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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![**Problem 47**: Given the functions \( z = x^2 y^3 \), \( x = r \cos(\theta) \), and \( y = r \sin(\theta) \), find the partial derivatives \(\frac{\partial z}{\partial r}\) and \(\frac{\partial z}{\partial \theta}\) when \( r = 2 \) and \( \theta = \frac{\pi}{6} \).
To find \(\frac{\partial z}{\partial r}\) and \(\frac{\partial z}{\partial \theta}\), we need to perform the following steps:
1. **Substitute \( x \) and \( y \) in terms of \( r \) and \( \theta \)**:
\[
z = (r \cos(\theta))^2 (r \sin(\theta))^3
\]
Simplify the expression:
\[
z = r^2 \cos^2(\theta) \times r^3 \sin^3(\theta)
\]
Combine the terms involving \( r \):
\[
z = r^5 \cos^2(\theta) \sin^3(\theta)
\]
2. **Differentiate \( z \) with respect to \( r \)**:
\[
\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} (r^5 \cos^2(\theta) \sin^3(\theta))
\]
Since \(\cos^2(\theta) \sin^3(\theta)\) is independent of \( r \), we treat it as a constant:
\[
\frac{\partial z}{\partial r} = 5r^4 \cos^2(\theta) \sin^3(\theta)
\]
3. **Evaluate \(\frac{\partial z}{\partial r}\) at \( r = 2 \) and \( \theta = \frac{\pi}{6}\)**:
- \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
- \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
\[
\frac{\partial z}{\partial r} \bigg|_{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18107b7c-8d8e-412b-a2cd-6addf4186be2%2F58d8d1f3-ce88-4ba1-a41d-ded3db1809bc%2Fosbuhe9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 47**: Given the functions \( z = x^2 y^3 \), \( x = r \cos(\theta) \), and \( y = r \sin(\theta) \), find the partial derivatives \(\frac{\partial z}{\partial r}\) and \(\frac{\partial z}{\partial \theta}\) when \( r = 2 \) and \( \theta = \frac{\pi}{6} \).
To find \(\frac{\partial z}{\partial r}\) and \(\frac{\partial z}{\partial \theta}\), we need to perform the following steps:
1. **Substitute \( x \) and \( y \) in terms of \( r \) and \( \theta \)**:
\[
z = (r \cos(\theta))^2 (r \sin(\theta))^3
\]
Simplify the expression:
\[
z = r^2 \cos^2(\theta) \times r^3 \sin^3(\theta)
\]
Combine the terms involving \( r \):
\[
z = r^5 \cos^2(\theta) \sin^3(\theta)
\]
2. **Differentiate \( z \) with respect to \( r \)**:
\[
\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} (r^5 \cos^2(\theta) \sin^3(\theta))
\]
Since \(\cos^2(\theta) \sin^3(\theta)\) is independent of \( r \), we treat it as a constant:
\[
\frac{\partial z}{\partial r} = 5r^4 \cos^2(\theta) \sin^3(\theta)
\]
3. **Evaluate \(\frac{\partial z}{\partial r}\) at \( r = 2 \) and \( \theta = \frac{\pi}{6}\)**:
- \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
- \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
\[
\frac{\partial z}{\partial r} \bigg|_{
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