Chapter 15: Problem 28
Use a CAS to evaluate the line integrals along the given curves. $$ \begin{array}{l}{\text { (a) } \int_{C} x^{7} y^{3} d s} \\ {\quad C: x=\cos ^{3} t, \quad y=\sin ^{3} t \quad(0 \leq t \leq \pi / 2)} \\ {\text { (b) } \int_{C} x^{5} z d x+7 y d y+y^{2} z d z} \\ {\quad C: \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\ln t \mathbf{k} \quad(1 \leq t \leq e)}\end{array} $$
Short Answer
Step by step solution
Define the integral and parametrize curve C (a)
Determine the differential arc length (ds)
Simplify and evaluate the integral (a)
Define the integral and parametrize curve C (b)
Calculate the derivatives and substitute into the integral (b)
Evaluate the simplified integrals using CAS (b)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
- 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 \( t \).
Differential Arc Length
- 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.
Vector Function
- In part (b) of the original exercise, the vector function given is: \[ \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \ln t \mathbf{k} \]
- \( \frac{d}{dt}(t) = 1 \)
- \( \frac{d}{dt}(t^2) = 2t \)
- \( \frac{d}{dt}(\ln t) = \frac{1}{t} \)