A cylindrical container of fixed length 90 cm is being pressure tested, and the radious id increasing at a constant rate of .01 cm/min. WHen the redious is 25 cm, find the rate of the change of:
(a) the total surface area
(b) the Volume
A cylindrical container of fixed length 90 cm is being pressure tested, and the radious id increasing at a constant rate of .01 cm/min. WHen the redious is 25 cm, find the rate of the change of:
(a) the total surface area
(b) the Volume
So we have:
$\displaystyle V = (90)(\pi)(r^2)$
and
$\displaystyle SA = 2\pi r^2+2\pi r (90) $
For volume we have:
$\displaystyle \frac{dV}{dt} = (180)(\pi)(r)\frac{dr}{dt}$
At the values you gave we have:
$\displaystyle \frac{dV}{dt} = (180)(\pi)(25)(0.01)$
Thats the rate of change of the volume, can you figure it out the same way for surface area?