Derivative of the Volume of a Cone


by chompysj
Tags: cone, derivative, volume
chompysj
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#1
Nov25-07, 02:50 PM
P: 19
Hi, I'm working on a related rates problem, and I need to find the derivative of the volume of a cone.

So the equation is:

V = (1/3) (pi) (r^2) (h)

I'm not sure how to find the derivative. Would the whole thing turn out to be 0? Or do I need to use the product rule?

Please help!

Thanks
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mathman
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#2
Nov25-07, 03:39 PM
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Your problem statement is incomplete. You have 3 variables - V, r, and h. Which is the dependent variable and which is independent variable and which if any is constant? If none are constant you may need a second equation, depending on the problem statement.
chompysj
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#3
Nov25-07, 03:43 PM
P: 19
I'm supposed to find the rate at which the volume of a cone is changing, when r and h are a certain amount (the r and h are both changing at a constant rate).
So except for pi there are no constants.

Office_Shredder
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#4
Nov25-07, 04:30 PM
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Derivative of the Volume of a Cone


You have a relationship for r in terms of h or vice versa, so make it a function of one variable and take the derivative with respect to that
HallsofIvy
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#5
Nov25-07, 07:15 PM
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Quote Quote by chompysj View Post
I'm supposed to find the rate at which the volume of a cone is changing, when r and h are a certain amount (the r and h are both changing at a constant rate).
So except for pi there are no constants.
Then you need to use the chain rule for functions of 2 variables:
If f(x,y) and x and y are functions of t, then
[tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]

Here, V= [itex](1/3)\pi r^2 h[/itex] so
[tex]\frac{dV}{dt}= \frac{1}{3}\pi\left( 2rh\frac{dr}{dt}+ r^2\frac{dh}{dt}\right)[/tex]
chompysj
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#6
Nov25-07, 09:19 PM
P: 19
Thanks for all your help!
Flieger
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#7
Oct23-09, 08:34 PM
P: 3
Hey, chompysj, I think what HallsofIvy said may be right, but I'm not sure...He said chain rule, but showed the product rule. I think it is correct to use the product rule, differentiating it implicitly, as you do in related rates. A rate is what taking the derivative of the cone's volume formula would give you.

When:


The derivative (with respect to time, t), I THINK would be:


Then simplified to:


You will have, now, a related rate for the volume of a cone. This is correct to the best of my knowledge, and I note the fact that I took the derivative of the radius, r, because it, too, is not constant (as you can obviously imagine, as it changes depending on how high or low you go in the cone). Had this been a cylinder, you would not treat r as a variable, but rather as a constant, like (π/3) or 2 per se, because the radius does not change in a cylinder.

I hope this helps, and correct me if I'm wrong!
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HallsofIvy
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#8
Oct24-09, 05:31 AM
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No, that is NOT the product rule, it is the "chain rule for functions of two variables" as I said.
Flieger
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#9
Oct24-09, 01:13 PM
P: 3
Oh ok, I'm sorry. You have to understand that I'm only a high schooler in AP Calculus, so my knowledge of Calculus is limited anyway. But I am sure that what I have learned tells me to call that the product rule. What I was taught the product rule was, I used to solve that. So I really think this is a matter of semantics here...

Product rule (that I learned):

f(x)g(x) = f(x)g'(x) + g(x)f'(x)
HallsofIvy
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#10
Oct24-09, 01:18 PM
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Yes, that is the product rule. But if you look closely, you will see that that is NOT what I wrote.
Flieger
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#11
Oct24-09, 01:41 PM
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Yes, I see, but we came up with the same answers, did we not? Haha! Perhaps I am just misnaming it, or just did it a different way. Anyway, thanks for the response.


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