I need help with Related Rates

[Craw]

Its one of those first year calc classic related rates questions...

The volume V of a cone ( V = 1/3 pi rsquared h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.

I am on the last part of the problem and can't remember how to do it:

At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height h?

I know you're looking for dA/dh...

I remembered how to get dA/dt which is 3pi and dh/dt which is 8 (asked in the previous parts of the question)

I can show my work for those parts if needed...

Can anyone help me find dA/dh?

I got 3pi/8 as my answer by using the chain rule, but I don't know if that is right...

 

[Deleted member 4993]

Its one of those first year calc classic related rates questions...

The volume V of a cone ( V = 1/3 pi rsquared h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.

I am on the last part of the problem and can't remember how to do it:

At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height h?

I know you're looking for dA/dh...

I remembered how to get dA/dt which is 3pi and dh/dt which is 8 (asked in the previous parts of the question)

I can show my work for those parts if needed...

Can anyone help me find dA/dh?

I got 3pi/8 as my answer by using the chain rule, but I don't know if that is right...

Hint:

\(\displaystyle A \, = \, \frac{3V}{h}\)

 

[Craw]

Thanks for the responses guys. I think I see where you guys are going with it. I will go try to work it out right now...

I am still getting 3pi/8

I hope this is right. Thanks again.

 

[Deleted member 4993]

Its one of those first year calc classic related rates questions...

The volume V of a cone ( V = 1/3 pi rsquared h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second.

I am on the last part of the problem and can't remember how to do it:

At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height h?

I know you're looking for dA/dh...

I remembered how to get dA/dt which is 3pi and dh/dt which is 8 (asked in the previous parts of the question)

I can show my work for those parts if needed...

Can anyone help me find dA/dh?

I got 3pi/8 as my answer by using the chain rule, but I don't know if that is right...

The interesting point of this exercise is that V, A (or r) and h - all are function of 't'.

However, the ratio between the radius and the height does not change.

\(\displaystyle \frac {h}{r} = \frac{4}{3}\)

so

\(\displaystyle A \, = \, \pi \cdot \frac{9h^2}{16}\)

\(\displaystyle \frac{dA}{dh} \, = \, \pi \cdot \frac{9h}{8}\)

when h = 4

\(\displaystyle \frac{dA}{dh} \, = \, 4.5 \cdot \pi\)

This rate is only shape dependent.