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### Water Draining from a Conical Reservoir

```
Date: 12/20/95 at 13:57:54
From: Anonymous
Subject: reservoir

A circular conical reservoir, vertex down, has depth 20 ft and radius
of the top 10 ft. Water is leaking out so that the surface is falling at
the rate of 1/2 ft/hr. The rate, in cubic feet per hour, at which the
water is leaving the reservoir when the water is 8 ft deep is?

a. 4  b. 8  c. 16  d. 1/4  e. 1/8
```

```
Date: 6/13/96 at 20:34:32
From: Doctor Jerry
Subject: Re: reservoir

Tony,

My answer is -8pi cubic feet per hour. The answer is negative since
the volume is decreasing. I think this means that (b) is the correct
response. Here's how I thought about the problem.

Draw a right triangle representing a cross-section of the cone. The
horizontal side is 10 and the vertical side is 20.  Starting at the
bottom of the vertical side, go up a distance y (the depth of the water
at any time); let the radius of the cone of water of depth y be x.
From similar triangles, y = 2x.

The volume of the water at this time is V=(pi/3)*x^2 *y.  Rewrite
this in terms of y, obtaining V=(pi/12)*y^3.  Both V and y are
functions of time t. Differentiate this equation with respect to y.
You should  obtain dV/dt=(pi/4)*y^2*(dy/dt).

We want dV/dt when y = 8; it is given that dy/dt = -1/2. This rate is
negative since y is decreasing with t. So  dV/dt = -8pi.

-Doctor Jerry,  The Math Forum

```
Associated Topics:
High School Calculus

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