Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 13 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 17 feet high

Answer 1

Answer:

0.00636 ft/min

Step-by-step explanation:

considering 'r' as radius of base at time t

    'h' as height at time t

    V = volume of cone at time t

 

Given:  dV/dt = 10  ft³/min

Find: dh/dt when h = 17 ft

 

As we know that equation of volume is defined as

V = (1/3)πr²h  

[The diameter of the base of the cone is approximately three times the altitude.therefore, 2r = 3h, r = (3/2)h  ]

 

V = (1/3)π[(3/2)h]2h

 

  = (3/4)πh³  

 

= ()πh²()

 

13 = () π (17)²(dh/dt)

$\frac{13.4}{9.289.\pi }$ = $\frac{dh}{dt}$

 

$\frac{dh}{dt}$ = $\frac{52}{2601\pi }$

$\frac{dh}{dt}$ = 0.00636 ft/min

therefore, the pile is changing at the rate of  0.00636 ft/min