Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
1.1.6
![**Problem Statement**
Find the arc length (s) of the curve \(\vec{r}(t) = \langle 4t + 2, -5t + 3, -2t - 5 \rangle\) for \(-3 \leq t \leq 2\).
**Solution**
To find the arc length \(s\), evaluate the integral of the magnitude of the derivative of \(\vec{r}(t)\) with respect to \(t\) over the given interval. The formula for the arc length of a vector function is:
\[ s = \int_{a}^{b} \|\vec{r}'(t)\| \, dt \]
**Steps**
1. Calculate \(\vec{r}'(t)\) by differentiating each component:
\[
\vec{r}'(t) = \left\langle \frac{d}{dt}(4t + 2), \frac{d}{dt}(-5t + 3), \frac{d}{dt}(-2t - 5) \right\rangle = \langle 4, -5, -2 \rangle
\]
2. Find the magnitude \(\|\vec{r}'(t)\|\):
\[
\|\vec{r}'(t)\| = \sqrt{4^2 + (-5)^2 + (-2)^2} = \sqrt{16 + 25 + 4} = \sqrt{45}
\]
3. Integrate over the interval \([-3, 2]\):
\[
s = \int_{-3}^{2} \sqrt{45} \, dt = \sqrt{45} \cdot \int_{-3}^{2} \, dt = \sqrt{45} \cdot [t]_{-3}^{2}
\]
4. Evaluate the integral:
\[
s = \sqrt{45} \cdot (2 - (-3)) = \sqrt{45} \cdot 5
\]
5. Simplify the expression:
\[
s = 5\sqrt{45} = 15\sqrt{5}
\]
Therefore, the arc length \(s\) is \(15\sqrt{5}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b463452-b960-4b00-bce0-3a69d9f467e2%2Fe200325b-a0a8-40f2-8f6f-9b5afb4777e0%2Fr9epmu_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find the arc length (s) of the curve \(\vec{r}(t) = \langle 4t + 2, -5t + 3, -2t - 5 \rangle\) for \(-3 \leq t \leq 2\).
**Solution**
To find the arc length \(s\), evaluate the integral of the magnitude of the derivative of \(\vec{r}(t)\) with respect to \(t\) over the given interval. The formula for the arc length of a vector function is:
\[ s = \int_{a}^{b} \|\vec{r}'(t)\| \, dt \]
**Steps**
1. Calculate \(\vec{r}'(t)\) by differentiating each component:
\[
\vec{r}'(t) = \left\langle \frac{d}{dt}(4t + 2), \frac{d}{dt}(-5t + 3), \frac{d}{dt}(-2t - 5) \right\rangle = \langle 4, -5, -2 \rangle
\]
2. Find the magnitude \(\|\vec{r}'(t)\|\):
\[
\|\vec{r}'(t)\| = \sqrt{4^2 + (-5)^2 + (-2)^2} = \sqrt{16 + 25 + 4} = \sqrt{45}
\]
3. Integrate over the interval \([-3, 2]\):
\[
s = \int_{-3}^{2} \sqrt{45} \, dt = \sqrt{45} \cdot \int_{-3}^{2} \, dt = \sqrt{45} \cdot [t]_{-3}^{2}
\]
4. Evaluate the integral:
\[
s = \sqrt{45} \cdot (2 - (-3)) = \sqrt{45} \cdot 5
\]
5. Simplify the expression:
\[
s = 5\sqrt{45} = 15\sqrt{5}
\]
Therefore, the arc length \(s\) is \(15\sqrt{5}\).
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