Problem 7
Question
A 20-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the level pavement directly away from the building at 1 foot per second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
Step-by-Step Solution
Verified Answer
The top of the ladder moves down at \(-\frac{\sqrt{15}}{15}\) feet per second.
1Step 1: Understand the Problem
We have a 20-foot ladder leaning against a wall. The bottom of the ladder moves away from the wall, and we need to find how fast the top of the ladder moves down the wall when the bottom is 5 feet away from the wall.
2Step 2: Set Up the Variables
Let's denote the distance from the wall to the bottom of the ladder as \( x(t) \), and the height of the top of the ladder on the wall as \( y(t) \). Since it's a right triangle, we have the relationship: \( x^2 + y^2 = 20^2 \). We also know \( \frac{dx}{dt} = 1 \text{ ft/s} \) as the bottom moves away.
3Step 3: Differentiate the Equation
Differentiate the equation \( x^2 + y^2 = 400 \) with respect to time \( t \): \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \). Simplifying gives: \( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \).
4Step 4: Substitute Known Values
When \( x = 5 \), substitute \( \frac{dx}{dt} = 1 \), \( x = 5 \), and find \( y \) using \( x^2 + y^2 = 400 \). We get \( 25 + y^2 = 400 \), so \( y = \sqrt{375} = 5\sqrt{15} \). Now substitute in the differentiated equation: \( 5 \times 1 + 5\sqrt{15} \frac{dy}{dt} = 0 \).
5Step 5: Solve for \( \frac{dy}{dt} \)
Re-arrange to solve for \( \frac{dy}{dt} \): \( 5\sqrt{15} \frac{dy}{dt} = -5 \). This simplifies to \( \frac{dy}{dt} = -\frac{5}{5\sqrt{15}} = -\frac{1}{\sqrt{15}} \). Simplify further: \( \frac{dy}{dt} = -\frac{\sqrt{15}}{15} \).
Key Concepts
Calculus DifferentiationRight Triangle GeometryDerivatives with Respect to Time
Calculus Differentiation
Calculus differentiation is a fundamental concept in understanding how quantities change with respect to one another. In this problem, we're dealing with a changing triangle, where the ladder's position shifts over time. Differentiation helps us find rates of change in various components of this system.
To differentiate the relationship between the distances formed by the ladder, we looked at the equation for the right triangle: \(x^2 + y^2 = 400\). This describes how the distance from the wall (\(x\)) and the height the ladder reaches on the building (\(y\)) relate. Differentiating this equation with respect to time \(t\) helps us see how these distances change as time progresses.
Differentiating gives us: \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\). By having this, we can input the known rates of change and solve for the unknown, helping us figure out how fast the top of the ladder moves down the wall.
To differentiate the relationship between the distances formed by the ladder, we looked at the equation for the right triangle: \(x^2 + y^2 = 400\). This describes how the distance from the wall (\(x\)) and the height the ladder reaches on the building (\(y\)) relate. Differentiating this equation with respect to time \(t\) helps us see how these distances change as time progresses.
Differentiating gives us: \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\). By having this, we can input the known rates of change and solve for the unknown, helping us figure out how fast the top of the ladder moves down the wall.
Right Triangle Geometry
In this exercise, right triangle geometry plays a crucial role in understanding the movement of the ladder.
The ladder, the ground, and the building form a right triangle with the ladder as the hypotenuse. By using the Pythagorean theorem, \(x^2 + y^2 = 400\), we dynamically model the triangle's changing sides as the ladder shifts.
The ladder, the ground, and the building form a right triangle with the ladder as the hypotenuse. By using the Pythagorean theorem, \(x^2 + y^2 = 400\), we dynamically model the triangle's changing sides as the ladder shifts.
- \(x\) is the distance from the wall to the bottom of the ladder.
- \(y\) is the height of the ladder on the wall.
- The ladder's length is constant at 20 feet.
Derivatives with Respect to Time
This concept involves understanding rates of change in a dynamic system by examining derivatives with respect to time. In related rates problems, such as the ladder scenario, different parts of a geometric setup change over time.
In our context, the derivative \(\frac{dx}{dt} = 1 \text{ ft/s}\) is the rate at which the base of the ladder moves away from the wall. Using calculus, we determine \(\frac{dy}{dt}\), the rate at which the ladder falls along the wall.
We accomplish this by plugging known values into our differentiated equation: \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\). Given \(x = 5\), \(y = 5\sqrt{15}\), and \(\frac{dx}{dt}\), we solve for \(\frac{dy}{dt}\), which results in \(\frac{dy}{dt} = -\frac{\sqrt{15}}{15}\).
Understanding these rates provides insight into how quickly real-world phenomena change, emphasizing how calculus can model dynamic systems efficiently.
In our context, the derivative \(\frac{dx}{dt} = 1 \text{ ft/s}\) is the rate at which the base of the ladder moves away from the wall. Using calculus, we determine \(\frac{dy}{dt}\), the rate at which the ladder falls along the wall.
We accomplish this by plugging known values into our differentiated equation: \(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\). Given \(x = 5\), \(y = 5\sqrt{15}\), and \(\frac{dx}{dt}\), we solve for \(\frac{dy}{dt}\), which results in \(\frac{dy}{dt} = -\frac{\sqrt{15}}{15}\).
Understanding these rates provides insight into how quickly real-world phenomena change, emphasizing how calculus can model dynamic systems efficiently.
Other exercises in this chapter
Problem 7
$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{\pi}{x} $$
View solution Problem 7
In Problems 1-20, find \(D_{x} y\). $$ y=(3 x-2)^{2}\left(3-x^{2}\right)^{2} $$
View solution Problem 8
In Problems 1-18, find \(D_{x} y\). $$ y=\cot x=\frac{\cos x}{\sin x} $$
View solution Problem 8
In Problems \(1-8\), find \(d^{3} y / d x^{3}\). $$ y=\frac{3 x}{1-x} $$
View solution