Mathematics

Problem

The concentration of a chemical in a vertical cylindrical tank is modeled by

C(V)=\frac{100}{V^{2}+1},

where

C

is measured in grams per liter and

V

is the volume of liquid in the tank, measured in cubic meters. The tank has a constant circular cross-section of radius

2\text{ m}

, so the volume satisfies

V=4\pi h

, where

h(t)

(in meters) is the depth of the liquid at time

t

(in seconds). At the instant when

h=3\text{ m}

and the depth is decreasing at a rate of

0.10\text{ m/s}

, what is

\dfrac{dC}{dt}

to the nearest

5\times10^{-4}\;

g/L·s? a) Approximately

\;4.7\times10^{-2}

b) Approximately

\;4.7\times10^{-3}

c) Approximately

\;-4.7\times10^{-3}

d) Approximately

\;4.7\times10^{-4}