please help asap.. water is being drained from a container which has the shape of an inverted right circular cone. the container has a radius of 6.00 inches at the top and a height of 10.0 inches. at the instant when the water in the container is 9.00 inches deep, the surface level is falling at a rate 1.2 inches per second. find the rate at which water is being drained from the container.

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please help asap.. water is being drained from a container which has the shape of an inverted right circular cone. the container has a radius of 6.00 inches at the top and a height of 10.0 inches. at the instant when the water in the container is 9.00 inches deep, the surface level is falling at a rate 1.2 inches per second. find the rate at which water is being drained from the container.

Calculus1
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use similar triangles to put the volume formula for the cone into a function of just height
differentiate, and solve for dV/dt
I have no idea what im doing on this problem or what goes where?

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Other answers:

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Do you know the formula for the volume of a cone as a function of its height V=f(h)
solve for r, and put that into the volume formula so it is just in terms of h.
Givens - h=9, dh/dt = 1.2
and I saw this |dw:1438103242401:dw|
\[V = \frac{ 1 }{ 3 }\pi*r^2*h\] from similar triangles... \[r = \frac{ 3 }{ 5 }h\] \[V(h) = \frac{ 1 }{ 3 }\pi*(\frac{ 3 }{ 5 }h)^2*h\]
simplify that, then differentiate both sides w.r.t time, you are given dh/dt
when h = 9
\[V = \frac{ 3 }{ 25 }\pi*h^3\]
chain rule \[\frac{ dV }{ dt }=[\frac{ 3 }{ 25 }\pi]*3h^2*\frac{ dh }{ dt }\]
you are given h and dh/dt, find dV/dt

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