Related Rates calculus

Question

Sand is being dropped on the ground at a steady rate of 9pi cm^3 per second, and forms a conical pile whose height remains equal to the radius of the base. At what rate is the height increasing at the eight-second mark?

Answer 1

Since the height is equal to the radius of the base, the volume is:

V = (1/3) π r²h = (1/3) π h³

Now, taking the derivative with respect to time, we have:

dV/dt = π h² dh/dt

We know that dV/dt = 9π cm³/s, but we need to know h. At the 8s mark, the volume of the pile is 72π cm³, so we plug back into our original volume equation to get:

72π = (1/3) π h³

h³ = 216

h = 6

Now, we plug everything into our derivative equation to get:

9π = π (6)² dh/dt

dh/dt = 9/(6)^2 = 1/4

Therefore, the height is increasing at a rate of 1/4 cm/s.