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I have been working on this problem for about 2 hours and I can't seem to get it, here is exactly what the question reads. "Water is poured into the top of a conical tank at the constant rate of 1 cubic inch per second and flows out of an opening at the3 bottom at a rate of .5 cubic inches per second. The tank has a height of 4 inches, and a radius of 2 inches at the top. How fast is the water level changing when the water is 2 inches high?"

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1 Answer

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Think about what is happening. The water (volume) is being poured in at a constant rate. This relates to how the water level ($h$) changes and how the width of the water in the tank at that level ($r$) changes. Further, $r$ and $h$ are related.

The volume of a cone is

$$V = \frac13 \pi \, r^2 \, h$$

How is $h$ related to $r$? You know that, at the top, the radius is $2$ and $h=4$. Because this is a cone, we can say that $r = h/2$ at all levels of the cone. Thus,

$$V(h) = \frac{1}{12} \pi \, h^3$$

We may then differentiate with respect to time; use the chain rule here

$$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$

You are given $dV/dt$ and the height $h$ at which to evaluate; solve for $dh/dt$.

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