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### Section 3.11 : Related Rates

10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10 meters?

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Here is a sketch of the cross section of the tank and it is not even remotely to scale as I found it easier to reuse an old image that I had lying around. I can be a little lazy sometimes. At least I was less lazy with the image in the problem statement….

We want to determine $$h'$$ when $$r = 10$$ and we know that $$V' = 12$$.

Show Step 2

We’ll need the equation for the volume of a cone.

$V = {\textstyle{1 \over 3}}\pi {r^2}h$

This is a problem however as it has both $$r$$ and $$h$$ in it and it would be best to have only $$h$$ since we need $$h'$$. We can use similar triangles to fix this up. Based on similar triangles we get the following equation which can be solved for $$r$$.

$\frac{r}{h} = \frac{{26}}{8}\hspace{0.5in} \Rightarrow \hspace{0.5in}r = {\textstyle{{13} \over 4}}h$

Plugging this into the volume equation gives,

$V = {\textstyle{{169} \over {48}}}\pi {h^3}$ Show Step 3

Next, let’s differentiate this with respect to $$t$$.

$V' = {\textstyle{{169} \over {16}}}\pi {h^2}h'$ Show Step 4

To finish off this problem all we need to do is determine the value of $$h$$ for the time we are interested in. This can easily be done from the similar triangle equation and the fact that we know $$r = 10$$.

$h = {\textstyle{4 \over {13}}}r = {\textstyle{4 \over {13}}}\left( {10} \right) = {\textstyle{{40} \over {13}}}$

The rate of change of the height of the water is then,

$12 = {\textstyle{{169} \over {16}}}\pi {\left( {{\textstyle{{40} \over {13}}}} \right)^2}h' = 100\pi h'\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{h' = {\textstyle{3 \over {25\pi }}}}}$