Thank you for reporting, we will resolve it shortly
Q.
The radius of a sphere initially at zero increases at the rate of 5 cm/sec. Then its volume after 1 sec is increasing at the rate of :
Application of Derivatives
Solution:
Let ‘r’ be the radius and V be the volume of the sphere.
Given : Radius increases at the rate of 5cm/sec.
$\therefore \, \frac{dr}{dt} = 5cm /\sec$
Now, $V = \frac{4}{3} \pi r^3$
$\therefore \, \frac{dV}{dt} = \frac{4}{3} \pi (3r^2 ) \frac{dr}{dt} = 4 \pi r^2 (5) = 20 \pi r^2$
Now, after one second, r = 5
$\therefore \frac{dV}{dt}$ after $1 \sec = 20 \pi(5)^2 = 500 \pi $