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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Find dr/dtheta for theta^(-7/9) + r^(-7/9) = -5
![**Calculus Problem on Derivatives**
**Problem Statement:**
13. Find \(\frac{dr}{d\theta}\) for \(\theta^{3/4} + r^{4/3} = -5\).
**Solution:**
Given the equation:
\[ \theta^{3/4} + r^{4/3} = -5, \]
we are required to find the derivative of \( r \) with respect to \( \theta \), i.e., \(\frac{dr}{d\theta}\).
The function involves implicit differentiation due to the mixed terms of \( \theta \) and \( r \).
### Steps:
1. Differentiate the equation with respect to \( \theta \).
\[ \frac{d}{d\theta}\left(\theta^{3/4} + r^{4/3}\right) = \frac{d}{d\theta}(-5) \]
2. Apply the chain rule to each term.
\[ \frac{d}{d\theta}(\theta^{3/4}) + \frac{d}{d\theta}(r^{4/3}) = 0 \]
\[ \frac{3}{4}\theta^{-1/4} + \frac{4}{3}r^{1/3} \cdot \frac{dr}{d\theta} = 0 \]
3. Solve for \(\frac{dr}{d\theta}\):
\[ \frac{4}{3}r^{1/3} \cdot \frac{dr}{d\theta} = -\frac{3}{4}\theta^{-1/4} \]
\[ \frac{dr}{d\theta} = -\frac{3}{4} \cdot \frac{4}{3} \cdot r^{-1/3} \cdot \theta^{-1/4} \]
\[ \frac{dr}{d\theta} = -\frac{\theta^{-1/4}}{r^{1/3}} \]
\[ \frac{dr}{d\theta} = -\frac{1}{\theta^{1/4} r^{1/3}} \]
Therefore, the derivative of \( r \) with respect to \( \theta \) is:
\[ \frac{dr}{d\theta} = -\frac{1}{\theta^{1/4} r^{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d31b151-adde-452e-9f9e-014f79fed582%2Fd4610ee6-456d-420f-9c3e-3da50c8c11c3%2Fthk7lhf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculus Problem on Derivatives**
**Problem Statement:**
13. Find \(\frac{dr}{d\theta}\) for \(\theta^{3/4} + r^{4/3} = -5\).
**Solution:**
Given the equation:
\[ \theta^{3/4} + r^{4/3} = -5, \]
we are required to find the derivative of \( r \) with respect to \( \theta \), i.e., \(\frac{dr}{d\theta}\).
The function involves implicit differentiation due to the mixed terms of \( \theta \) and \( r \).
### Steps:
1. Differentiate the equation with respect to \( \theta \).
\[ \frac{d}{d\theta}\left(\theta^{3/4} + r^{4/3}\right) = \frac{d}{d\theta}(-5) \]
2. Apply the chain rule to each term.
\[ \frac{d}{d\theta}(\theta^{3/4}) + \frac{d}{d\theta}(r^{4/3}) = 0 \]
\[ \frac{3}{4}\theta^{-1/4} + \frac{4}{3}r^{1/3} \cdot \frac{dr}{d\theta} = 0 \]
3. Solve for \(\frac{dr}{d\theta}\):
\[ \frac{4}{3}r^{1/3} \cdot \frac{dr}{d\theta} = -\frac{3}{4}\theta^{-1/4} \]
\[ \frac{dr}{d\theta} = -\frac{3}{4} \cdot \frac{4}{3} \cdot r^{-1/3} \cdot \theta^{-1/4} \]
\[ \frac{dr}{d\theta} = -\frac{\theta^{-1/4}}{r^{1/3}} \]
\[ \frac{dr}{d\theta} = -\frac{1}{\theta^{1/4} r^{1/3}} \]
Therefore, the derivative of \( r \) with respect to \( \theta \) is:
\[ \frac{dr}{d\theta} = -\frac{1}{\theta^{1/4} r^{
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