another rates of change question

a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

show when the depth of the water = h, V= (pi x h^3) / 12

dv/dt = 4

dh/dt = dv/dt x dh/dv

dh/dv = 2/3 x pi x r x h

I said r = h/2
and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

anyone able to show me where i went wrong
thanks

h/2 ?

Yes.

i subd that in and differentiated and got the answer, but why dont you have to time by 4?

Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

Remember to square both the numerator and denominator.

Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

Remember to square both the numerator and denominator.

there is a second part, find in terms of pi, the rate of rise of the water when the depth is 2cm I got

dh/dt = dv/dt x dh/dv

dh/dv = pih^3 / 12

h=2

answer = 8pi/3, does this look right?

...

I have to agree with TenOfThem here.

There's no point using the chain rule when you don't understand the basic rules of differentiation.

Ask your teacher for help/clarification ASAP.