# Related rates of increase

1. Mar 5, 2006

### jimbo007

hey guys,
just wondering if i did this correctly

a spherical balloon is to be filled with water so that there is a constant increase in the rate of its surface area of 3cm2/sec .
a) Find the rate of increase in the radius when the radius is 3cm.
b) Find the volume when the volume is increasing at a rate of 10cm3=sec.

A=surface area, V=volume
a) i thought was fairly easy and got 1/(8Pi) cm/sec
b) was a bit more trickier here is what i did

(1) $$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$$
(2) $$\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}$$
from the question we have
$$\frac{dV}{dt}=10$$
$$\frac{dA}{dt}=3$$
$$\frac{dV}{dr}=4 \pi r^2$$
$$\frac{dA}{dr}=8 \pi r$$
from (2)
$$3=8 \pi r \frac{dr}{dt}$$
so $$\frac{dr}{dt}=\frac{3}{8 \pi r}$$
sub into (1) we get that r=20/3
so $$V=\frac{4}{3}\pi r^3=\frac{4 \pi}{3}(20/3)^3$$

2. Mar 6, 2006

### HallsofIvy

Staff Emeritus
I did it slightly differently, using
$$\frac{dV}{dA}= \frac{\frac{dV}{dt}}{\frac{dA}{dt}}= \frac{r}{2}= \frac{10}{3}$$
to find r and then V but I got exactly the same answer!