Question

The radius of a spherical balloon increases at the rate of 0.3 cm/sec. Find the rate of increase of its surface area, when the radius is 5 cm.

Answer 1

Dear Student:

We know,

$\frac{dr}{dt} =0.3cm/s$

Let Surface area is S  

$S=4\pi r^{2}$

$\frac{dS}{dt} =8\pi *5*0.3$

$\frac{dS}{dt} =12\pi cm^2/s$

Hope it helps

Thanks

With Regards

Answer 2

radius of spherical balloon increases at the rate , dr/dt = 0.3 cm/sec .

we know, surface area of sphere , A = 4πr²
hence, A(r) is a function depends on value of r.
so, A(r) = 4πr²
differentiate with respect to time both sides,
dA(r)/dt = 8πr. dr/dt

it is given that radius of balloon, r = 5cm
so, dA(r)/dt = 8π × 5 × 0.3
= 12π cm²/sec

hence, rate of increase of its surface area is 12π cm²/sec