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I am in an intro calculus class and I have a problem which I am unsure about. It reads:

Water is being poured into a conical reservoir at a rate of pi cubic feet per second. The reservoir has a radius of 6 fees across the top and a height of 12 feet. At what rate is the depth of the water increasing when the depth is 6 feet?

Here's what I came up with for the problem: h = height of cone

r = radius of cone

V = volume of cone

$\frac{r}{h}$ = $\frac{1}{2}$

V = $ \frac{\pi r^2h}{3}$

$\frac{dV}{dt}$ = $\frac {2\pi r \frac{dr}{dt}h}{3} + $$\frac{1}{3} \pi r^2 \frac {dh}{dt}$ (Using product rule)

Because r = $\frac{h}{2}, $$\frac{dv}{dt}$ = $\frac{2\pi h}{2}\frac{dh}{dt}\frac{h}{6} + \frac{\pi (\frac{h}{2})^2}{3} \frac{dh}{dt}$

I solved for $\frac{dh}{dt}$

$\frac{dh}{dt}$ = $\frac{12\frac{dv}{dt}}{3\pi h^2}$

I plugged in $\pi$ for $\frac{dv}{dt}$ and 6 for h and got 1/9 or 0.111

Can anyone tell me if this is correct? Thanks if advance!

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  • $\begingroup$ Please note that in my answer I assumed $r$ was constant. It is not, so I deleted it. $\endgroup$ – Nażysław Zbyłutowicz May 1 '17 at 4:02
  • $\begingroup$ @NażysławZbyłutowicz that makes sense, thank you! $\endgroup$ – Jamman00 May 1 '17 at 4:07
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Since the dimension of the cone is such that $r = 6$ft and $h=12$ft, we have $\dfrac{r}{h} = \dfrac{6}{12}$, which means $r = \dfrac{h}{2}$. We must substitute this into the equation of the volume of a cone before differentiating. It follows that

$V=\pi r^2 \dfrac{h}{3}$

$V=\pi \bigg(\dfrac{h}{2}\bigg)^2 \dfrac{h}{3}$

$V=\pi \dfrac{h^2}{4} \dfrac{h}{3}$

$V= \pi \dfrac{h^3}{12}$

$\dfrac{dV}{dt} = \dfrac{\pi}{12} 3h^2 \dfrac{dh}{dt}$

$\pi = \dfrac{\pi}{12} 3(6)^2 \dfrac{dh}{dt}$

$\pi = 9\pi \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{1}{9}$

The height of the cone is increasing at a rate of $\dfrac{1}{9}$ feet per second .

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    $\begingroup$ Thanks, I didn't even think of substituting before differentiating, that made the algebra much simpler. Thanks for the response! $\endgroup$ – Jamman00 May 1 '17 at 4:06

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