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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The function s is called the arc length function for y=f(x)
![## Calculus Problem: Derived Equations from Given Hyperbola
### Problem Statement:
Given the equation of a hyperbola:
\[ 4x^2 - y^2 = 64 \]
Show that:
\[ ds^2 = \frac{4}{y^2}(5x^2 - 16)dx^2 \]
### Detailed Solution:
1. **Starting Equation**:
\[ 4x^2 - y^2 = 64 \]
2. **Implicit Differentiation**: Differentiate both sides of the equation with respect to \( x \) to find the relationship between \( \frac{dy}{dx} \).
Given \( 4x^2 - y^2 = 64 \),
take the derivative with respect to \( x \):
\[ \frac{d}{dx}(4x^2 - y^2) = \frac{d}{dx}(64) \]
\[ 8x - 2y \frac{dy}{dx} = 0 \]
3. **Solve for \( \frac{dy}{dx} \)**:
Rearranging the terms:
\[ 8x = 2y \frac{dy}{dx} \]
\[ \frac{dy}{dx} = \frac{8x}{2y} \]
\[ \frac{dy}{dx} = \frac{4x}{y} \]
4. **Square \( \frac{dy}{dx} \) to find \( \left(\frac{dy}{dx}\right)^2 \)**:
\[ \left(\frac{dy}{dx}\right)^2 = \left(\frac{4x}{y}\right)^2 \]
\[ \left(\frac{dy}{dx}\right)^2 = \frac{16x^2}{y^2} \]
5. **Expression for \( ds^2 \)**:
By definition of the differential arc length \( ds \) in terms of \( dx \) and \( dy \):
\[ ds^2 = dx^2 + dy^2 \]
But,
\[ dy = \frac{dy}{dx} dx \]
So,
\[ ds^2 = dx^2 + \left(\frac{dy}{dx}dx\right)^2 \]
Substituting](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a617c73-16c8-405c-b9c4-72dd4741d892%2Fe1484600-8627-4df6-9f94-31c04e6d2c75%2Fvbchbaj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Calculus Problem: Derived Equations from Given Hyperbola
### Problem Statement:
Given the equation of a hyperbola:
\[ 4x^2 - y^2 = 64 \]
Show that:
\[ ds^2 = \frac{4}{y^2}(5x^2 - 16)dx^2 \]
### Detailed Solution:
1. **Starting Equation**:
\[ 4x^2 - y^2 = 64 \]
2. **Implicit Differentiation**: Differentiate both sides of the equation with respect to \( x \) to find the relationship between \( \frac{dy}{dx} \).
Given \( 4x^2 - y^2 = 64 \),
take the derivative with respect to \( x \):
\[ \frac{d}{dx}(4x^2 - y^2) = \frac{d}{dx}(64) \]
\[ 8x - 2y \frac{dy}{dx} = 0 \]
3. **Solve for \( \frac{dy}{dx} \)**:
Rearranging the terms:
\[ 8x = 2y \frac{dy}{dx} \]
\[ \frac{dy}{dx} = \frac{8x}{2y} \]
\[ \frac{dy}{dx} = \frac{4x}{y} \]
4. **Square \( \frac{dy}{dx} \) to find \( \left(\frac{dy}{dx}\right)^2 \)**:
\[ \left(\frac{dy}{dx}\right)^2 = \left(\frac{4x}{y}\right)^2 \]
\[ \left(\frac{dy}{dx}\right)^2 = \frac{16x^2}{y^2} \]
5. **Expression for \( ds^2 \)**:
By definition of the differential arc length \( ds \) in terms of \( dx \) and \( dy \):
\[ ds^2 = dx^2 + dy^2 \]
But,
\[ dy = \frac{dy}{dx} dx \]
So,
\[ ds^2 = dx^2 + \left(\frac{dy}{dx}dx\right)^2 \]
Substituting
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