Mathematics

Problem

A storage tank is shaped like a frustum of a right circular cone. The tank is

10\text{ m}

tall with a bottom radius of

2\text{ m}

and a top radius of

4\text{ m}

. The tank is draining at a constant rate of

6\text{ m}^3\!\!/\!\text{min}

. The water depth

h

is measured from the bottom. When

h=5\text{ m}

, what is the rate of change of the water depth

\dfrac{dh}{dt}

? (The volume of a frustum of a cone is

V=\tfrac{1}{3}\pi h\bigl(r_1^{2}+r_1r_2+r_2^{2}\bigr)

, where

r_1

and

r_2

are the radii of the lower and upper faces, respectively.) a)

-\dfrac{3}{8\pi}\,\text{m/min}

-\dfrac{2}{3\pi}\,\text{m/min}

\dfrac{2}{3\pi}\,\text{m/min}

-\dfrac{3}{2\pi}\,\text{m/min}