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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![**Assume x and y are functions of t. Evaluate dy/dt.**
12) \( x^3 + y^3 = 9 \); \( \frac{dx}{dt} = 4 \), \( x = 1 \)
13) \( xy + x = 12 \); \( \frac{dx}{dt} = -3 \), \( x = 2 \), \( y = 5 \)
### Solution Steps:
To find \( \frac{dy}{dt} \), we'll use implicit differentiation with respect to \( t \). This means we differentiate each equation while treating \( x \) and \( y \) as functions of \( t \).
#### For equation 12:
1. Start with:
\( x^3 + y^3 = 9 \).
2. Differentiate both sides with respect to \( t \):
\[
\frac{d}{dt}(x^3 + y^3) = \frac{d}{dt}(9) \implies 3x^2 \frac{dx}{dt} + 3y^2 \frac{dy}{dt} = 0.
\]
3. Plug in the known values:
\[
3(1)^2 (4) + 3y^2 \frac{dy}{dt} = 0 \implies 12 + 3y^2 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{12}{3y^2} \implies \frac{dy}{dt} = -\frac{4}{y^2}.
\]
4. Solve for \( \frac{dy}{dt} \):
To find \( y \), substitute \( x = 1 \) in the original equation:
\[ 1^3 + y^3 = 9 \implies 1 + y^3 = 9 \implies y^3 = 8 \implies y = 2. \]
Now, substitute \( y = 2 \) back into the differentiated equation:
\[
\frac{dy}{dt} = -\frac{4}{2^2} = -\frac{4}{4} = -1.
\]
So, \( \frac{dy}{dt} \) is -1 for equation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b6dc6c4-2828-4529-a1c7-3495f006d86b%2F1937ba1d-8313-4479-acb0-fee94b6b15fe%2Fktp129g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Assume x and y are functions of t. Evaluate dy/dt.**
12) \( x^3 + y^3 = 9 \); \( \frac{dx}{dt} = 4 \), \( x = 1 \)
13) \( xy + x = 12 \); \( \frac{dx}{dt} = -3 \), \( x = 2 \), \( y = 5 \)
### Solution Steps:
To find \( \frac{dy}{dt} \), we'll use implicit differentiation with respect to \( t \). This means we differentiate each equation while treating \( x \) and \( y \) as functions of \( t \).
#### For equation 12:
1. Start with:
\( x^3 + y^3 = 9 \).
2. Differentiate both sides with respect to \( t \):
\[
\frac{d}{dt}(x^3 + y^3) = \frac{d}{dt}(9) \implies 3x^2 \frac{dx}{dt} + 3y^2 \frac{dy}{dt} = 0.
\]
3. Plug in the known values:
\[
3(1)^2 (4) + 3y^2 \frac{dy}{dt} = 0 \implies 12 + 3y^2 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{12}{3y^2} \implies \frac{dy}{dt} = -\frac{4}{y^2}.
\]
4. Solve for \( \frac{dy}{dt} \):
To find \( y \), substitute \( x = 1 \) in the original equation:
\[ 1^3 + y^3 = 9 \implies 1 + y^3 = 9 \implies y^3 = 8 \implies y = 2. \]
Now, substitute \( y = 2 \) back into the differentiated equation:
\[
\frac{dy}{dt} = -\frac{4}{2^2} = -\frac{4}{4} = -1.
\]
So, \( \frac{dy}{dt} \) is -1 for equation
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