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A sphere with radius 10 cm is being filled at a rate of 1 cm3 per minute. What rate is the depth of the water increasing when it is 5 cm?

So what I did was:

$$r=10$$ $$\frac{dV}{dt} = 1$$ $$d = 5$$ $$V= \frac{4}{3}\pi r^2$$ $$\frac{dV}{dt} = \frac{8}{3}\pi r \frac{dr}{dt}$$

However, what I'm really looking for isn't $\frac{dr}{dt}$ but instead $\frac{dD}{dt}$, and I can't think of any way to relate volume to depth or radius. How can I do this?

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2 Answers

It seems that you're using $V$ for two different things -- both the volume of the entire sphere, and the volume of water yet in the sphere.

What you need to do is choose a variable for for the volume of water in the sphere, and then use the formula for a spherical cap to relate it to the depth of water. Only after you have this relation will it make sense to start doing calculus on the situation.

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We need to find first derivative of spherical cap formula: $V'_t=(\frac{\pi h}{6}(3a^2+h^2))'_t$ , where $a^2=h^2-2hr$ ; Since $r=5 \Rightarrow a^2=h^2-10h$ ,

If we substitute this into derivation formula we get

$V'_t=\frac{\pi}{3}(2h^3-15h^2)'_t \Rightarrow V'_t=2\pi (h^2-5h)\frac{dh}{dt} \Rightarrow 1=2\pi (h^2-5h)\frac{dh}{dt}$

from last equation we have: $\frac{dh}{dt}=\frac{1}{2\pi (h^2-5h)}\frac{cm}{min}$

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