# Thread: Related Rates Involving A Cone

1. ## Related Rates Involving A Cone

A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

So far, I've got that r/h = 1/4, so r = h/4.

And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

However, I'm stumped on how to find the height when the cone is half filled.

Any help is appreciated.

2. Originally Posted by Shapeshift
A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

So far, I've got that r/h = 1/4, so r = h/4.

And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

However, I'm stumped on how to find the height when the cone is half filled.

Any help is appreciated.
volume of full cone ... $V = \frac{\pi}{48} \cdot 40^3$

half full ... $V = \frac{\pi}{96} \cdot 40^3 = \frac{\pi}{48} h^3$

$h^3 = \frac{40^3}{2}$

$h = \frac{40}{\sqrt[3]{2}}$

3. Originally Posted by Shapeshift
A cone is 40 cm in height and 20 cm in diameter at widest point. If it is being filled at a rate of 10pi mL/min, how fast is its height changing when it is half filled?

So far, I've got that r/h = 1/4, so r = h/4.

And I've got that dV/dt = pi/48 x 3h^2 x dh/dt

However, I'm stumped on how to find the height when the cone is half filled.

Any help is appreciated.
..

half filled means volume is halved.