Calculus help! :) (urgent)

Question

A volcanic cone spews fine particles of rock which land in the surrounding area and cause the size of the cone to increase.

(a) Suppose that a volcanologist takes measurements that show that when the cone is 100 meters wide and 50 meters tall, its height is increasing at 0.5 meters/day. If the assumption is made that the cone will retain its height-width proportion, what is the rate at which the volume of rock is being released by the volcano?

 

1250pi m³/day

(b) Suppose that the volcanologist takes more measurements and realizes the original assumption was poorly made, because the width is increasing at 0.6 meters/day. What would be the new finding for the rate at which the volume of rock is being released by the volcano?

 10471.98 is not right idk what i am doing wrong. help

 m³/day

Answer 1

The assumption is that by width of the cone is meant the diameter of the base. R=1/2 D and H= 1/2 D, H=R

V = (1/3) π R²H

For part b) you are given dD/dt and dH/dt at the point in time when D=100 and H = 50. The assumption that the ratio between width and height remains constant is rejected.

V = (1/3) π (D/2)²(H) = 1/12 π D²H

dV/dt = 1/12 π [D² dH/dt + 2DH dD/dt] (note the use of the product rule)

At the point in time where D = 100:

dV/dt = 1/12 π [10000 (1/2) + 2(100)(50) (3/5)]

dV/dt = 1/12 π [5000 + 6000]

dV/dt = 11000 π /12 = 2750π/3 ≈ 2879.79 meters³/day