Question

A ladder is leaning against a wall, and the floor and slipping. If the bottom of the ladder is slipping at `30 cms^(-1)` then how fast is the top of the ladder sliding down the wall when the ladder is at `45^o`?

Answer 1

It is sliding down at a rate (speed) of `30 \ ms^-1`.

Explanation:

Let us define the following variables:

# { (x, "Distance of bottom of ladder from the wall " (cms^-1)), (y, "Distance of top of ladder from the floor "(cm)), (t, "Time "(s)), (theta, "Angle between Ladder and floor "angle ABO " (radians)") :} #

We are told that `dx/dt=30 \ cms^(-1)` (constant)
We aim to find `dy/dt` when `theta=45^o=pi/4`

The ladder is a fixed length, `l\ (m)` say, so by Pythagoras;

`x^2+y^2=l^2`

Differentiating Implicitly wrt `t` we get:

`2xdx/dt + 2ydy/dt=0 => xdx/dt + ydy/dt=0`

`dx/dt = 30 => 30x + ydy/dt=0`

`:. dy/dt = -30x/y` ..... [A]

Using trigonometry, we have:

`tan \ angle ABO = y/x => tantheta = y/x`
`theta = pi/4 => y/x=1 => x/y =1`

With `x/y=1` in Eq[A] then:

`=> dy/dt = -30`

The minus sign denotes that the ladder is sliding down, i.e., the height `y` is decreasing. It is sliding down at a rate (speed) of `30 \ ms^-1`.