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Water is flowing into a large spherical tank at a constant rate. Let $V\left(t\right)$ be the volume of water in the tank at time $t$, and $h\left(t\right)$ of the water level at time $t$.

Is $\frac{dV}{dt}$ positive, negative, or zero when the tank is one quarter full? Is $\frac{dh}{dt}$ positive, negative, or zero when the tank is one quarter full?

My answer

I believe that the rate of change of the volume is increasing till 1/2 full, then decreasing. I believe that the rate of change of the height is decreasing till 1/2 full, then increasing.

THis is because of the widest part of the sphere is when h = r (the radius of the sphere), and if you think of it as many slices of circles upon each other, then the area change would follow the patter of each slice.

Am I correct?

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wait maybe I'm being dumb, but if you're adding water, doesn't the volume and water level both go up, i.e. both positive always? (until overflow or something, then constant?) –  uncookedfalcon Nov 1 '12 at 4:31
    
You are thinking rate of rate of change. The question asks for rate of change only. –  E.O. Nov 1 '12 at 4:41
    
@e.o. yes, I realize my error –  yiyi Nov 1 '12 at 5:34
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1 Answer

up vote 2 down vote accepted

Uncookedfalcon is correct. You're adding water, which means that both the volume and water depth are increasing (that is, $\frac{dV}{dt}$ and $\frac{dh}{dt}$ are positive) until it's full. In fact, $\frac{dV}{dt}$ is constant until it's full.

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