A 14-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall? The top of the ladder is moving down at a rate of feet/second when the foot of the ladder is 5 feet from the wall. (Round to the nearest thousandth as needed.)
A 14-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall? The top of the ladder is moving down at a rate of feet/second when the foot of the ladder is 5 feet from the wall. (Round to the nearest thousandth as needed.)
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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Round to the hundredths
![### Related Rates Problem: Ladder Sliding Down
#### Problem Statement:
A 14-foot ladder is leaning against a building. The bottom of the ladder is sliding along the pavement directly away from the building at a rate of 2 feet per second.
**Question**: How fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
*(Round your answer to the nearest thousandth as needed.)*
#### Method:
This problem involves the concept of related rates in calculus. You will need to use the Pythagorean theorem to relate the distances and differentiate to find the rate at which the top of the ladder is moving down.
Here’s the step-by-step method to solve this problem:
1. **Define Variables**:
- Let \( x \) be the distance from the foot of the ladder to the wall.
- Let \( y \) be the height of the ladder from the ground to the top of the ladder.
- Given the ladder length \( L \) is constant at 14 feet, the relationship between \( x \), \( y \), and \( L \) is given by:
\[
x^2 + y^2 = L^2
\]
2. **Differentiate with Respect to Time (\( t \))**:
- Differentiating both sides of the equation with respect to \( t \):
\[
\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(L^2)
\]
- Since \( L \) is constant, \( \frac{d}{dt}(L^2) = 0 \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
3. **Substitute Given Values and Solve for \( \frac{dy}{dt} \)**:
- Distance from the wall when \( x = 5 \) feet.
- Rate of change of \( x \) is \( \frac{dx}{dt} = 2 \) feet per second.
- Using the Pythagorean theorem:
\[
y = \sqrt{L^2 - x^2} = \sqrt{14^2 - 5^2} = \sqrt{196 - 25} = \sqrt{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F02b4a1f4-9e35-42db-aa2a-674bb26df94b%2Fbed411da-d95f-43f5-99f1-aa16dec5e292%2F4wco34f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Related Rates Problem: Ladder Sliding Down
#### Problem Statement:
A 14-foot ladder is leaning against a building. The bottom of the ladder is sliding along the pavement directly away from the building at a rate of 2 feet per second.
**Question**: How fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
*(Round your answer to the nearest thousandth as needed.)*
#### Method:
This problem involves the concept of related rates in calculus. You will need to use the Pythagorean theorem to relate the distances and differentiate to find the rate at which the top of the ladder is moving down.
Here’s the step-by-step method to solve this problem:
1. **Define Variables**:
- Let \( x \) be the distance from the foot of the ladder to the wall.
- Let \( y \) be the height of the ladder from the ground to the top of the ladder.
- Given the ladder length \( L \) is constant at 14 feet, the relationship between \( x \), \( y \), and \( L \) is given by:
\[
x^2 + y^2 = L^2
\]
2. **Differentiate with Respect to Time (\( t \))**:
- Differentiating both sides of the equation with respect to \( t \):
\[
\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(L^2)
\]
- Since \( L \) is constant, \( \frac{d}{dt}(L^2) = 0 \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
3. **Substitute Given Values and Solve for \( \frac{dy}{dt} \)**:
- Distance from the wall when \( x = 5 \) feet.
- Rate of change of \( x \) is \( \frac{dx}{dt} = 2 \) feet per second.
- Using the Pythagorean theorem:
\[
y = \sqrt{L^2 - x^2} = \sqrt{14^2 - 5^2} = \sqrt{196 - 25} = \sqrt{
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