Question

Sand is pouring from a pipe at the rate of 12 cm^3 /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 the height is 4 cm?

Answer 1

Let r is the radius , h is height , V is the volume of cone.
Given, h = r/6 => r = 6h
and dV/dt =12cm³/s -------(1)
h = 4cm ------(2)
now, volume of cone , V = 1/3 πr²h
V = 1/3 × π(6h)²h
V = 1/3 × π × 36h³ = 12πh³
V = 12πh³
now, differentiate it with respect to time t,
dV/dt = 36πh².dh/dt
12 = 36π(4)² × dh/dt
12 = 36π × 16 × dh/dt
1 = 3π × 16 × dh/dt
1/48π = dh/dt
hence, dh/dt = 1/(48π) cm/s

Therefore, the height of the sand cone increasing when the height is 4 cm is 1/(48π) cm/s

Answer 2

HELLO DEAR,
At any instant t, let r be the radius, h the height and V the volume of the cone.

then , h = r/6
r = 6h.

therefore, V = 1/3πr²h = 1/3π(6h)²h.
V = 12πh³.
now, dV/dt = DV/dh × dh/dt

12 = d(12πh³)/dh × dh/dt

12 = 36πh² × dh/dt

dh/dt = 1/(3πh²)

$\bold{\frac{dh}{dt}_{h = 4cm} = \frac{1}{3*\pi *4*4}cm/sec}$

$\bold{\frac{1}{48\pi}}$cm/sec.

HENCE, the rate of increase of the height of the sand cond at the instant whene h = 4cm is $\bold{\frac{1}{48\pi}}$cm/sec..

I HOPE ITS HELP YOU DEAR,
THANKS