \frac{8}{\pi} \cdot \frac{\pi}{2} = 4$

(b) $a_n = \frac{2}{\pi} \int_{0}^{\pi/2} 4 \cos nx , dx = \frac{8}{\pi} \left[ \frac{\sin nx}{n} \right]_{0}^{\pi/2} = \frac{8}{\pi n} \sin \frac{n\pi}{2}$

Since $\sin \frac{n\pi}{2} = 0$ for $n$ even, $= 1$ for $n = 1, 5, 9, \ldots$, and $= -1$ for $n = 3, 7, 11, \ldots$:

• $a_n = 0$ (n even)
• $a_n = \frac{8}{\pi n}$ (n = 1, 5, 9, ...)
• $a_n = -\frac{8}{\pi n}$ (n = 3, 7, 11, ...)

(c) $b_n = 0$ since $f(x)$ is even.

Required series: $$f(x) = 2 + \frac{8}{\pi} \left{ \cos x - \frac{1}{3} \cos 3x + \frac{1}{5} \cos 5x - \frac{1}{7} \cos 7x + \cdots \right}$$

Theorem 2 — Fourier Series for Odd Functions

If $f(x)$ is an odd function defined over the interval $-\pi < x < \pi$, then the Fourier series for $f(x)$ contains sine terms only.

Proof:

Since $f(x)$ is odd, $\int_{-\pi}^{0} f(x) , dx = -\int_{0}^{\pi} f(x) , dx$.

(a) $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) , dx = 0$ (since $f(x)$ is odd)

(b) $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx , dx$. Since $f(x)$ is odd and $\cos nx$ is even, the product $f(x)\cos nx$ is odd. Therefore: $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\text{odd function}) , dx = 0$$

(c) $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx$. Since $f(x)$ and $\sin nx$ are each odd, the product $f(x)\sin nx$ is even. Therefore: $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\text{even function}) , dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin nx , dx$$

So for an odd function: $a_0 = 0$, $a_n = 0$, and $b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin nx , dx$. The Fourier series contains sine terms only.

Example — Odd function: $$f(x) = -6 \text{ for } -\pi < x < 0, \quad f(x) = 6 \text{ for } 0 < x < \pi, \quad f(x) = f(x+2\pi)$$

This is an odd function, so $a_0 = 0$ and $a_n = 0$.

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx , dx$$

Since $f(x)\sin nx$ is the product of two odd functions, it is even:

$$b_n = \frac{2}{\pi} \int_{0}^{\pi} 6 \sin nx , dx = \frac{12}{\pi} \left[ -\frac{\cos nx}{n} \right]_{0}^{\pi} = \frac{12}{\pi n} [\cos 0 - \cos n\pi] = \frac{12}{\pi n} (1 - \cos n\pi)$$

$$b_n = 0 \text{ (n even)}, \quad b_n = \frac{24}{\pi n} \text{ (n odd)}$$

Series: $$f(x) = \frac{24}{\pi} \left{ \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \right}$$

💡 Why this matters: If a function is neither odd nor even, we must compute $a_0$, $a_n$, and $b_n$ in full without any simplification.

⭐ Key Takeaways

The most critical things to remember are: (1) Even functions satisfy $f(-x)=f(x)$ and have Fourier series with only cosine terms (including $a_0$), while odd functions satisfy $f(-x)=-f(x)$ and have Fourier series with only sine terms and $a_0=0$. (2) For even functions, $a_n = \frac{2}{\pi} \int_0^\pi f(x) \cos nx , dx$ and $b_n=0$; for odd functions, $b_n = \frac{2}{\pi} \int_0^\pi f(x) \sin nx , dx$ and $a_0=a_n=0$. (3) The product rules are: even×even=even, odd×odd=even, and odd×even=odd. (4) For integrals: even functions satisfy $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$ and odd functions satisfy $\int_{-a}^a f(x)dx = 0$. (5) If a function is neither odd nor even, all Fourier coefficients must be computed in full without simplification.

🧠 Quick Revision Questions

1. What are the defining properties of even and odd functions, and what does their symmetry look like graphically?

2. State the product rules for odd and even functions: what is the result when multiplying two even functions, two odd functions, and one odd with one even function?

3. For an even function defined on $(-\pi, \pi)$, write simplified formulas for $a_0$, $a_n$, and $b_n$ in terms of integrals from $0$ to $\pi$.

4. For an odd function defined on $(-\pi, \pi)$, what are the values of $a_0$ and $a_n$, and write the simplified formula for $b_n$?

5. In the first example where $f(x) = -x$ on $(-\pi, 0)$ and $f(x) = 0$ on $(0, \pi)$, is this function odd, even, or neither? How can you tell from the Fourier series obtained?

📘 Lecture 42 — Examples

📖 Overview: This lecture presents practical examples of Fourier series calculations for functions with period 2π, including functions that are neither odd nor even, functions with discontinuities, and half-range series expansions. It demonstrates how to compute Fourier coefficients, handle points of discontinuity, and use symmetry to simplify calculations.

🗂️ Topics Covered

The lecture covers a complete worked example of finding Fourier series for a piecewise function that is neither odd nor even, the behavior of Fourier series at points of finite discontinuity, half-range cosine and sine series for functions defined over (0, π), and the application of odd/even function properties to simplify calculations.

📝 Lecture Summary

Example (First Example)

Determine the Fourier series for the function shown which is neither odd nor even, defined as f(x) = (2/π)x for 0 < x < π, and f(x) = 2 for π < x < 2π.

The general Fourier series is: f(x) = ½a₀ + Σ_{n=1}^{∞} {aₙ cos nx + bₙ sin nx}

Finding a₀: a₀ = (1/π) ∫₀²π f(x)dx = (1/π) {∫₀π (2/π)x dx + ∫π²π 2 dx} = (1/π) { [x²/π]₀π + [2x]π²π } = (1/π) {π + 4π - 2π} a₀ = 3

Finding aₙ: aₙ = (1/π) ∫₀²π f(x) cos nx dx = (1/π) {∫₀π (2/π)x cos nx dx + ∫π²π 2 cos nx dx} = (2/π²) { (1/n)[x sin nx]₀π - (1/n) ∫₀π sin nx dx } + (2/π) ∫π²π cos nx dx After integration: aₙ = (2/(π²n²)) {cos nπ - 1} aₙ = 0 (n even); aₙ = -4/(π²n²) (n odd)

Finding bₙ: bₙ = (1/π) ∫₀²π f(x) sin nx dx = (1/π) {∫₀π (2/π)x sin nx dx + ∫π²π 2 sin nx dx} = (2/π²) { (-x cos nx)/n + (1/n) ∫ cos nx dx }₀π + (2/π) ∫π²π sin nx dx After integration: bₙ = (2/π) {(-1/n) cos nπ + 0 - (1/n)(cos 2πn - cos nπ)} = (2/π) {-(1/n) cos 2nπ} Since cos 2nπ = 1: bₙ = -2/(πn)

🔑 Definition — Fourier series: A representation of a periodic function as an infinite sum of sine and cosine functions.

📐 Formula: f(x) = ½a₀ + Σ_{n=1}^{∞} {aₙ cos nx + bₙ sin nx} where a₀, aₙ, bₙ are Fourier coefficients.

📌 Example: The required Fourier series is: f(x) = 3/2 - (4/π²){cos x + (1/9) cos 3x + (1/25) cos 5x + ...} - (2/π){sin x + (1/2) sin 2x + (1/3) sin 3x + (1/4) sin 4x + ...}

Sum of a Fourier series at a point of discontinuity

At a point of finite discontinuity or "jump" of a function y = f(x), the function has two distinct values: y₁ = f(x₁ - 0) approaching from below, and y₂ = f(x₁ + 0) approaching from above. When x = x₁ is substituted into the Fourier series, it converges to the average value ½{f(x₁ - 0) + f(x₁ + 0)} = ½(y₁ + y₂).

🔑 Definition — Point of finite discontinuity: A point where a function has a jump, with different limiting values from the left and right.

📌 Example: Consider f(x) = 0 for -π < x < 0, f(x) = a for 0 < x < π, period 2π. a₀ = a, aₙ = 0, bₙ = 2a/(nπ) for n odd, bₙ = 0 for n even The series: f(x) = a/2 + (2a/π){sin x + (1/3) sin 3x + (1/5) sin 5x + ...} At x = 0 (a discontinuity), all sine terms vanish, giving f(0) = a/2, which is the average of 0 and a.

Half-Range Series

Sometimes a function of period 2π is defined only over the range 0 to π. We can choose to represent it as an even function (giving a cosine series) or an odd function (giving a sine series), known as half-range series.

🔑 Definition — Half-range series: A Fourier series representing a function defined only on (0, π) by extending it as either an even or odd function over (-π, 0).

Half-Range Cosine Series Example: For f(x) = 2x, 0 < x < π, period 2π, to obtain a cosine series, assume the function is even. a₀ = (2/π) ∫₀π 2x dx = 2π aₙ = (2/π) ∫₀π 2x cos nx dx = (4/(πn²))(cos nπ - 1) aₙ = 0 (n even); aₙ = -8/(πn²) (n odd) bₙ = 0 (since function is even) f(x) = π - (8/π){cos x + (1/9) cos 3x + (1/25) cos 5x + ...}

Half-Range Sine Series Example: For f(x) = 1 + x, 0 < x < π, period 2π, to obtain a sine series, assume the function is odd. a₀ = 0, aₙ = 0 bₙ = (2/π) ∫₀π (1+x) sin nx dx = (2/(πn)){1 - (1+π) cos nπ} bₙ = -2/n (n even); bₙ = (4+2π)/(πn) (n odd) f(x) = (4/π + 2){sin x + (1/3) sin 3x + (1/5) sin 5x + ...} - 2{(1/2) sin 2x + (1/4) sin 4x + (1/6) sin 6x + ...}

💡 Why this matters: Using half-range series and knowing odd/even properties saves significant computational work by eliminating unnecessary coefficient calculations.

⭐ Key Takeaways

The Fourier series at a point of discontinuity converges to the average of the left and right limiting values, not to either value individually. For functions defined only on (0, π), half-range series allow choosing between cosine series (even extension) or sine series (odd extension), which eliminates the need to compute the corresponding coefficients. Understanding odd and even function properties (aₙ = 0 for odd functions, bₙ = 0 for even functions) dramatically simplifies Fourier series calculations. The determination of Fourier coefficients requires careful integration by parts for products like x cos nx and x sin nx. When computing bₙ, the integral of sin nx over a full period [0, 2π] is zero, but the integral of x sin nx over [0, π] requires integration by parts.

🧠 Quick Revision Questions

1. At a point of discontinuity x = 0, where f(0⁻) = 0 and f(0⁺) = a, what value does the Fourier series converge to?
2. For a half-range cosine series, what assumption is made about the function between x = -π and x = 0?
3. What is the general formula for aₙ when f(x) = x over (0, π) and the function is extended as an even function?
4. In the first example of this lecture, why is bₙ = -2/(πn) and not dependent on n being odd or even?
5. For a function defined only on (0, π), what are the two possible half-range series representations called?

📘 Lecture 43 — Functions With Periods Other Than 2π

📖 Overview: This lecture extends Fourier series analysis to functions with arbitrary period T, which is crucial for real-world applications like engineering and physics where functions rarely have a period of exactly 2π. It covers the general Fourier series for period T, the calculation of Fourier coefficients with the new variable, and practical examples including half-range series and half-wave rectifiers.

🗂️ Topics Covered

The lecture covers functions with period T, converting the independent variable using angular velocity ω, deriving Fourier coefficients for general period, worked examples of periodic functions, half-range sine and cosine series for even and odd functions, and the Fourier series of a half-wave rectifier.

📝 Lecture Summary

Functions With Period T

If y = f(x) is defined over an interval of length T (period T), we can convert this to the standard 2π interval by changing the units of the independent variable. For physical oscillations, the independent variable is often time (t), with period T seconds, so f(t) = f(t + T). The frequency is f = 1/T hertz, and the angular velocity ω = 2πf, so ω = 2π/T and T = 2π/ω. The Fourier series in terms of t is: f(t) = 1/2 a₀ + Σ_{n=1}^{∞} [aₙ cos(nωt) + bₙ sin(nωt)] which can also be written as f(t) = 1/2 A₀ + Σ_{n=1}^{∞} Bₙ sin(nωt + φₙ).

Fourier Coefficients

With the new variable, the Fourier coefficients for a function of period T are:

• a₀ = (2/T) ∫₀ᵀ f(t) dt = (ω/π) ∫₀^{2π/ω} f(t) dt
• aₙ = (2/T) ∫₀ᵀ f(t) cos(nωt) dt = (ω/π) ∫₀^{2π/ω} f(t) cos(nωt) dt
• bₙ = (2/T) ∫₀ᵀ f(t) sin(nωt) dt = (ω/π) ∫₀^{2π/ω} f(t) sin(nωt) dt

The limits can be any interval covering a complete period, such as 0 to T, -T/2 to T/2, -π/ω to π/ω, or 0 to 2π/ω.

🔑 Definition — Angular Velocity: ω = 2π/T, where T is the period and ω is measured in radians per second. 📐 Formula: ω = 2π/T → The angular frequency relates the period to the standard circular frequency.

Example (Periodic Function)

Determine the Fourier series for f(t) = 2(1+t) for -1 < t < 0 and f(t) = 0 for 0 < t < 1, with f(t) = f(t+2). Given: T = 2, so ω = 2π/T = π. Coefficients:

• a₀ = (2/T) ∫{-T/2}^{T/2} f(t) dt = (2/2) ∫{-1}^{1} f(t) dt = ∫{-1}^{0} 2(1+t) dt + ∫₀¹ 0 dt = [2t + t²]{-1}^{0} = 1
• aₙ = (2/T) ∫{-1}^{1} f(t) cos(nωt) dt = ∫{-1}^{0} 2(1+t) cos(nπt) dt Using integration by parts: aₙ = 2/n²ω² (1 - cos(nπ)) So aₙ = 0 (n even) and aₙ = 4/n²ω² (n odd)
• bₙ = (2/T) ∫{-1}^{1} f(t) sin(nωt) dt = ∫{-1}^{0} 2(1+t) sin(nπt) dt Using integration by parts: bₙ = -2/nω (since sin(nπ) = 0)

📌 Example: The first few terms of the series give: f(t) = 1/2 + (4/ω²)[cos(ωt) + (1/9)cos(3ωt) + (1/25)cos(5ωt) + ...] - (2/ω)[sin(ωt) + (1/2)sin(2ωt) + (1/3)sin(3ωt) + ...]

Half-Range Series

The theory of half-range sine and cosine series applies with the new variable.

(a) Even function — Half-range cosine series: For y = f(t) with 0 < t < T/2, extended as an even function (symmetrical about y-axis), we have bₙ = 0. f(t) = 1/2 a₀ + Σ_{n=1}^{∞} aₙ cos(nωt) where a₀ = (4/T) ∫₀^{T/2} f(t) dt and aₙ = (4/T) ∫₀^{T/2} f(t) cos(nωt) dt

(b) Odd function — Half-range sine series: For y = f(t) with 0 < t < T/2, extended as an odd function (symmetrical about origin), we have a₀ = 0 and aₙ = 0. f(t) = Σ_{n=1}^{∞} bₙ sin(nωt) where bₙ = (4/T) ∫₀^{T/2} f(t) sin(nωt) dt

Example — Half-Range Cosine Series

A function f(t) = 4 - t, 0 < t < 4. Form a half-range cosine series. Given: T = 8, extended as even function, T/2 = 4, ω = 2π/T = π/4.

• a₀ = (4/8) ∫₀⁴ (4 - t) dt = (1/2)[4t - t²/2]₀⁴ = (1/2)[16 - 8] = 4
• aₙ = (4/8) ∫₀⁴ (4 - t) cos(nωt) dt = (1/2) ∫₀⁴ (4 - t) cos(nπt/4) dt Integration by parts yields: aₙ = (1/2)[-2 sin(4nω)/(nω) - (1/n²ω²)(cos(4nω) - 1)] With ω = π/4, sin(nπ) = 0, cos(nπ) = 1 (n even) or -1 (n odd), so aₙ = 0 (n even) and aₙ = 1/n²ω² (n odd)

📌 Example: The half-range cosine series is: f(t) = 2 + (1/ω²)[cos(ωt) + (1/9)cos(3ωt) + (1/25)cos(5ωt) + ...] where ω = π/4

Example — Half-Range Sine Series

A function f(t) = 3 + t, 0 < t < 2, with f(t) = f(t+4). Obtain half-range sine series. Given: T = 4, extended as odd function, T/2 = 2, ω = 2π/T = π/2.

• bₙ = (4/4) ∫₀² (3 + t) sin(nωt) dt = ∫₀² (3 + t) sin(nπt/2) dt Integration by parts gives: bₙ = (1/nω)(3 - 5 cos(2nω)) + (1/n²ω²) sin(2nω) With ω = π/2, sin(nπ) = 0, cos(nπ) = 1 (n even) or -1 (n odd), so bₙ = -2/nω (n even) and bₙ = 8/n²ω² (n odd)

📌 Example: The half-range sine series is: f(t) = (2/ω)[(4/1)sin(ωt) - (1/2)sin(2ωt) + (4/3)sin(3ωt) - (1/4)sin(4ωt) + ...] where ω = π/2

💡 Why this matters: Half-range series allow representation of functions defined only on half a period, which is common in engineering problems involving piecewise or asymmetric signals.

Half-Wave Rectifier

A sinusoidal voltage E sin(ωt) is passed through a half-wave rectifier that clips the negative portion. Find the Fourier series. Given: u(t) = 0 for -T/2 < t < 0, u(t) = E sin(ωt) for 0 < t < T/2, with T = 2π/ω.

• a₀ = (2/T) ∫₀^{T/2} E sin(ωt) dt = (ω/π) ∫₀^{π/ω} E sin(ωt) dt = 2E/π
• aₙ = (2/T) ∫₀^{π/ω} E sin(ωt) cos(nωt) dt = (ωE/2π) ∫₀^{π/ω} [sin(1+n)ωt + sin(1-n)ωt] dt For n = 1, a₁ = 0. For n = 2, 3, ...: aₙ = 0 (n odd), aₙ = 2E/(π(1-n²)) (n even)
• bₙ = (2/T) ∫₀^{π/ω} E sin(ωt) sin(nωt) dt = -(ωE/2π) ∫₀^{π/ω} [cos(1+n)ωt - cos(1-n)ωt] dt For n = 1: b₁ = E/2. For n ≠ 1: bₙ = 0.

📌 Example: The Fourier series for the half-wave rectifier is: u(t) = E/π + (E/2) sin(ωt) - (2E/π)[(1/1·3)cos(2ωt) + (1/3·5)cos(4ωt) + ...]

💡 Why this matters: Half-wave rectification is fundamental in power electronics, converting AC to pulsating DC, and its Fourier series reveals the harmonic content.

⭐ Key Takeaways

The key concept is adapting Fourier series to any period T by introducing angular velocity ω = 2π/T, with coefficients aₙ and bₙ computed using integrals over one period. Half-range series (cosine for even, sine for odd) simplify representation for functions defined on half a period. The half-wave rectifier example demonstrates how to handle piecewise-defined periodic functions with discontinuities. Mastery of these techniques is essential for analyzing oscillatory systems in electrical engineering, signal processing, and mechanical vibrations.

🧠 Quick Revision Questions

1. What is the formula for the angular velocity ω in terms of the period T, and how is it used to convert a function of period T to the standard Fourier series form?
2. In the example of the function f(t) = 2(1+t) for -1<t<0, f(t)=0 for 0<t<1, with T=2, what are a₀ and the expressions for aₙ and bₙ?
3. How do the formulas for a₀ and aₙ differ between a half-range cosine series and a full-range Fourier series for a periodic function?
4. For the half-wave rectifier with input E sin(ωt), what are the values of a₁ and b₁, and why is a₁ zero?
5. In the half-range sine series example for f(t) = 3+t (0<t<2, T=4), why are the bₙ coefficients different for odd and even n?

📘 Lecture 44 — Laplace transforms

📖 Overview: This lecture formally introduces the Laplace transform, defines its standard results for elementary functions (exponential, polynomial, trigonometric, and hyperbolic), and demonstrates how to find both transforms and inverse transforms. It also covers complex numbers and Euler’s formula, which are used to derive the transforms of sine and cosine. Understanding these fundamentals is essential for solving differential equations and analyzing systems in engineering and physics.

🗂️ Topics Covered

The lecture covers the formal definition of the Laplace transform, the derivation of transforms for constant functions, exponentials, and polynomials (tⁿ). It introduces complex number operations and Euler’s formula as tools to find the Laplace transforms of sin at and cos at. Standard results are extended to hyperbolic functions sinh at and cosh at. The lecture concludes with examples of linear combinations and a table of standard Laplace transforms.

📝 Lecture Summary

Laplace Transform

The Laplace Transform of a function F(t) is denoted by L{F(t)} and is defined as the integral of F(t) e⁻ˢᵗ between the limits t=0 and t = ∞. ∞ L{F(t)} = ∫₀ e⁻ˢᵗ F(t) dt In all cases, the constant parameter s is assumed to be positive and large enough to ensure that the product F(t)e⁻ˢᵗ converges to zero as t→∞. The result of the transform is always a function of s.

🔑 Definition — Laplace Transform: L{F(t)} = ∫₀∞ e⁻ˢᵗ F(t) dt

Laplace Transform of F(t) = a (constant)

∞ ∞ L(a) = ∫₀ a e⁻ˢᵗ dt = a ∫₀ e⁻ˢᵗ dt = a [e⁻ˢᵗ / -s]₀∞ = -a/s [0 - 1] = a/s

🔑 Definition — Laplace Transform of a constant: L{a} = a/s

Laplace Transform of eᵃᵗ

∞ L(eᵃᵗ) = ∫₀ eᵃᵗ e⁻ˢᵗ dt = ∫₀∞ e⁻⁽ˢ⁻ᵃ⁾ᵗ dt = [e⁻⁽ˢ⁻ᵃ⁾ᵗ / -(s-a)]₀∞ = -1/(s-a) [1/e⁽ˢ⁻ᵃ⁾ᵗ]₀∞ = -1/(s-a)[0 - 1] = 1/(s-a), for s > a

🔑 Definition — Laplace Transform of eᵃᵗ: L{eᵃᵗ} = 1/(s-a)

Complex Numbers Power of i

Every time a factor i⁴ occurs, it can be replaced by the factor 1, so that the power of i is reduced to one of four results: i⁰=1, i¹=i, i²=-1, i³=-i. i⁹ = (i⁴)² i = 1² i = i i²⁰ = (i⁴)⁵ = 1⁵ = 1 i³⁰ = (i⁴)⁷ i² = 1⁷ (-1) = -1 i¹⁵ = (i⁴)³ i³ = 1³ (-i) = -i

Complex Numbers

z = 3 + 5i is a complex number where 3 is the real part and 5 is the imaginary part. In general, z = a + bi, where a is the real part and b is the imaginary part.

Conjugate Complex Numbers

For a complex number a + i b, the complex number a - i b is called its conjugate. Conjugate complex numbers are identical except for the sign of the imaginary part. 🔑 Remember: The product of a complex number and its conjugate is always entirely real. (3+4i)(3-4i) = 9 + 16 = 25. (a+bi)(a-bi) = a² + b².

Euler's Formula

The series expansion of eˣ, cos x, and sin x are given. Replacing x by (it) yields: e^(it) = cos t + i sin t The real part R(e^(it)) = cos t and the imaginary part I(e^(it)) = sin t.

🔑 Definition — Euler's Formula: e^(iθ) = cos θ + i sin θ

The Laplace Transform of F(t) = sin at

L(sin at) = L(I(e^(iat))) = I[∫₀∞ e^(iat) e⁻ˢᵗ dt] = I[∫₀∞ e⁻⁽ˢ⁻ⁱᵃ⁾ᵗ dt] = I{1/(s-ia)} = I[(s+ia)/(s²+a²)] = I[s/(s²+a²) + i a/(s²+a²)] = a/(s²+a²)

🔑 Definition — Laplace Transform of sin at: L{sin at} = a/(s² + a²)

The Laplace Transform of F(t) = cos at

L(cos at) = L(R(e^(iat))) = R[1/(s-ia)] = R[(s+ia)/(s²+a²)] = R[s/(s²+a²) + i a/(s²+a²)] = s/(s²+a²)

🔑 Definition — Laplace Transform of cos at: L{cos at} = s/(s² + a²)

The Transform of F(t) = tⁿ where n is a positive integer

Using integration by parts and a reduction formula, we get: L(tⁿ) = n/s L(tⁿ⁻¹) Repeatedly applying this reduction formula gives: Iₙ = n/s * (n-1)/s * (n-2)/s * ... * 2/s * 1/s * I₀ Since I₀ = L{1} = 1/s, we get L(tⁿ) = n! / sⁿ⁺¹

🔑 Definition — Laplace Transform of tⁿ: L{tⁿ} = n! / sⁿ⁺¹ 📐 Formula: L{tⁿ} = n! / sⁿ⁺¹ → The transform of t raised to a positive integer power n is n factorial divided by s to the power of (n+1). 📌 Example: L{t³} = 3! / s⁴ = 6/s⁴

Laplace Transform of F(t) = sinh at and F(t) = cosh at

Starting from the exponential definitions: sinh at = ½(eᵃᵗ - e⁻ᵃᵗ) and cosh at = ½(eᵃᵗ + e⁻ᵃᵗ)

a) L{sinh at} = L[½(eᵃᵗ - e⁻ᵃᵗ)] = ½[L(eᵃᵗ) - L(e⁻ᵃᵗ)] = ½[1/(s-a) - 1/(s+a)] = ½[(s+a - (s-a))/(s²-a²)] = ½[2a/(s²-a²)] = a/(s²-a²)

b) L{cosh at} = L[½(eᵃᵗ + e⁻ᵃᵗ)] = ½[L(eᵃᵗ) + L(e⁻ᵃᵗ)] = ½[1/(s-a) + 1/(s+a)] = ½[(s+a + s-a)/(s²-a²)] = ½[2s/(s²-a²)] = s/(s²-a²)

🔑 Definition — Laplace Transform of sinh at: L{sinh at} = a/(s² - a²) 🔑 Definition — Laplace Transform of cosh at: L{cosh at} = s/(s² - a²)

Several Standard Results

L{a} = a/s ; L{eᵃᵗ} = 1/(s-a) ; L{tⁿ} = n!/sⁿ⁺¹ L{sin at} = a/(s²+a²) ; L{cos at} = s/(s²+a²) L{sinh at} = a/(s²-a²) ; L{cosh at} = s/(s²-a²)

These transforms can be combined by adding or subtracting as necessary, but they must not be multiplied together to form the transform of a product.

📌 Example a: Find L{2 sin 3t + cos 3t} L{2 sin 3t + cos 3t} = 2L(sin 3t) + L(cos 3t) = 2 * [3/(s²+9)] + [s/(s²+9)] = (6 + s)/(s²+9) = (s+6)/(s²+9)

📌 Example b: Find L{4e²ᵗ + 3 cosh 4t} L{4e²ᵗ + 3 cosh 4t} = 4L(e²ᵗ) + 3L(cosh 4t) = 4*[1/(s-2)] + 3*[s/(s²-16)] = 4/(s-2) + 3s/(s²-16)

⭐ Key Takeaways

(1) The Laplace transform is defined as an integral L{F(t)} = ∫₀∞ e⁻ˢᵗ F(t) dt, and its result is always a function of s. (2) The standard transforms for constants, exponentials, powers of t, sine, cosine, hyperbolic sine, and hyperbolic cosine form a critical "toolkit" for the course; these are derived and must be memorized. (3) Euler's formula e^(iθ) = cos θ + i sin θ is a powerful tool for deriving the transforms of sine and cosine by taking the imaginary or real parts, respectively, of the transform of e^(iat). (4) The product of a complex number with its conjugate always yields a real number (a²+b²), which is essential in simplifying the transforms of sine and cosine. (5) The transform is a linear operator, meaning L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}. Transforms can be added or subtracted in linear combinations but must NOT be multiplied to find the transform of a product.

🧠 Quick Revision Questions

1. State the formal definition of the Laplace transform L{F(t)}.
2. What is the Laplace transform of eᵃᵗ? Under what condition on s does this transform hold?
3. Derive the Laplace transform of sin(at) using Euler's formula.
4. What is the reduction formula used to find L{tⁿ}, and what is the final result?
5. State the standard Laplace transforms for both sinh(at) and cosh(at). How are they different from the transforms of sin(at) and cos(at)?

📘 Lecture 45 — Theorems

📖 Overview: This lecture covers three fundamental theorems of Laplace transforms: the First Shift Theorem, Multiplication by t, and Division by t. It then introduces Inverse Laplace transforms, explaining how to recover the original time function from a given transform using partial fractions and a table of standard forms, and concludes by applying these techniques to solve ordinary differential equations.

🗂️ Topics Covered

The lecture begins with the First Shift Theorem, which shows how multiplying a function by e^(-at) shifts the Laplace variable s to s+a. It then covers the theorem for multiplying F(t) by t, which involves differentiating the transform, and the theorem for dividing F(t) by t, which involves integrating the transform. The second half of the lecture thoroughly discusses the inverse Laplace transform process, including rules for partial fraction decomposition, a table of standard inverse transforms, and formulas for transforms of derivatives. The lecture culminates with the step-by-step solution of differential equations using the Laplace transform method, including a detailed example with a second-order ODE.

📝 Lecture Summary

Theorem 1 The First Shift Theorem

The First Shift theorem states that if L{F(t)} = f(s), then L{e^(-at)F(t)} = f(s+a). The transform of the exponentially damped function is the same as the transform of the original function with ‘s’ replaced by (s+a).

🔑 Definition — First Shift Theorem: L{e^(-at)F(t)} = f(s+a)

📐 Formula: L{e^(-at)F(t)} = f(s+a) → The Laplace transform of an exponential multiple of a function is the transform of the function itself with 's' shifted by 'a'.

📌 Example: Given L{sin 2t} = 2/(s² + 4). Then L{e^(-3t) sin 2t} = 2/((s+3)² + 4) = 2/(s² + 6s + 13). Another example: L{t²} = 2/s³, so L{t² e^(4t)} = 2/(s – 4)³.

Theorem 2 Multiplying by t

If L{F(t)} = f(s), then L{t F(t)} = -d/ds {f(s)}. This theorem extends the range of functions we can deal with, and in general, L{tⁿF(t)} = (-1)ⁿ dⁿ/dsⁿ {f(s)}.

🔑 Definition — Multiplying by t Theorem: L{tF(t)} = -d/ds {f(s)}

📐 Formula: L{tF(t)} = -d/ds f(s) → Multiplying a time function by 't' corresponds to taking the negative derivative of its Laplace transform with respect to 's'.

📌 Example: L{sin 2t} = 2/(s² + 4). Then L{t sin 2t} = -d/ds [2/(s² + 4)] = 4s/(s² + 4)². Another example: L{t cos 3t} = -d/ds [s/(s² - 9)] = (s² + 9)/(s² - 9)². This can be taken a step further: L{t² cos 3t} = -d/ds [(s² + 9)/(s² - 9)²] = 2s(s² + 27)/(s² - 9)³.

💡 Why this matters: This theorem allows us to find transforms of functions multiplied by tⁿ, which is crucial for solving differential equations with polynomial coefficients.

Theorem 3 Dividing by t

If L{F(t)} = f(s), then L{F(t)/t} = ∫ₛ^∞ f(s) ds, provided the limit of F(t)/t as t approaches 0 exists.

🔑 Definition — Dividing by t Theorem: L{F(t)/t} = ∫ₛ^∞ f(u) du

📐 Formula: L{F(t)/t} = ∫ₛ^∞ f(s) ds → Dividing a time function by 't' corresponds to integrating its Laplace transform from 's' to infinity.

📌 Example: Determine L{sin(at)/t}. As L{sin at} = a/(s² + a²), then L{sin at/t} = ∫ₛ^∞ a/(s² + a²) ds = [tan⁻¹(s/a)]ₛ^∞ = π/2 - tan⁻¹(s/a) = tan⁻¹(a/s).

📌 Example: Determine L{(1 - cos 2t)/t}. As L{1 - cos 2t} = 1/s - s/(s² + 4), then L{(1 - cos 2t)/t} = ∫ₛ^∞ (1/s - s/(s² + 4)) ds = [ln s - ½ ln(s² + 4)]ₛ^∞ = [½ ln(s²/(s² + 4))]ₛ^∞ = 0 - ½ ln(s²/(s² + 4)) = ½ ln((s² + 4)/s²).

Standard Forms

A table of standard Laplace transforms is essential for both forward and inverse transforms.

Inverse Transforms

Inverse Laplace transforms are the reverse process: given a Laplace transform, we find the function of t to which it belongs. The symbol L⁻¹ indicates the inverse transform.

🔑 Definition — Inverse Laplace Transform: Given a function f(s), the inverse transform L⁻¹{f(s)} finds the original function F(t) such that L{F(t)} = f(s).

📌 Example: L⁻¹{1/(s – 2)} = e^(2t) L⁻¹{s/(s² + 25)} = cos 5t L⁻¹{4/s} = 4 L⁻¹{12/(s² – 9)} = 4 sinh 3t

Often, a transform must be broken into simpler parts using partial fractions. For instance, to find L⁻¹{3s/(s² – s – 6)}, we first decompose 3s/(s² – s – 6) into 1/(s + 2) + 2/(s – 3). Then, L⁻¹{3s/(s² – s – 6)} = L⁻¹{1/(s + 2)} + L⁻¹{2/(s – 3)} = e^(-2t) + 2e^(3t).

Rules of Partial Fractions

1. The numerator must be of a lower degree than the denominator. If not, divide out first.
2. Factorise the denominator into its prime factors.
3. A linear factor (s + a) gives a partial fraction of the form A/(s + a).
4. A repeated factor (s + a)² gives A/(s + a) + B/(s + a)².
5. Similarly, (s + a)³ gives A/(s + a) + B/(s + a)² + C/(s + a)³.
6. A quadratic factor (s² + ps + q) gives (Ps + Q)/(s² + ps + q).
7. A repeated quadratic factor (s² + ps + q)² gives (Ps + Q)/(s² + ps + q) + (Rs + T)/(s² + ps + q)².

📌 Example: Determine L⁻¹{(5s + 1)/(s² – s – 12)}. a) The numerator (1st degree) is of lower degree than the denominator (2nd degree). b) Factorise: (s – 4)(s + 3). c) Write identity: (5s + 1)/(s² – s – 12) = A/(s – 4) + B/(s + 3). d) Multiply by denominator: 5s + 1 = A(s + 3) + B(s – 4). e) Substitute s = 4: 21 = A(7) → A = 3. f) Substitute s = -3: -14 = B(-7) → B = 2. g) Therefore, L⁻¹{(5s + 1)/(s² – s – 12)} = L⁻¹{3/(s – 4) + 2/(s + 3)} = 3e^(4t) + 2e^(-3t).

📌 Example: Determine L⁻¹{(9s – 8)/(s² – 2s)}. a) Decompose: (9s – 8)/[s(s – 2)] = A/s + B/(s – 2). b) Multiply by s(s – 2): 9s – 8 = A(s – 2) + B(s). c) Put s = 0: -8 = A(-2) → A = 4. d) Put s = 2: 10 = B(2) → B = 5. e) Therefore, L⁻¹{4/s + 5/(s – 2)} = 4 + 5e^(2t).

Table of Inverse Transforms

A standard table is used for finding inverse transforms. The table shows common f(s) and their corresponding F(t).

Transforms of Derivatives

The Laplace transforms of derivatives are crucial for solving differential equations. Let F'(t), F''(t), etc., denote derivatives.

🔑 Transform of First Derivative: L{F'(t)} = sL{F(t)} – F(0) 🔑 Transform of Second Derivative: L{F''(t)} = s²L{F(t)} – sF(0) – F'(0) 🔑 Transform of Third Derivative: L{F'''(t)} = s³L{F(t)} – s²F(0) – sF'(0) – F''(0) 🔑 Transform of Fourth Derivative: L{Fᶤᵛ(t)} = s⁴L{F(t)} – s³F(0) – s²F'(0) – sF''(0) – F'''(0)

Differential Equation And Its Solution

The process of solving a differential equation using Laplace transforms involves four distinct stages: (a) Rewrite the equation in terms of Laplace transforms. (b) Insert the given initial conditions. (c) Rearrange the equation algebraically to give the transform of the solution. (d) Determine the inverse transform to obtain the particular solution.

📌 Example: Solve the first-order ODE dx/dt – 2x = 4, given x(0) = 1.

1. Take Laplace transform: L{dx/dt} – 2L{x} = L{4} → sL{x(t)} – x(0) – 2L{x(t)} = 4/s.
2. Substitute initial condition x(0)=1: (s – 2)L{x(t)} – 1 = 4/s → (s – 2)L{x(t)} = (4 + s)/s.
3. Solve for L{x(t)}: L{x(t)} = (s + 4)/[s(s – 2)].
4. Use partial fractions to invert: (s+4)/[s(s-2)] = A/s + B/(s-2). Solving gives A = -2, B = 3.
5. Therefore, x(t) = L⁻¹{-2/s + 3/(s-2)} = -2 + 3e^(2t).

📌 Example: Solve the second-order ODE d²x/dt² – 3 dx/dt + 2x = 2e^(3t), given x(0) = 5 and x'(0) = 7.

1. Rewrite: L{x''} – 3L{x'} + 2L{x} = 2L{e^(3t)}.
2. Apply derivative formulas: [s²L{x} – s x(0) – x'(0)] – 3[sL{x} – x(0)] + 2L{x} = 2/(s – 3).
3. Substitute initial conditions x(0)=5, x'(0)=7: (s²L{x} – 5s – 7) – 3sL{x} + 15 + 2L{x} = 2/(s – 3).
4. Rearrange: (s² – 3s + 2)L{x} = 2/(s – 3) + 5s – 8 = (2 + (5s – 8)(s – 3))/(s – 3) = (5s² – 23s + 24)/(s – 3).
5. So L{x} = (5s² – 23s + 24)/[(s – 1)(s – 2)(s – 3)].
6. Partial fraction decomposition gives L{x} = 3/(s – 1) + 2/(s – 2) + 0/(s – 3).
7. Take inverse Laplace: x(t) = 3e^t + 2e^(2t).

⭐ Key Takeaways

The three core theorems—First Shift, Multiplying by t, and Dividing by t—are powerful tools for manipulating Laplace transforms. Mastery of partial fraction decomposition is absolutely critical for finding inverse Laplace transforms of rational functions. The transforms of derivatives directly incorporate initial conditions, which makes the Laplace transform method exceptionally efficient for solving initial value problems of ordinary differential equations, reducing them to algebraic manipulations. The step-by-step process of transform, substitute, rearrange, and invert is a systematic and powerful technique for solving ODEs.

🧠 Quick Revision Questions

1. State the First Shift Theorem. If L{F(t)} = 2/(s – 3), what is L{e^(-5t)F(t)}?
2. Explain the theorem for L{tF(t)}. Use it to find L{t cos 2t}.
3. What is the formula for L{F(t)/t}? Use it to find L{sin(3t)/t}.
4. Describe the steps to find the inverse Laplace transform of 1/[s(s+1)].
5. Write the Laplace transform formula for the second derivative L{F''(t)} and outline how it is used to solve a second-order initial value problem.