Hi all,

I am fairly confident that on this occasion, my answer is correct and my maths book is wrong, but I have a niggling doubt, and thought I better take it to the experts just in case.

If the volume of a sphere increases at the rate of $6 cm^3/sec$ , find the rate of increase of the surface area of the sphere at the instant when its radius is 4 cm.

We are given $ \frac{dV}{dt} = 6 cm^3/s$. We have to find $\frac{dS}{dt}$ when r = 4 cm.

We know that for a sphere, $V = \frac{4 \pi r^3}{3}$ and $S = 4 \pi r^2$ therefore $V = \frac{S.r}{3}$

Also $\frac{dV}{dt} = \frac{dV}{dS} . \frac{dS}{dt}$

$\frac{dV}{dS} = \frac{r}{3}$

$6 = \frac{r}{3} \times \frac{dS}{dt}$

$\frac{dS}{dt} = \frac{18}{r}$ and $r = 4$

therefore $\frac{dS}{dt} = \frac{18}{4} = 4.5 cm^3/s$

My book says that the rate is $6 cm^3/s$