Calculus Differentiation Question

Question

A conical pile of rocks increases in volume at a rate of 16pi/3 cubic feet per minute. When the height of the pile is 30 feet, the radius is 10 feet. What is the rate of change of the height of the pile when the height is 9 feet?

Answer 1

Assume the ratio H/R between the height H and the radius R of the cone remains the same (in our case it is H/R=30ft/10ft=3)

therefore

H=3R or R=H/3 (1)

Now,

Let's use (1) in the formula for volume V of the cone with hight H and radius R

V=(π R² H)/3 (2)

to express V as a function of H only:

V=(π (H/3)²H)/3=(π/27)H³ (3)

where V and H are functions of time t

From (3) we can express rate of change of volume V through the rate of change of height H:

dV/dt=d/dt((π/27)H³)=(π/9)H²(dH/dt). (4)

Therefore, (dH/dt)=(dV/dt)/((π/9)H²) (5)

Then use (5) with dV/dt=16π/3(ft³/min) ; H=9ft

to get the result for dH/dt in ft/min

Hope it is useful

Dr.Ariel B.