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Chapter 14: Problem 31

Use a scalar line integral to find the length of the following curves. $$\mathbf{r}(t)=\left\langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4}, \frac{t}{2}\right\rangle, \text { for } 0 \leq t \leq 2$$

Short Answer

Expert verified
Answer: The length of the curve is 20.

Step by step solution

01

Compute the derivative of the vector function with respect to t

To find the derivative of $$\mathbf{r}(t)$$, we need to differentiate each component of the vector function with respect to t. We have: $$\mathbf{r'}(t) = \frac{d}{dt} \langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4}, \frac{t}{2} \rangle = \langle \frac{d(20 \sin \frac{t}{4})}{dt}, \frac{d(20 \cos \frac{t}{4})}{dt}, \frac{d(\frac{t}{2})}{dt} \rangle$$ Using the chain rule, we can find the derivative of each component: $$\mathbf{r'}(t) = \langle 20 \cos \frac{t}{4} \cdot \frac{1}{4}, -20 \sin \frac{t}{4} \cdot \frac{1}{4}, \frac{1}{2} \rangle$$
02

Find the magnitude of the derivative vector

The magnitude of the derivative vector can be found using the following formula: $$||\mathbf{r'}(t)|| = \sqrt{(\frac{1}{4} \cdot 20 \cos \frac{t}{4})^2 + (- \frac{1}{4} \cdot 20 \sin \frac{t}{4})^2 + \frac{1}{4}}$$. Now, we simplify the expression: $$||\mathbf{r'}(t)|| = \sqrt{25 (\cos^2 \frac{t}{4} + \sin^2 \frac{t}{4}) + \frac{1}{4}} = \sqrt{25 + \frac{1}{4}} = \sqrt{100}$$ Thus, we have: $$||\mathbf{r'}(t)|| = 10$$
03

Set up the scalar line integral for the curve's length

To find the length of the curve, we need to set up the scalar line integral: $$L = \int_{0}^{2} ||\mathbf{r'}(t)|| dt$$. We have already found $$||\mathbf{r'}(t)|| = 10$$, so we can substitute that value into the integral: $$L = \int_{0}^{2} 10 dt$$
04

Evaluate the integral to find the length of the curve

Now we need to evaluate the integral to find the length of the curve: $$L = \int_{0}^{2} 10 dt = 10 \int_{0}^{2} dt = 10[t]_{0}^{2} = 10(2-0) = 20$$ So, the length of the curve is 20.

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Most popular questions from this chapter

Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \cdot(\varphi \mathbf{F})=\nabla \varphi \cdot \mathbf{F}+\varphi \nabla \cdot \mathbf{F} \quad \text { (Product Rule) }$$

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