# Thread: Quick question. related rates

1. ## Quick question. related rates

I am doing the chain rule in calculus, and this question has got me stuck for a long time now. I am not sure how to approach or solve this, help would be appreciated

The amount of cleaning fluid in a partially filled, spherical tank is given by V = (pi)R(h^2)-pi(h^3)
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Where R is the radius of the rank, and h is the depth of the center of the tank. If cleaning fluid is being pumped into the tank, of radius 12m, at a rare of 8 m^3/min, how fast is the fluid rising when the tank is 3/4 full?

2. Hello, Sky12!

This problem doesn't involve the Chain Rule.

The amount of fluid in a partially-filled spherical tank
. . is given by: .$\displaystyle V \:= \:\pi Rh^2 -\frac{\pi}{3}h^3$
where $\displaystyle R$ is the radius of the tank, and $\displaystyle h$ is the depth at the center of the tank.

If fluid is pumped into the tank at a rate of 8 m³/min, and $\displaystyle R = 12$ m,
how fast is the fluid rising when the tank is 3/4 full?

We have: .$\displaystyle V \:= \:\pi Rh^2 -\frac{\pi}{3}h^3$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt}\:=\:24\pi h\!\left(\frac{dh}{dt}\right) - \pi h^2\!\left(\frac{dh}{dt}\right)$

Then: .$\displaystyle \frac{dV}{dt}\:=\:\pi h(24-h)\frac{dh}{dt}\quad\Rightarrow\quad\frac{dh}{dt} \:=\:\frac{\frac{dV}{dt}}{\pi h(24 - h)}$

. . We are given $\displaystyle \frac{dV}{dt} = 8$ m³/min . . . When the tank is $\displaystyle \frac{3}{4}$ full: $\displaystyle h \,\approx\,16$ m. **

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:\frac{8}{\pi\!\cdot\!16(24-16)}\:=\:\frac{1}{16\pi}$m/min

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** . No, I didn't make an error . . .

If the tank is three-fourth full, then: .$\displaystyle V \:=\:\pi\!\cdot\!12h^2 - \frac{\pi}{3}h^3\:=\:\frac{3}{4}\left(\frac{4}{3}\ pi\!\cdot\!12^3\right)$
. . and we have a cubic equation to solve: .$\displaystyle h^3 - 36h^2 + 5184 \:=\:0$

which has a root near $\displaystyle h = 16.17$

3. great help, thanks!