Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when height is 4 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.

We know that the Volume of a sphere is given by 1/3π r²h

It is given that,

h = 1/6 r

Hence, r = 6h

Therefore,

V = 1/3 π (6h)² h

= 12πh³

Thus,

dV/dt = 12π d/dt (h³) dh/dt

= 12π 3 h² dh/dt

= 36π h² dh/dt

We have,

dV/dt = 12 cm²/s

When h = 4 cm

Then,

On substituting the values, we get

12 = 36 π (4)² dh/dt

dh/dt = 12/(36π (16))

= 1/48π cm/s

Therefore, the rate at which the height of the sand cone increasing when height is 4 cm is 1/48π cm/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 14