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: Calculating the Derivative \(\frac{dz}{dt}\)**
Given the function \(z^2 = x^2 + y^2\), calculate \(\frac{dz}{dt}\) at the point \((x, y) = (-2, -1)\). The rates of change are given as \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\), with the condition \(z > 0\).
### Step-by-Step Solution:
1. **Differentiate with respect to \(t\):**
Start with the equation:
\[
z^2 = x^2 + y^2
\]
Differentiate both sides with respect to \(t\):
\[
\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)
\]
2. **Apply the chain rule:**
The chain rule states:
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
3. **Substitute the given values:**
- At point \((x, y) = (-2, -1)\), substitute \(x = -2\) and \(y = -1\).
- Also use \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\).
\[
2z \frac{dz}{dt} = 2(-2)(10) + 2(-1)(5)
\]
Simplify:
\[
2z \frac{dz}{dt} = -40 - 10 = -50
\]
4. **Solve for \(\frac{dz}{dt}\):**
If \(z > 0\), solve for \(\frac{dz}{dt}\):
\[
\frac{dz}{dt} = \frac{-50}{2z}
\]
5. **Find \(z\) using the equation \(z^2 = x^2 + y^2\):**
\[
z^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7dcbfcd-ec40-4419-b8e9-2a693a942e70%2F02377b3b-4904-40ac-8897-0ba77d815165%2Ff8c0dt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Calculating the Derivative \(\frac{dz}{dt}\)**
Given the function \(z^2 = x^2 + y^2\), calculate \(\frac{dz}{dt}\) at the point \((x, y) = (-2, -1)\). The rates of change are given as \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\), with the condition \(z > 0\).
### Step-by-Step Solution:
1. **Differentiate with respect to \(t\):**
Start with the equation:
\[
z^2 = x^2 + y^2
\]
Differentiate both sides with respect to \(t\):
\[
\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)
\]
2. **Apply the chain rule:**
The chain rule states:
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
3. **Substitute the given values:**
- At point \((x, y) = (-2, -1)\), substitute \(x = -2\) and \(y = -1\).
- Also use \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\).
\[
2z \frac{dz}{dt} = 2(-2)(10) + 2(-1)(5)
\]
Simplify:
\[
2z \frac{dz}{dt} = -40 - 10 = -50
\]
4. **Solve for \(\frac{dz}{dt}\):**
If \(z > 0\), solve for \(\frac{dz}{dt}\):
\[
\frac{dz}{dt} = \frac{-50}{2z}
\]
5. **Find \(z\) using the equation \(z^2 = x^2 + y^2\):**
\[
z^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5
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