Answer to Question #234826 in Calculus for Jese Junior

Question

Air is pumped into a spherical balloon such that the surface area of the balloon increases at a rate of 10 cm2 per second. Find the rate at which the radius is increasing when the radius is 4 cm. [ i.e. find when r = 4] [Note : surface area, A = 4πr2 ] 

Answer 1

Let the rate of area change be $\dfrac{dA}{dt} = 10cm^2/s$. Assuming that $A = 4\pi r^2$ and applying chain rule, obtain:

$$\dfrac{d}{dt}(4\pi r^2) = 4\pi\cdot 2r\cdot \dfrac{dr}{dt} = 10cm^2/s$$

where $\dfrac{dr}{dt}$ is the rate at which the radius is increasing. Expressing this rate and substituing $r = 4cm$ (current radius), obtain:

$$\dfrac{dr}{dt} = \dfrac{10cm^2/s}{8\pi r} = \dfrac{10cm^2/s}{8\pi \cdot 4cm} \approx 0.1cm/s$$

Answer. 0.1 cm/s.