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when the radius is decreasing and the height is increasing?

i have to calculate the partial derivatives, right? but then do i add the values?

like V = partial derivative height + partial derivative radius??

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    To find the rate of change as the height changes, solve the equation for volume of a cone ($\frac{\pi r^2 h}{3}$) for h, and find the derivative, using the given radius. For the rate of change as the radius changes - same idea.2012-11-11

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It would help if you gave example numbers of some sort, but here is the general solution. The radius $r$ is changing at the rate of $r'$, and the height $h$ is changing at the rate of $h'$. The volume $V$ has a rate of change of $V'$.

Now we know that $V = (\frac13\pi) r^2 h$. If you take the derivative of that, then you get (using product rule):

$V' = \frac13\pi\frac{d}{dt}(r^2h) = (\frac13\pi)(2rr'h + r^2h')$
All you have to do is plug in your current $r$ and $h$ values, and the rate of changes $r'$ and $h'$.

Just to make sure it is clear, $x'$ is the derivative of $x$.

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    2rr'h? i am not sure what you did. did you replace r and h by their partial derivatives?2012-11-11
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    @Matthew I didn't really use partial derivatives (or at least I don't think I did) but when you take the derivative of $r^2*h$ then you need to use both product rule and chain rule. Because of product rule, $\frac{d}{dt}(r^2h) = h*\frac{d}{dt}r^2 + r^2*\frac{d}{dt}h$ and because of chain rule, the derivative of $h$ isn't $1$, it is $1*h'$, and the derivative of $r^2$ isn't $2r$ but it is $(2r)r'$. Therefore, $\frac{d}{dt}(r^2h) = ((2r)r')h + (r^2)h' = 2rr'h + r^2h'$. Does that clear things up?2012-11-11
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    I would add: *Just to make sure it is clear, $x'$ is the derivative of* x *with respect to* t, *that is* - $$x'=\dfrac{dx}{dt}$$2013-05-07