Trouble finding dr/dt

[cjtaylor31]

A container has the shape f an open right circlular cone. 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 its depth h is changing at the constant rate of -3/10 cm/hr
(Note:The volume of a cone of height h and r is given by V=1/3 pi r^2 h)
a.) Find the volume V of water i the container when h=5 cm.
b.)Find the rate of change of the volume of water in the container, with respect to time, when h=5 cm.


I know that r=h/2 and i can just plug the values in to find the volume but it is in part b that I have trouble..
I have this so far.. dv/dt=pi/3 2r dr/dt h + r^2 dh/dt which becomes dv/dt=25pi/3 dr/dt - 1.875 but now I have two unknown rates and I'm stuck. Please help

 

[MarkFL]

What you want to do is first express the volume as a function of height (substitute for \(\displaystyle r\) using the relationship you found between \(\displaystyle r\) and \(\displaystyle h\)). Then, differentiate with respect to time, and plug in the given values for \(\displaystyle h\) and \(\displaystyle \dfrac{dh}{dt}\) to find the value of \(\displaystyle \dfrac{dV}{dt}\).