Question

A water tank is in the shape of a cone with vertical axis down. The tank has radius 3 meters and is 5 meters high. At first, the tank is full of water, but at time t=0 seconds, a small hole st the vertex is opened and the water begins to drain. When the height of the water in the tank has dropped to 3 meters, the water is flowing out at 2 meters cubed per second. At what rate is the water level dropping then?

Answer 1

\[V=\frac{\pi}{3}r^2h\]

The ratio between the radius and height is constant, so you can use that to eliminate r from the equation since you don't have any info on dr/dt.
\[\frac{r}{h}=\frac{3}{5}\]
\[V=\frac{9\pi}{75}h^3\]
Take an implicit derivative to relate dV/dt and dh/dt

Answer 2

@peachpi how did you get the volume formula?

Answer 3

the last one

Answer 4

You could always google the volume formula, or you could derive it, which i hardly doubt you would want to do.

Answer 5

solve for r
r = 3h/5

then substitute into the formula for volume of a cone

Answer 6

hi

Answer 7

I'm still lost

Answer 8

would you get \[V=\frac{ 27\pi }{ 75 }h^2\]

Answer 9

Hello?