Question

The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

Answer 1

`5` m

Explanation:

The following assumes that (a) the ground is horizontal (i.e is perpendicular to the wall) and (b) the ladder is of constant length.

Variables
Let `y` be the height up the wall to the top of the ladder.

Let `x` be the distance between the bottom of the wall and the bottom of the ladder.

Let `L` be the length of the ladder. (Note that `L` is the constant we seek.)

Rates of Change

The ladder is sliding down the wall, so `y` is decreasing at `0.15` m/s.

`dy/dt = -0.15` m/s

The bottom of the ladder is sliding away from the wall, so `x` is increasing at `0.2` m/s when `x = 3` m.

`dx/dt = 0.2` when `x = 3`

Equation relating the variables

`x^2+y^2=L^2` `" "` (Here we are using perpendicularity.)

Note that if we knew both `x` and `y` at some instant, then we could find `L`

Differentiate to find an equation relating the rates of change.

`2x dx/dt + 2y dy/dt = 0`

We know `dx/dx when`x=3`and we know`dy/dt`, so we can find`y`when`x = 3#

`x dx/dt +y dy/dt = 0`

`(3)(0.2)+y(-0.15) = 0`

`0.6 = 0.15y`

`y = 0.6/0.15 = 60/15 = 4`

When `x=3` m, we find that `y=4` m.

We also have `x^2+y^2=L^2.`

So, `L=5` m.