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
8 on paper please
![### Problem 8:
**Given the space curve** \( \mathbf{r}(t) = \langle \sin^2{t}, \cos^2{t}, \tan{t} \rangle \), **find an equation of the tangent line in parametric form to this curve when** \( t = \frac{\pi}{4} \).
---
In order to solve this problem, we need to:
1. **Evaluate** \( \mathbf{r}(t) \) at \( t = \frac{\pi}{4} \).
2. **Find the derivative** \( \mathbf{r}'(t) \), which gives the direction of the tangent line.
3. **Evaluate** \( \mathbf{r}'(t) \) at \( t = \frac{\pi}{4} \).
4. **Formulate the parametric equations** for the tangent line using the point from step 1 and the direction vector from step 3.
#### Step-by-Step Solution:
1. **Evaluate** \( \mathbf{r}(t) \) at \( t = \frac{\pi}{4} \):
\[
\mathbf{r}\left( \frac{\pi}{4} \right) = \left\langle \sin^2{\left( \frac{\pi}{4} \right)}, \cos^2{\left( \frac{\pi}{4} \right)}, \tan{\left( \frac{\pi}{4} \right)} \right\rangle
\]
Given \( \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \) and \( \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \), we have:
\[
\mathbf{r}\left( \frac{\pi}{4} \right) = \left\langle \left( \frac{\sqrt{2}}{2} \right)^2, \left( \frac{\sqrt{2}}{2} \right)^2, 1 \right\rangle = \left\langle \frac{1}{2}, \frac{1}{2}, 1 \right\rangle
\]
2. **Find the derivative** \( \mathbf{r}'(t) \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48922dd6-b6f8-4fac-84c6-05a7db5750f6%2F4775e4af-8311-4b15-8ab8-f8eed02207d2%2Fou5f6di_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 8:
**Given the space curve** \( \mathbf{r}(t) = \langle \sin^2{t}, \cos^2{t}, \tan{t} \rangle \), **find an equation of the tangent line in parametric form to this curve when** \( t = \frac{\pi}{4} \).
---
In order to solve this problem, we need to:
1. **Evaluate** \( \mathbf{r}(t) \) at \( t = \frac{\pi}{4} \).
2. **Find the derivative** \( \mathbf{r}'(t) \), which gives the direction of the tangent line.
3. **Evaluate** \( \mathbf{r}'(t) \) at \( t = \frac{\pi}{4} \).
4. **Formulate the parametric equations** for the tangent line using the point from step 1 and the direction vector from step 3.
#### Step-by-Step Solution:
1. **Evaluate** \( \mathbf{r}(t) \) at \( t = \frac{\pi}{4} \):
\[
\mathbf{r}\left( \frac{\pi}{4} \right) = \left\langle \sin^2{\left( \frac{\pi}{4} \right)}, \cos^2{\left( \frac{\pi}{4} \right)}, \tan{\left( \frac{\pi}{4} \right)} \right\rangle
\]
Given \( \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \) and \( \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \), we have:
\[
\mathbf{r}\left( \frac{\pi}{4} \right) = \left\langle \left( \frac{\sqrt{2}}{2} \right)^2, \left( \frac{\sqrt{2}}{2} \right)^2, 1 \right\rangle = \left\langle \frac{1}{2}, \frac{1}{2}, 1 \right\rangle
\]
2. **Find the derivative** \( \mathbf{r}'(t) \
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