# Thread: Related Rates with a cone

1. ## Related Rates with a cone

The volume of a cone of radius r and height h is given by v=1/3*pi*r^2*h. If the radius and height both increase at a constant rate of 2 cm per second, at which rate in cubic centimeters per second is the volume increasing when the height is 8 cm and the radius is 6 cm.

I'm kinda puzzled by this problem. Should I do the product rule with the 1/3*pi*r^2*h part? Thanks a lot.

2. Spoiler:
Write down what you are given, what you are required to find, etc.

Given:

$\displaystyle \dfrac{dr}{dt} = 2$

$\displaystyle \dfrac{dh}{dt} = 2$

$\displaystyle V = \dfrac13 \pi r^2h$

Required:

$\displaystyle \dfrac{dV}{dt}$ at h = 8, and r = 6.

You will need to express either r or h in terms of h and r respetively. To find this relation, make a quick sketch of the cone, and you'll see that at the instant asked for:

$\displaystyle \dfrac{h}{r} = \dfrac86$

$\displaystyle h = \dfrac{4r}{3}$

This way, you can use the chain rule:

$\displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

Substitute h in the formula for V, find the derivative and used the chain rule with what you already know

EDIT: Sorry, I just realised that this is a type of question I never encountered as both r and h are changing and I see that the chain rule doesn't apply here. Sorry for the misleading.

3. Originally Posted by zhangxupage2
The volume of a cone of radius r and height h is given by v=1/3*pi*r^2*h. If the radius and height both increase at a constant rate of 2 cm per second, at which rate in cubic centimeters per second is the volume increasing when the height is 8 cm and the radius is 6 cm.

I'm kinda puzzled by this problem. Should I do the product rule with the 1/3*pi*r^2*h part? Thanks a lot.
sounds like a good plan ... carry it out.

4. Is Unknown008's plan workable? Why do I get different answers with the product rule and the substitution method?

5. Originally Posted by zhangxupage2
Is Unknown008's plan workable? Why do I get different answers with the product rule and the substitution method?
How can we know when you haven't shown any working?

6. Sorry for not posting my work.

This is what I did:

dV/dt=pi/3*(2r*dr/dt*h+dh/dt*r^2)

Plug in the numbers: pi/3 (2*6*2*8+2*6*6)

Simplify: pi/3 * 264 = 88pi

Is this correct?

Thanks a lot!

7. I see how my method was wrong now. It involved considering the rate of change of one parameter only (ie either r or h, not both at the same time). I now realise my mistake and I'll not make the same mistake again.

8. Originally Posted by zhangxupage2
Sorry for not posting my work.

This is what I did:

dV/dt=pi/3*(2r*dr/dt*h+dh/dt*r^2) Mr F says: This is correct.

Plug in the numbers: pi/3 (2*6*2*8+2*6*6)

Simplify: pi/3 * 264 = 88pi

Is this correct?

Thanks a lot!
Assuming your arithmetic is correct (I don't check arithmetic) you will have the correct answer. But you should include the unit too ....