Helium is pumped into a spherical balloon at a rate of 25m^3/min. At what rate is the radius of the balloon changing when the diameter is 4m?

Question

campbell_st · Answer

well it looks like a related rates question 

the volume of the sphere is

$$V = \frac{4}{3} \pi r^3$$

find dV/dr, the rate of change in the volume with respect to the radius

you know 

$$\frac{dV}{dt} = 25 $$

and you need to find 

$$\frac{dr}{dt}$$

so using related rates

$$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$$

make 

$$\frac{dr}{dt} ~~the ~~subject$$

then substitute r = 2 

hope it helps