Posted by **Anonymous** on Tuesday, March 15, 2011 at 9:45pm.

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.]

- Calculus -
**MathMate**, Tuesday, March 15, 2011 at 10:34pm
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).

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