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 Statement
Write an equation for \(\frac{dy}{dx}\) on the ellipse \(xy = y^2 + x^2 - 3\), showing ALL your work. Then find the horizontal tangent lines of this ellipse.
### Solution Steps
1. **Implicit Differentiation**:
To find \(\frac{dy}{dx}\), we need to differentiate both sides of the given equation with respect to \(x\).
Given: \(xy = y^2 + x^2 - 3\)
Differentiate both sides with respect to \(x\):
\[
\begin{aligned}
\frac{d}{dx}(xy) &= \frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) - \frac{d}{dx}(3) \\
x\frac{dy}{dx} + y &= 2y\frac{dy}{dx} + 2x
\end{aligned}
\]
Rearrange the terms to solve for \(\frac{dy}{dx}\):
\[
\begin{aligned}
x\frac{dy}{dx} + y &= 2y\frac{dy}{dx} + 2x \\
y - 2x &= 2y\frac{dy}{dx} - x\frac{dy}{dx} \\
y - 2x &= (2y - x) \frac{dy}{dx} \\
\frac{dy}{dx} &= \frac{y - 2x}{2y - x}
\end{aligned}
\]
2. **Finding Horizontal Tangents**:
For horizontal tangent lines, \(\frac{dy}{dx} = 0\). Setting the numerator equal to zero gives us:
\[
y - 2x = 0
\]
Solving for \(y\):
\[
y = 2x
\]
Substitute \(y = 2x\) back into the original equation to find the corresponding \(x\) values:
\[
x(2x) = (2x)^2 + x^2 - 3 \\
2x^2 = 4x^2 + x^2 - 3 \\
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3d3ac4a-874a-47f2-996d-ede45746d6c1%2F46c8a82d-de0f-4117-9ed5-0ec9475e5949%2Fkdwgkjd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Write an equation for \(\frac{dy}{dx}\) on the ellipse \(xy = y^2 + x^2 - 3\), showing ALL your work. Then find the horizontal tangent lines of this ellipse.
### Solution Steps
1. **Implicit Differentiation**:
To find \(\frac{dy}{dx}\), we need to differentiate both sides of the given equation with respect to \(x\).
Given: \(xy = y^2 + x^2 - 3\)
Differentiate both sides with respect to \(x\):
\[
\begin{aligned}
\frac{d}{dx}(xy) &= \frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) - \frac{d}{dx}(3) \\
x\frac{dy}{dx} + y &= 2y\frac{dy}{dx} + 2x
\end{aligned}
\]
Rearrange the terms to solve for \(\frac{dy}{dx}\):
\[
\begin{aligned}
x\frac{dy}{dx} + y &= 2y\frac{dy}{dx} + 2x \\
y - 2x &= 2y\frac{dy}{dx} - x\frac{dy}{dx} \\
y - 2x &= (2y - x) \frac{dy}{dx} \\
\frac{dy}{dx} &= \frac{y - 2x}{2y - x}
\end{aligned}
\]
2. **Finding Horizontal Tangents**:
For horizontal tangent lines, \(\frac{dy}{dx} = 0\). Setting the numerator equal to zero gives us:
\[
y - 2x = 0
\]
Solving for \(y\):
\[
y = 2x
\]
Substitute \(y = 2x\) back into the original equation to find the corresponding \(x\) values:
\[
x(2x) = (2x)^2 + x^2 - 3 \\
2x^2 = 4x^2 + x^2 - 3 \\
2
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