Problem 28 - Calculus Early Transcendentals

Question

Use a CAS to evaluate the line integrals along the given curves. $$ \begin{array}{l}{\text { (a) } \int_{C} x^{7} y^{3} d s} \\ {\quad C: x=\cos ^{3} t, \quad y=\sin ^{3} t \quad(0 \leq t \leq \pi / 2)} \\ {\text { (b) } \int_{C} x^{5} z d x+7 y d y+y^{2} z d z} \\ {\quad C: \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\ln t \mathbf{k} \quad(1 \leq t \leq e)}\end{array} $$

Step-by-Step Solution

Answer

Use a CAS to evaluate both integrals after setup.

Step 1: Define the integral and parametrize curve C (a)

For part (a), the line integral to evaluate is \( \int_C x^7 y^3 \, ds \). The curve \( C \) is given by the parametric equations \( x = \cos^3 t \) and \( y = \sin^3 t \) with \( 0 \leq t \leq \frac{\pi}{2} \). Use these equations to substitute \( x \) and \( y \) in the integral.

Step 2: Determine the differential arc length (ds)

For the curve parametrized as \( x = \cos^3 t \) and \( y = \sin^3 t \), calculate \( ds \) using \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt \). First, compute \( dx/dt = -3\cos^2 t \sin t \) and \( dy/dt = 3\sin^2 t \cos t \). Then calculate \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} = \sqrt{9\cos^4 t \sin^2 t + 9\sin^4 t \cos^2 t} \, dt = 3\cos t \sin t \, dt \).

Step 3: Simplify and evaluate the integral (a)

Substitute \( x = \cos^3 t \) and \( y = \sin^3 t \) into the integral: \( \int_0^{\pi/2} (\cos^3 t)^7 (\sin^3 t)^3 \, (3 \cos t \sin t) \, dt \). This simplifies to \( 3 \int_0^{\pi/2} \cos^{22} t \sin^7 t \, dt \). Use a CAS to evaluate this integral, yielding the result.

Step 4: Define the integral and parametrize curve C (b)

For part (b), we need to evaluate \( \int_C x^5 z \, dx + 7y \, dy + y^2 z \, dz \). The curve \( C \) is given by the parametric vector function \( \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + \ln t\mathbf{k} \), with \( 1 \leq t \leq e \). Use these expressions for \( x = t \), \( y = t^2 \), and \( z = \ln t \) to substitute into the line integral.

Step 5: Calculate the derivatives and substitute into the integral (b)

Find the derivatives, \( dx = 1 \, dt \), \( dy = 2t \, dt \), and \( dz = 1/t \, dt \). Substitute these into the integral: \( \int_{1}^{e} (t^5 \ln t)(1) \, dt + (7 t^2)(2t) \, dt + (t^4 \ln t)(1/t) \, dt \). Simplify to: \( \int_{1}^{e} t^5 \ln t \, dt + 14t^3 \, dt + t^3 \ln t \, dt \).

Step 6: Evaluate the simplified integrals using CAS (b)

Use a CAS to evaluate each part: 1. \( \int_{1}^{e} t^5 \ln t \, dt \), 2. \( \int_{1}^{e} 14t^3 \, dt \),3. \( \int_{1}^{e} t^3 \ln t \, dt \).Sum these results to obtain the value of the line integral.

Key Concepts

Parametric Equations

Parametric equations are a powerful tool for describing curves in a coordinate system. They express the coordinates of points on the curve as functions of a variable, often called a parameter. This parameter is usually denoted by the letter "t". By using parametric equations, more complex curves can be studied, even if they cannot easily be described by a single algebraic equation.

• For example, consider the parametric equations used in the exercise:
  • In part (a), we have \[ x = \cos^3 t, \quad y = \sin^3 t \] These equations describe a curvy path in the plane. As the parameter \( t \) changes, the point \( (x, y) \) traces out the curve. Each position on this path is associated with a specific value of \( t \).
  • In part (b), there's a vector function given by \[ \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \ln t \mathbf{k} \] Here, \( t \) is still the parameter, but now it's describing motion in three dimensions. The vector \( \mathbf{r}(t) \) points to a position in 3D space for each value of \( t \).

Parametric equations are versatile, allowing not only mathematical ingenuity but also practical applications, such as robotics and computer graphics, where defining paths is essential.

Differential Arc Length

Differential arc length refers to a small segment of a curve, and is essential for computing the length of curves parametrized by variables. Essentially, it tells us how much the curve's path "stretches" over a tiny interval of the parameter \( t \). To find this, we use calculus to calculate an expression in terms of \( dt \). For a curve given by parametric equations \( x(t) \) and \( y(t) \), the differential arc length \( ds \) is calculated by: \[ ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \] This formula measures the speed at which the point moves along the curve. Think of \( ds \) as walking along the curve; the faster you walk in some sections, the larger \( ds \) will be for a set \( dt \).

• In part (a) of the exercise, we computed: \[ ds = 3 \cos t \sin t \, dt \] This addresses how both \( dx/dt = -3\cos^2 t \sin t \) and \( dy/dt = 3\sin^2 t \cos t \) contribute to the curve's length, by blending their contributions into one expression for the speed along the curve.

Understanding differential arc lengths is key when performing tasks like arc length calculations or when working with vector fields. It also broadens one’s ability to analyze complex curves effectively.

Vector Function

A vector function is a function where the input is one or more variables and the output is a vector. It's a way of presenting a curve in multi-dimensional spaces, combining separate components into a single entity. Each component depends on a shared parameter, and they collectively describe the trajectory of points in space.

• In part (b) of the original exercise, the vector function given is: \[ \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \ln t \mathbf{k} \]

This vector function maps the parameter \( t \) into a three-dimensional space with the basis vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). Each component \( t, t^2, \ln t \) signifies movement along the x, y, and z directions, respectively. As \( t \) transitions through its range, the output vector sweeps across a path in 3D space.By using vector functions, you can find derivatives of each component separately:

• \( \frac{d}{dt}(t) = 1 \)
• \( \frac{d}{dt}(t^2) = 2t \)
• \( \frac{d}{dt}(\ln t) = \frac{1}{t} \)

Vector functions are significant in physics and engineering, particularly when studying motion along a path, electromagnetic fields, or when designing curved surfaces. They form the bedrock for many applications, simplifying the representation and analysis of movement and fields in multiple dimensions.