# A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10 cm

**Solution:**

Derivatives are used to find the rate of changes of a quantity with respect to the other quantity. By using the application of derivatives we can find the approximate change in one quantity with respect to the

change in the other quantity.

They are used in many situations like finding maxima or minima of a function, finding the slope of the curve, and even inflection points.

We know that V = 4/3π r^{2}

Therefore,

dV/dt = d/dt (4/3π r^{2})

= 4/3π (3r^{2})

= 4π r^{2}

When radius, r = 10 cm

Then,

dV/dt = 4π (10)^{2}

= 400 π

Thus, the volume of the balloon is increasing at the rate of 400π cm^{3}/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 9

## A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10 cm

**Summary:**

The rate at which its volume is increasing with the radius when the latter is 10 cm is 400π cm^{3}/s. By using the application of derivatives we can find the approximate change in one quantity with respect to the change in the other quantity