Question

Sand is falling into a conical pile at the rate of 10 m^3/s such that height of the pile is always half the diameter of the base of the pile find the rate at which the height of the pile is changing when the pile is 5 m. high.

Answer 1

Answer:

The rate at which the height of the pile is $\dfrac{2}{5\pi}$.

Explanation:

Given that,

Height = radius = 5 m

Rate = 10 m³/s

We need to calculate the rate at which the height of the pile

Using formula of volume

$V=\dfrac{1}{3}\pi\times r^2\times h$

r = h,

On differentiating

$\dfrac{dV}{dt}=\pi\times h^2\times\dfrac{dh}{dt}$

Put the value into the formula

$10=\pi\times25\times\dfrac{dh}{dt}$

$\dfrac{dh}{dt}=\dfrac{10}{25\pi}$

$\dfrac{dh}{dt}=\dfrac{2}{5\pi}$

Hence, The rate at which the height of the pile is $\dfrac{2}{5\pi}$.

Answer 2

Answer:

0.127 m/s

Explanation:

Let the diameter of the conical pile is d.

height of the conical pile = d/2

dV/dt = 10m^3/s

h = 5 m

then, d = 2 x h = 2 x 5 = 10 m

Volume of cone is given by

$V=\frac{1}{3}\pi r^{2}h$

where, r is the radius of the pile.

r = d/2 and h = d/2 , it means r = h

$V=\frac{1}{3}\pi \times h^{3}$

Differentiate with respect to t on both the sides

$\frac{dV}{dh}=\frac{1}{3}\pi \times 3h^{2}\times \frac{dh}{dt}$

by substituting the values

$10=\frac{1}{3}\pi \times 3\times 5\times 5\times \frac{dh}{dt}$

dh/dt = 0.127 m/s

Thus, the rate of change of height is 0.127 m/s.