1. ## related rates help

The Question is:
Water drains from the conical tank shown in the figure at the rate of 5ft^3/min.
How fast is the water level dropping from when h=6ft?

ok the relation between the variables h and r is: 5r/2=h

And i know the answer to the question is .276 ft/min, but i don't know how you get that.

What i did was do the derivative of the cone volume formula: (1/3)pi(r^2)h using the product rule and i differentiate the h as well. Plus i even substitute r with 2h/5 (since i want to know r in terms of h) after that i replace everything with the appropriate numbers and my answer is different. Can you guys show me how to do this step by step? Thank you.

EDIT: there is an upside down cone with a base radius of 4in and height of 10 in and there is an r and h which represents the height of the water and the base radius of it as well. Also the question before that says: What is the relation between the variables h and r? And that is every part of the question.

The Question is:
Water drains from the conical tank shown in the figure at the rate of 5ft^3/min.
How fast is the water level dropping from when h=6ft?

ok the relation between the variables h and r is: 5r/2=h

And i know the answer to the question is .276 ft/min, but i don't know how you get that.

What i did was do the derivative of the cone volume formula: (1/3)pi(r^2)h using the product rule and i differentiate the h as well. Plus i even substitute r with 2h/5 (since i want to know r in terms of h) after that i replace everything with the appropriate numbers and my answer is different. Can you guys show me how to do this step by step? Thank you.
since $\displaystyle h = \frac{5r}{2}$ , $\displaystyle r = \frac{2h}{5}$

$\displaystyle V = \frac{\pi}{3} \left(\frac{2h}{5}\right)^2 h$

$\displaystyle V = \frac{4\pi}{75} h^3$

you know $\displaystyle \frac{dV}{dt} = -5 \, ft^3/min$ ... take the time derivative of the above equation and determine $\displaystyle \frac{dh}{dt}$ when $\displaystyle h = 6$