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March 1, 2015

March 1, 2015

Posted by **Anonymous** on Thursday, March 29, 2012 at 7:26pm.

I have no idea how to even start with this problem. Help!

- CALCULUS -
**Steve**, Friday, March 30, 2012 at 11:16amtanθ = h/√(625-h^2)

secθ = 25/√(625-h^2)

sec^2θ dθ/dt = 625/(625-h^2)^3/2 dh/dt

dθ/dt = 1/√(625-h^2) dh/dt

so, when dh/dt = -r, plug and chug.

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