At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 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 feet high?

Dh/dt = 0,141 ft/min

Step-by-step explanation:

Volume of a cone is:

V(c)  = 1/3 *π*r²*h      (1)

r is radius of the base,  and h is the height.

As the diameter ( 2*r) is three times h

2r = 3h    ⇒  r = (3/2)*h

We have DV(t)/dt  = 1 ft³

And from equation (1)

V(c)  = 1/3 *π*r²*h      ⇒  V(t)  = 1/3 *π* (3/2*h)² * h

V(t)  = 1/3 *π*9/4 *h³    ⇒   V(t)  = 3/4*π*h³

Taking derivatives on both sides of the equation

DV(t)/dt  = 3/4*π* ( 3 h² Dh/dt)

By substitution we get

1 = 9/4*π*h²* Dh/dt   solving for Dh/dt    when h = 1 ft

1 = 9/4 *π*(1)²*Dh/dt

Dh/dt = ( 4/9)/π

Dh/dt = 0,141 ft/min

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.

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