Problem 69 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

A 10 -foot-long ladder leans against the side of a wall. The top of the ladder is sliding down the wall at the rate of \(3 \mathrm{ft} / \mathrm{sec}\). How fast is the foot of the ladder moving away from the building when the top is 6 feet above the ground?

Step-by-Step Solution

Answer

The foot of the ladder is moving away from the wall at ft/sec.

Step 1 - Identify and define variables

Let’s define the variables. Let’s assume the distance from the foot of the ladder to the wall is denoted by x (in feet), and the distance from the top of the ladder to the ground is denoted by y (in feet).

Step 2 - Use Pythagorean Theorem

Since the ladder forms a right triangle with the wall and ground, we use the Pythagorean theorem:

Step 3 - Differentiate with respect to time

Differentiate both sides of the equation with respect to time t, using implicit differentiation:

Step 4 - Plug in known values

Given values are: the length of the ladder (constant) = 10 feet, rate at which the top is sliding down,

Step 5 - Solve for the unknown

Substitute the values into the differentiated equation:

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function implicitly given by an equation. In our given problem, we have an equation defined by the Pythagorean theorem. Here's how implicit differentiation is used:

We start with the Pythagorean relationship: \[ x^2 + y^2 = 100 \] differentiating both sides with respect to time t, we apply the chain rule: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] This equation tells us how the rates of change of x and y (\(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)), which are dependent on time, relate to each other. Let's use this understanding to find the solution to our exercise.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, especially helpful in problems involving right triangles. The theorem states:

• In a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

For our problem, the ladder, the wall, and the ground form a right triangle. The theorem can be expressed as: \[ x^2 + y^2 = 10^2 = 100 \] Here, x is the distance from the foot of the ladder to the wall and y is the distance from the top of the ladder to the ground. This equation will help us relate the movement of the ladder’s foot and top.

Right Triangle

In our ladder problem, the right triangle is key to understanding the relationship between the ladder, the wall, and the ground. Remember, a right triangle has one angle of 90 degrees. Here is a breakdown:

• The ladder acts as the hypotenuse, the longest side.
• The distance from the wall to the ladder's foot forms one leg (x).
• The distance from the ground to the ladder's top forms the other leg (y).

Understanding this setup helps us use the Pythagorean theorem directly and simplifies the process of setting up the differential equations correctly.

Differential Equations

Differential equations involve functions and their derivatives. In this problem, we deal with the rates of change of distances in the triangle formed by the ladder. The differential equation derived from the Pythagorean theorem looks like this after implicit differentiation:

• \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\)

Given that the rate at which the top of the ladder slides down (\(\frac{dy}{dt}\)) is -3 ft/sec (negative because y decreases), and y = 6 feet at the instant we are interested in, we substitute to find \(\frac{dx}{dt}\):

Solving \[ 2x \frac{dx}{dt} + 2(6)(-3) = 0 \] gives us how fast the foot of the ladder is moving away from the wall. This shows how essential differential equations are in relating different rates of changes in related rates problems.