calculus
posted by Anonymous .
1) A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
Find the rate at which the are of the triangle is changing when the base of the ladder is 7 feet from the wall.

I did the Pythagorean to find the other side of the triangle and got 24. But I'm confused...am I looking for the dr/dt OR dv/dt? Or something else?
Help please!

Rate at which area is changing, dA/dt
where A= 1/2 b*height
dA/dt= 1/2 b dh/dt + 1/2 h db/dt 
Wow, thanks for the clarification!
Now I can just plug in b  which is 7 but what about the dh/dt and db/dt? For those unknowns, am I suppose to use the area formula to find it? 
No. You are given db/dt
Using the pyth theorm
c^2= b^2 + h^2
dc/dt=0= 2b db/dt + 2h dh/dt
solve for dh/dt in terms of b,h, db/dt
this is what related rates means.
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