Calculus: Rates of Change of Cone: height and radius

[ilovemynny]

Homework Statement

A water tank the shape of an inverted circular cone with a base radius of 2m and height of 4m. if water is being pumped into the tank at a rate of 2m^3/min, find the rate at which the water level is rising when the water is 3m deep.

dv/dt = 2m^3/min
h = 3m
r = h/2

I know how to find dh/dt and the answer is dh/dt = 8/9pi m/min

but the next part asks for the rate of change of the radius (dr/dt) at that moment (when h = 3m) and i don't know how to get get this

Homework Equations

V = (pi/3)(r^2)h

The Attempt at a Solution

i was thinking that i would find the derivative of the volume formula so:
V = (pi/3) * r^2 * h

dV/dt = (pi/3) * [r^2 * (dh/dt) + 2rh * (dr/dt)]

and plug in the values:
2m^3/min = (pi/3) * [2^2 * (8/9pi m/min) + 2(2)(3) * (dr/dt)]
and find dr/dt from there
is this right or am i completely wrong?
if i am wrong can someone please help me? I don't know if I'm even using the right variables.

 

[S.Mukherjee]

you should write volume in terms of "r"
V=1/3{(pi)r^2}*rcot(theta)
where
cot(theta)=4/2.
find dV/dt @ r=h/2.
About your answer, The expression you derived is absolutely correct but the value of "r" you put in your final answer is wrong.You should put r=3/2 instead of 2.