In this page rate of change question9 we are going to see solution of some practice question of the worksheet.
Gravel is being dumped from a conveyor belt at a rate of 30
ft³/min and its coarsened such that it forms a pile in the shape of cone
whose base diameter and height are always equal. How fast is the height
of the pile increasing when the pile is 10 ft high?
Let "V" be the volume of the cone at a time "t". Let r and h are the radius and height of the cone respectively.
Here we have a information that the base diameter and height are always equal. Form this we come to know that 2 r = h
volume of the gravel is being changed at the rate of 30 ft³/min
rate of change of volume (dV/dt) = 30
now we have to find how fast is the height of the pile increasing when the pile is 10 ft high
h = 10
Volume of cone (V) = (1/3) π r² h
V = (1/3) π (h/2)² h
V = (1/3) π (h³/4)
dV/dt = (1/3) π (3 h²/4) (dh/dt)
substitute h = 10 and dV/dt = 30
30 = (1/3) π [3 (10)²/4] (dh/dt)
(30 x 3 x 4)/300 π = (dh/dt)
6/5π = dh/dt
dh/dt = (6/5 π) ft/min
Therefore the height is increasing at the rate of (6/5 π) ft/min.