Prompt

Using R, provide the solution for any exercise in either Chapter 4 or Chapter 7 of the calculus textbook. If you are unsure of your solution, post your concerns.

Section 4.2: Exercise 9

A 24 ft. ladder is leaning against a house while the base is pulled away at a constant rate of 1 ft/s.

At what rate is the top of the ladder sliding down the side of the house when the base is:

Equation to relate the quantities of interest

Pythagorean theorem

\(x^2 + y^2 = 24^2\)

Take the derivative with respect to time of both sides of the equation. Remember the Chain Rule

\(\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(24^2)\)

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

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

\(\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}\)

A: 1 foot from the house?

# Using : https://www.matheno.com/calculus-1/related-rates/how-fast-is-the-ladders-top-sliding/
# x: horizontal distance from the wall
x <- 1
# y: vertical distance from the floor to the ladder's top
y <- sqrt(24^2 - x^2)

# dx/dt is 1 ft/s
dx_dt <- 1

# Solving for dy/dt
dy_dt <- -1 * (x/y) * dx_dt

Result: -0.0417029 ft/s.

Note: Value is negative because the y value is decreasing as ladder slides down the wall.

B: 10 feet from the house?

x <- 10
y <- sqrt(24^2 - x^2)

dy_dt <- -1 * (x/y) * dx_dt

Result: -0.4583492 ft/s.

C: 23 feet from the house?

x <- 23
y <- sqrt(24^2 - x^2)

dy_dt <- -1 * (x/y) * dx_dt

Result: -3.3548948 ft/s.

D: 24 feet from the house?

x <- 24
y <- sqrt(24^2 - x^2)

dy_dt <- -1 * (x/y) * dx_dt

Result: 0 ft/s. At this point, the ladder is completely flat on the ground, so there is no rate in which the top of the ladder is sliding down the side of the house.