Calculus
posted by Anonymous .
A 17 foot ladder is leaning on a 6 foot fence. The base of the ladder is being pulled away from the fence at the rate of 5 feet/minute. How fast is the top of the ladder approaching the ground when the base is 6 from the fence? [Note: The ladder is protruding over the top of the fence.]

Let:
h = the height of the fence is 6'.
x = distance of the base of the ladder from the fence
t = length of ladder between top of fence and the ground
= sqrt(h²+x²)
L = length of ladder
y = Height of top of ladder from the ground
From similar triangles,
h/t = y/L
or
y=Lh/t=Lh/sqrt(h²+x²) ..(1)
dx/dt=5'/min.
dy/dt=(dy/dx)*(dx/dt) ..(2)
So differentiate (1) to get dy/dx and substitute in (2).