Expert: Ahmed Salami - 6/29/2011

Question
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has a radius of 2 in? The depth of the soda decreases at a constant rate of 0.25in/second.

Answer
Hi John,
The volume of a cylinder, V = πr²h
where r is the radius and h is the height.
When soda is sucked out of the can, the volume and height change but the radius is constant. We can therefore find the rate of change of the volume with respect to h, which in turn can be used to evaluate the rate with respect to time.
By differentiating,
dV/dh = πr² = 4π sq-in
But,
dV/dt = dV/dh . dh/dt
= 4π . (-0.25)
= -π cu-in/s

Regards