Expert: Ahmed Salami - 1/2/2009

Question
Let V be the volume of a circular cylinder having a height h and a radius r, and assume h and r vary with time. At a certain instant, the height is 6in and is increasing at 1 in/s, while the radius is 10in and decreasing at 1 in/s. How fast is the colume changing at that instant and is the volume increasing or decreasing?

Answer
Hi Elena,
V = #r^2.h
where # represents pi
The total differential with respect to time;
dV/dt = (&V/&h)(dh/dt) + (&V/&r)(dr/dt)
but,
&V/&h = #r^2
&V/&r = 2#rh
At the said instant,
h = 6, r = 10, dh/dt = 1, dr/dt = -1 (decreasing)
&V/&h = #.10^2 = 100#
&V/&r = 2#.10.6 = 120#
Therefore,
dV/dt = (&V/&h)(dh/dt) + (&V/&r)(dr/dt)
     = 100#.1 + 120#.(-1)
     = 100# - 120#
     = -20# cubic in/s
The volume is therefore decreasing (negative rate!)

Regards