<s>[INST] Rate if change of surface area when volume of cylinder is increasing. I get the same answer as the text book for part a) of the following question but cannot agree with it for part b).
The radius of the base of a right circular cylinder is r cm and its height is 2r cm. Find a) the rate at which its volume is increasing, when the radius is 2 cm and is increasing at 0.25 cm/s. b) the rate at which the total surface area is increasing when the radius is 5 cm and the volume is increasing at $5\pi$ cubic cm per second.
I have said let V = volume and A = area.
a)
$V = 2 \pi r^3$ and $A = 4 \pi r^2$
$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$
$= \frac{6 \pi r^2}{4} = 6 \pi$ when r = 2.
b)
$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$
$5 \pi = 6 \pi r^2.\frac{dr}{dt}$
$\frac{dr}{dt} = \frac{5 \pi}{6 \pi r^2} = \frac{5}{6r^2}$
$dA/dt = \frac{dA}{dr}.\frac{dr}{dt}$
$dA/dt = 8 \pi r.\frac{5 \pi}{6 \pi r^2} = \frac{4 \pi}{3}$
But the book says $2 \pi$
Is the book wrong?
 [/INST] $A=6 \pi r^2$.  Four of them come from the cylindrical surface, which is $2 \times 2\pi$ and two from the ends, which are $\pi r^2$ each.
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