Volume, height and radius

Question

sand is leaking from.the back of a dump.truck.and forming a.conical.pile on the ground. the sand is leaking at the rate of 0.7m3 per hour. if the base radius of the pile is always 0.2 times the height, how fast is the base radius changing when the height is 1.1m

Answer 1

The volume V of the conical pile = (1/3)πr²h.

Since r = 0.2h, write h = 5r and V = (1/3)πr²(5r) or (5/3)πr³.

Next, D_tV =5πr² • D_tr. Given that D_tV = 0.7 (m³/hr), it follows that 0.7 (m³/hr) = 5πr² • D_tr.

When h = 1.1 m, r = (1.1/5) m and 0.7 (m³/hr) = 5π(1.1/5)² (m²)• D_tr (m/hr).

Then D_tr, the rate of change for the base radius, is given by 0.7(25)/((5π(1.1)²) or 3.5/(1.21π) (m/hr), equal to 0.9207310757 meters per hour.