# Thread: rate of change problem with a sphere.

1. ## rate of change problem with a sphere.

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 $6 cm^3/s$

2. ## Re: rate of change problem with a sphere.

You are incorrect. Your error is that when you differentiate $V= \frac{rS}{3}$ you are treating r as a constant- and it isn't. The volume and surface area cannot change without the radius changing.

You have, correctly, that $V= \frac{rS}{3}$. Differentiating, you should have $\frac{dV}{dt}= \frac{S}{3}\frac{dr}{dt}+ \frac{r}{3}\frac{dV}{dt}$. In order to get $\frac{dr}{dt}$, use $S= 4\pi r^2$ so that $\frac{dS}{dt}= 8\pi r^2\frac{dr}{dt}$. Since you are given that $\frac{dS}{dt}= 6$ and $r= 4$, $6= 8\pi (16)\frac{dr}{dt}$ so $\frac{dr}{dt}= \frac{3}{64\pi}$.

3. ## Re: rate of change problem with a sphere.

Many thanks for your answer, there is a slight typo $\frac{dV}{dt} = \frac{S}{3}\frac{dr}{dt} + \frac{r}{3}\frac{d\color{red}S}{dt}$ and $\frac{dS}{dt} = 6$ should be $\frac{dV}{dt} = 6$. I think this is one of the hazards of working with LaTex, but your answer was more than enough for me to figure out the problem, so once again Sir, thanks a lot!

4. ## Re: rate of change problem with a sphere.

You are welcome and thanks for pointing out the error.

5. ## Re: rate of change problem with a sphere.

$V = \dfrac{1}{3}(rS)$

$\dfrac{dV}{dt}=\dfrac{1}{3}\left(r \cdot \dfrac{dS}{dt} + S \cdot \dfrac{dr}{dt}\right)$

$18 = 4 \cdot \dfrac{dS}{dt} + 64\pi \cdot \dfrac{dr}{dt}$

note ... there is no way $\dfrac{dS}{dt} = 6$ in the above equation

$\dfrac{dV}{dt} = 4\pi r^2 \cdot \dfrac{dr}{dt}$

when $r=4 \, cm$ ...

$6 \, cm^3/sec\, = 4\pi(16 \, cm^2) \cdot \dfrac{dr}{dt}$

$\dfrac{dr}{dt}=\dfrac{3}{32\pi} \, cm/sec$

$\dfrac{dS}{dt}=8\pi r \dfrac{dr}{dt} = 3 \, cm^2/sec$