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A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.

I am not how to do this problem, but I've tried this using the volume formula for cone:

$$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$

I am pretty sure that I am wrong.

Could someone help me?

Thanks

The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet

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You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$

$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$.

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  • $\begingroup$ I added a few more details. Is there anything I should elaborate on? $\endgroup$ – Quinn Greicius Mar 10 '16 at 0:56
  • $\begingroup$ Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? $\endgroup$ – didgocks Mar 10 '16 at 1:00
  • $\begingroup$ Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. $\endgroup$ – Quinn Greicius Mar 10 '16 at 1:01
  • $\begingroup$ Thank you, it was very helpful! $\endgroup$ – didgocks Mar 10 '16 at 1:02
  • $\begingroup$ I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone $\endgroup$ – didgocks Mar 13 '16 at 14:26
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Retain symbols till the last plugin step

$$ r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha $$

$$ V = \pi h^3/3\, \tan ^2\alpha $$

$$ dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha $$

Plug in given values to get answer tallying with text.

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