A geometry/derivative word problem

Quote:

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a container has the shape of an open right circular cone, as shown in the figure above. The height of the container is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that it depth h is changing at the constant rate of -3/10 cm/hr.
The volume of a cone of height h and radius r is given by V=1/3 pi r^2 h
a) Find the volume V of water in the container when h=5 cm.
b) find the rate of change of volume of water in the container, with respect to time, when h=5 cm.

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I know you suppose to find an equation and take the derivative, but how to do it i don't know.

1)V=1/3 pi r^2 h

ookay. the cone has a radius of 10 and a height of 10. in order to find the radius when the height is 5, we need a proportion.
10/10 = 5/r
we can conclude r=5 cm as well, and we can solve for Volume!!

2)dV/dt can be found by differentiating the equation for volume..

it was given that dh/dt=-3/10.

using a proportion similiar to the one above, we know
10/10=h/r
therefore h=r

we can eliminate r in the volume equation by plugging in h:
V=1/3 Pi r^2 h r=h
V=1/3 Pi h^3

differentiate that:
dV/dt = 1/3 Pi 3h^2 * dh/dt
solve!