The top end of a ladder slides down a vertical wall as its bottom end slides across a horizontal floor. At a certain instant, the speed of the top end is k times the speed of the bottom end. Find the acute angle (ie. between 0 and 1/2) that the ladder makes with the ground at this instant.
The top end of a ladder slides down a vertical wall as its bottom end slides across a horizontal floor. At a certain instant, the speed of the top end is k times the speed of the bottom end. Find the acute angle (ie. between 0 and 1/2) that the ladder makes with the ground at this instant.
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:**
The top end of a ladder slides down a vertical wall as its bottom end slides across a horizontal floor. At a certain instant, the speed of the top end is \( k \) times the speed of the bottom end. Find the acute angle (i.e., between \( 0 \) and \( \frac{\pi}{2} \)) that the ladder makes with the ground at this instant.
**Detailed Explanation:**
This problem involves understanding the relationship between the rates of change of the top and bottom ends of a sliding ladder and finding a specific angle at which this relationship holds true. We are given that the speed of the top end is \( k \) times the speed of the bottom end and asked to determine the acute angle that the ladder makes with the horizontal floor at this specific moment.
This scenario can be analyzed with the principles of related rates in calculus. Suppose the length of the ladder is constant and denoted as \( L \). Let \( x(t) \) be the distance of the bottom end of the ladder from the wall at time \( t \), and let \( y(t) \) be the distance of the top end of the ladder from the ground at time \( t \). By the Pythagorean theorem,
\[
x(t)^2 + y(t)^2 = L^2
\]
Differentiating both sides with respect to time \( t \), we get
\[
2x(t) \frac{dx(t)}{dt} + 2y(t) \frac{dy(t)}{dt} = 0
\]
Simplifying, we find
\[
x(t) \frac{dx(t)}{dt} + y(t) \frac{dy(t)}{dt} = 0
\]
Given that the speed of the top end is \( k \) times the speed of the bottom end, we have
\[
\left| \frac{dy(t)}{dt} \right| = k \left| \frac{dx(t)}{dt} \right|
\]
Substituting this into our differentiated equation:
\[
x(t) \left| \frac{dx(t)}{dt} \right| + y(t) \left( k \left| \frac{dx(t)}{dt} \right| \right) = 0
\]
Solving for the ratio of \( x(t) \) and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9efe910-fb5f-41b2-a82d-eda9e45aba44%2F71e65dfb-b1f7-4350-83ba-1daa3aff4417%2Fbzqxvyf_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The top end of a ladder slides down a vertical wall as its bottom end slides across a horizontal floor. At a certain instant, the speed of the top end is \( k \) times the speed of the bottom end. Find the acute angle (i.e., between \( 0 \) and \( \frac{\pi}{2} \)) that the ladder makes with the ground at this instant.
**Detailed Explanation:**
This problem involves understanding the relationship between the rates of change of the top and bottom ends of a sliding ladder and finding a specific angle at which this relationship holds true. We are given that the speed of the top end is \( k \) times the speed of the bottom end and asked to determine the acute angle that the ladder makes with the horizontal floor at this specific moment.
This scenario can be analyzed with the principles of related rates in calculus. Suppose the length of the ladder is constant and denoted as \( L \). Let \( x(t) \) be the distance of the bottom end of the ladder from the wall at time \( t \), and let \( y(t) \) be the distance of the top end of the ladder from the ground at time \( t \). By the Pythagorean theorem,
\[
x(t)^2 + y(t)^2 = L^2
\]
Differentiating both sides with respect to time \( t \), we get
\[
2x(t) \frac{dx(t)}{dt} + 2y(t) \frac{dy(t)}{dt} = 0
\]
Simplifying, we find
\[
x(t) \frac{dx(t)}{dt} + y(t) \frac{dy(t)}{dt} = 0
\]
Given that the speed of the top end is \( k \) times the speed of the bottom end, we have
\[
\left| \frac{dy(t)}{dt} \right| = k \left| \frac{dx(t)}{dt} \right|
\]
Substituting this into our differentiated equation:
\[
x(t) \left| \frac{dx(t)}{dt} \right| + y(t) \left( k \left| \frac{dx(t)}{dt} \right| \right) = 0
\]
Solving for the ratio of \( x(t) \) and
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