Expert: Alon Mandes - 12/7/2009

Question
This one I feel like I'm getting right but i'd like to verify with your answer.

A ladder is slipping down a vertical wall. If the ladder is 15feet long and the top of it is slipping at the constant rate of 4feet per second, how fast is the bottom of the ladder moving along the ground when the bottom is 12feet from the wall?

Answer
Let's set v to be the velocity or rate of change in the ladder bottom. Ho is the initial
height of the top of the ladder, %26 Xo is the initial distance between the bottom of the
ladder %26 the wall .
According to Pythagoras : x(t)^2+h(t)^2=15 at all time . Where x(t) is the distance between
the bottom of the ladder %26 the wall. And h(t) is the height of the top of the ladder .
We can also write this equation as : (Xo+vt)^2+(Ho-4t)^2=15 .
Let's derive this equation :
2(Xo+vt)v+2(Ho-4t)(-4)=0  . Now let's substitute our variables : We know that Xo+vt=12, so we get 12v-4h=0 . We also know that 12^2+h^2=15 , therefore : h=9 . So,
12v-4*9=0 --> v=3 . That means that the bottom of the ladder is moving away from the wall
at speed of 3feet per sec .

Alon.