# Math

• A ladder 20 feet long leans against a vertical building. If the bottom of the ladder slides away from the building horizontally at a rate of 3 feet/sec.
• how fast is the ladder sliding down the building when the top of the ladder is 8 feet from the ground?

Let `x` be the distance from the wall, and let `y` be the height at the top of the ladder.

Then from the pythagorean theorem we have `x^2+y^2=20^2` .

We are given that the rate of change of the bottom of the ladder away from the wall is 3ft/sec, so `(dx)/(dt)=3` .

When the height of the ladder is 8 ft., `y=8` , from the pythagorean theorem we get `x=4sqrt(21)` .

We differentiate `x^2+y^2=20^2` with respect to `t` :

`2x(dx)(dt)+2y(dy)/(dt)=0` . We want the change in height, so solve for `(dy)/(dt)` :

`(dy)/(dt)=(-x(dx)/(dt))/y` . Plug in the known values to compute `(dy)/(dt):`

`(dy)/(dt)=((-4sqrt(21))(3))/8~~-6.87` ft/sec.

So the speed at which the top of the ladder is moving is approximately -6.87 ft/sec (6.87 ft/sec towards the ground)