Problem 19 - Calculus Early Transcendentals

Question

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}(x+2 y) d x+(x-y) d y} \\ {C: x=2 \cos t, y=4 \sin t \quad(0 \leq t \leq \pi / 4)}\end{array} $$

Step-by-Step Solution

Answer

The value of the line integral is \( 4 + 2 \sqrt{2} - \frac{\pi}{2} \).

Step 1: Parametrize the Curve

The curve \( C \) is already parametrized with \( x = 2 \cos t \) and \( y = 4 \sin t \), where \( 0 \leq t \leq \frac{\pi}{4} \). This gives us a parametrization in terms of \( t \).

Step 2: Find Differential Components

Calculate the differentials \( dx \) and \( dy \) based on the given parameterization. \( dx = \frac{d}{dt}(2 \cos t) \, dt = -2 \sin t \, dt \) and \( dy = \frac{d}{dt}(4 \sin t) \, dt = 4 \cos t \, dt \).

Step 3: Substitute into the Line Integral

Substitute \( x = 2 \cos t \), \( y = 4 \sin t \), \( dx = -2 \sin t \, dt \), and \( dy = 4 \cos t \, dt \) into the line integral. The integral becomes \[ \int_{0}^{\pi/4} [(2 \cos t + 2(4 \sin t))(-2 \sin t) + (2 \cos t - 4 \sin t)(4 \cos t)] \, dt. \]

Step 4: Simplify the Integral Expression

Simplify the expression inside the integral: \((2 \cos t + 8 \sin t)(-2 \sin t) + (2 \cos t - 4 \sin t)(4 \cos t)\). This simplifies to \(-4 \cos t \sin t - 16 \sin^2 t + 8 \cos^2 t - 16 \cos t \sin t\).

Step 5: Combine Like Terms

Combine and rearrange like terms: \( (-16 \cos t \sin t) + (-16 \sin^2 t) + 8 \cos^2 t \).

Step 6: Further Simplify using Trigonometric Identities

Use the identity \( \sin^2 t = 1 - \cos^2 t \) to replace \( \sin^2 t \). This leads to \( -16 \cos t \sin t + 8(1 - \cos^2 t) - 16 \cos t \sin t\), which further simplifies to \(-32 \cos t \sin t - 8 \cos^2 t + 8 \).

Step 7: Integrate over the Interval [0, π/4]

Integrate the simplified expression \(-32 \cos t \sin t - 8 \cos^2 t + 8 \) with respect to \( t \) from \( 0 \) to \( \frac{\pi}{4} \). Compute each term separately.

Step 8: Evaluate the Definite Integral

Compute each integral: 1) Integrate \(-32 \cos t \sin t \) using the substitution \( u = \sin t \), 2) Integrate \(-8 \cos^2 t \) using \( \cos 2t = 2 \cos^2 t - 1 \), and 3) Integrate the constant \( 8 \). The result is the calculated sum of all integrated terms over \( [0, \frac{\pi}{4}] \).

Key Concepts

Parametrization

Parametrization is a technique used to express a curve in terms of a single variable, often denoted as \( t \). This form of expression simplifies calculations, including integrals, by reducing the dimensions of analysis.

• For the given exercise, the curve \( C \) is parametrized using the equations \( x = 2 \cos t \) and \( y = 4 \sin t \).
• This is an example of using trigonometric functions, sine, and cosine, to represent the coordinates of a curve.
• Parametrization allows us to easily find derivatives, like \( dx \) and \( dy \), needed for line integrals.

In our example, as \( t \) varies from 0 to \( \pi/4 \), the curve traces a specific path on the coordinate plane. This cohesive parameter approach makes solving curves straightforward.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for all values of the variables involved. They are essential for simplifying and solving integrals in calculus.

• In calculus problems involving trigonometry, identities like \( \sin^2 t = 1 - \cos^2 t \) and \( \cos 2t = 2 \cos^2 t - 1 \) are particularly useful.
• These identities help transform complex trigonometric expressions into simpler ones, making integration more feasible.
• In our problem, using these identities allowed us to reduce and simplify the line integral function.

Understanding how and when to apply these identities accelerates problem-solving and enhances accuracy in calculus problems involving trigonometric expressions.

Definite Integral

A definite integral is used to find the total accumulation of a function over a given interval. It provides the solution in the form of a numerical value, signifying the area under a curve or accumulation along a path.

• To compute a definite integral, you evaluate a function's primitive (antiderivative) at the endpoints of the interval and subtract them.
• In line integrals, you're evaluating the function along a curve between two points, integrating over a parameterization like \( t \).
• In the solution provided, the integral of the function is conducted from \( 0 \) to \( \pi/4 \), involving detailed consideration for each component of the simplified expression.

This process requires careful integration of each term, and clear understanding of how to set initial boundaries and perform calculations to acquire a precise result.