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

We know that , surface area of spherical balloon $(A)=4\pi r^2$

Differentiating both sides w.r.t $t$ , we get

$\frac{dA}{dt}=\frac{d}{dt}(4\pi r^2) \\\frac{dA}{dt}=4\pi\frac{d}{dt}( r^2) \\\frac{dA}{dt}=8\pi r\frac{dr}{dt}...(1)$

Now, it is given that

$\frac{dA}{dt}=10\ cm^2/s$ and we need to find $\frac{dr}{dt}$ at $r=4$ .

So, from equation (1), we get

$10=8\pi \times 4\times \frac{dr}{dt} \\10=8\times\frac{22}{7} \times 4\times \frac{dr}{dt} \\\frac{dr}{dt}=\frac{35}{352}=0.994 \ cm/s$