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I have a question for calculus which I'm having trouble with. I've calculated the volume of the frustum with the formula $V=\frac\pi3(R^2+Rr+r^2).$ However, I'm having difficulty deriving this equation in order to find the rate of change of the volume. Would appreciate advice, here is the problem.

"A reservoir containing water has the shape of a frustum of a right circular cone of altitude 10 feet, a lower base of radius 10 feet, and an upper base radius 15 feet. How fast is the volume of the water increasing when the water is 6 feet deep and rising at a rate of 2 ft./hr?"

EDIT: Visual Representation

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  • $\begingroup$ Correct altitude $ h=10,\, $ or $ h+H== 10 \, ?$ $\endgroup$
    – Narasimham
    Commented Oct 23, 2017 at 2:21
  • $\begingroup$ The "upper" base is at the bottom of the figure and the "lower" base is above it. Which way is really up? But also note that with the numbers given in the figure, $h = \frac{10}{3}$ ft, so the entire depth of the reservoir is less than $6$ feet. No wonder you had difficulty with this question. $\endgroup$
    – David K
    Commented Feb 22, 2021 at 4:13

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The numbers in the figure are not consistent with the numbers in the setup.

As I read it, $h = 10$, and $H$ or $H+h$ is unknown.

$r = \frac 23 R$ which implies $h = \frac 13 (h+H).$ That is if we extened the theoretical cone above the height tank, the cone would be 30 ft tall.

What we really care about is the radius at the water level. If we say $h$ is actually the height to the surface of the water, and $r$ is the radius of the tank at this height, we can say,

$r = 15 - \frac 12 h$

That way when the tank is full $h = 10$, and $r = 10.$
When the $h = 6, r = 12$

$V = \frac 13\pi (R^2 + Rr + r^2) h$

$\frac {dv}{dh} = \frac 13\pi (R^2 + Rr + r^2) + \frac 13\pi (R^2 + Rr + r^2)'h$

We use the product rule because $r$ is a function of $h$

$\frac {dv}{dh} = \frac 13\pi (R^2 + Rr + r^2) + \frac 13\pi (R\frac {dr}{dh} + 2r \frac {dr}{dh})h$

We have $h, R, r,$ above. We can find $\frac {dr}{dh}$ from $r = 10 - \frac 12 h$ and that is everything to find $\frac {dV}{dh}$.

The chain rule says $\frac {dV}{dt} = \frac {dV}{dh}\frac {dh}{dt}.$

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  • $\begingroup$ For reasons given in other answers and comments, $h=10$ is the only plausible reading of the problem and therefore the figure is wrong. If the larger radius really is “upper” then we have $r=10+\frac12 h(t)=13$. But that would imply that the diagram is upside down in addition to being mislabeled. $\endgroup$
    – David K
    Commented Sep 3, 2023 at 18:22
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You can simply model it by finding the rate at which the volume of the subtracted cone changes. This is true because the Volume of the whole reservoir remains constant.

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Implicit differentiation! Take the derivative of both sides with respect to $t$ like so:

$$\begin{align*} V=\frac{\pi}{3} h \left( R^2+Rr+r^2 \right) &\implies \frac{d}{dt}(V) =\frac{\pi}{3}\frac{d}{dt} \left( h R^2+ hRr+ hr^2 \right) \end{align*}$$ $$\implies \frac{dV}{dt} = \frac{\pi}{3} \left( \frac{dh}{dt} R^2 + 2Rh\frac{dR}{dt} + \frac{dh}{dt}Rr+h\frac{dR}{dt}r+hR\frac{dr}{dt} + \frac{dh}{dt}r^2+2rh\frac{dt}{dt} \right)$$

Essentially implicit differentiation is doing the chain rule over and over but treating everything in the expression as an unknown function of $t$, hence why you're seeing $\frac{d}{dt}$ everywhere.

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