Discussion 13

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 at various distances from the house:

I solved this problem entirely based on this excellent video from Professor Dave Explains - https://www.youtube.com/watch?v=j6I3EXiKB2A

This is a tutorial on calculating related rates by implicit differentiation. If interested, you can skip the first 4 minute and start watching how he solved this exact problem using very simple explanation.

First, we need to set up an equation, i.e. Pythagorean theorem

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

Then, apply the Chain Rule to get the derivative of y with respect to time (or rate of change/sliding down per second). Once we have set up the formula, we can easily use the Pythagorean theorem to insert values of x and y to solve the problem.

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

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

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

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

# constant, i.e. 1 ft/s
dx_dt = 1

ladder = 24
x = c(1, 10, 23, 24)
y = lapply(x, function(x) sqrt(24^2 - x^2)) %>% unlist

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

for(i in 1:length(x)){
    print(glue(if(!is.infinite(dy_dt[i])){
        "The rate of sliding down when the base is {x[i]} ft from the house is approximately {round(dy_dt[i], 3)} per second"
        } else {
            "The rate of sliding down when the base is {x[i]} ft from the house is 0, and it's because the ladder is already on the ground"
        }))
}
## The rate of sliding down when the base is 1 ft from the house is approximately -0.042 per second
## The rate of sliding down when the base is 10 ft from the house is approximately -0.458 per second
## The rate of sliding down when the base is 23 ft from the house is approximately -3.355 per second
## The rate of sliding down when the base is 24 ft from the house is 0, and it's because the ladder is already on the ground

We see negative values here and it’s because the y value is decreasing/sliding down the wall.