Calc Word Problem

[Guest]

A conical tank with vertex down has a vertex angle of 60.0 degrees. Water flows from the tank at a rate of 5.00 cm [cubed]/min. At what rate is the inner surface of the tank being exposed when the water is 6.00 cm deep?


I really hope you can help me.

Thanks

 

[Gene]

The volume of water =
V = (1/3)*base*Depth =
pi/3*(D/sqrt(3))^2*D =
pi*D^3/9
dV/dt = 5
Find dV/dD
Find dD/dt = (dV/dt)/(dV/dD)
Almost there.
Water covers an area = {(the circumfrence at water level) / (circumfrence of the circle the cone (at that depth) was made from)} * (the area of the circle the cone (at that depth) was made from) =
A =
(2*pi*D/sqrt(3))/(2*pi*(2D/sqrt(3)))*(pi*(2D/sqrt(3))^2)
Simplify and find dA/dD
dA/dt = (dA/dD)*(dD/dt)
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