Problem 9 Ex 4.2 page 187

Problem

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:

  1. 1 foot from the house?
  2. 10 feet from the house?
  3. 23 feet from the house?
  4. 24 feet from the house?

Solutions:

1 foot from the house

The ladder and house makes up a right triangle: \(x^2+y^2=24^2\)

x= the horizontal distance the ladder is from the wall

y= the vertical distance the ladder is up the wall

Taking the derivative with respect to time gives us:

\(2x dx/dt + 2y dy/dt = 0\)

\(\implies dy/dt = -x/y dx/dt\)

x <- 1
y <- sqrt(24^2-x^2)
dxdt <- 1
dydt <- -x*dxdt/y
dydt
## [1] -0.04170288

The required rate is dy/dt = -0.0417 ft/sec

10 feet from the house

Using the derivation in (a) and plugging in the variables

x <- 10
y <- sqrt(24^2-x^2)
dxdt <- 1
dydt <- -x*dxdt/y
dydt
## [1] -0.4583492

The required rate is dy/dt = -.458 ft/sec

23 feet from the house

Using the derivation in (a) and plugging in the variables

x <- 23
y <- sqrt(24^2-x^2)
dxdt <- 1
dydt <- -x*dxdt/y
dydt
## [1] -3.354895

The required rate is dy/dt = -3.35 ft/sec

24 feet from the house

The ladder is completely horizontal and there is no vertical height.

x <- 24
y <- sqrt(24^2-x^2)
dxdt <- 1
dydt <- -x*dxdt/y
dydt
## [1] -Inf

As the vertical height approaches zero, dy/dt approaches infinity.