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3 years ago

6

At a sand and gravel plant, sand is falling off a conveyor, and onto a conical pile at a rate of 10 cubic feet per minute. The d

iameter 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 15 feet high?
I know this is a Calc 1 related rates problem,

1 answer:

kherson [118]3 years ago

5 0

Answer:

dh/dt=9.82x10^-3 ft/min

Step-by-step explanation:

1. You have that the rate is10 ft³/min. Then:

dV/dt=10

2. The formula for calculate the volume of a cone, is:

V=1/3(πr²h)

"r" is the radius and "h" is the height.

3. The diameter of the base of the cone is approximately 3 times the altitude. Then, the radius is:

r=diameter/2

diameter=3h

r=3h/2

4. When you susbstitute r=3h/2 into the formula V=πr²h/3, you have:

V=1/3(πr²h)

V=1/3(π(3h/2)²(h)

V=1/3(π9h²/4)(h)

V=9πh³/12

5. Therefore:

dV/dt=(9πh²/4)dh/dt

h=12

dV/dt=10

6. When you substitute the values of dV/dt and h into dV/dt=(9π(12)²/4)dh/dt, you have:

dV/dt=(9π(12)²/4)dh/dt

10=(1017.876)

7. Finally, you obtain:

dh/dt=10/1017.876

dh/dt=9.82x10^-3 ft/min

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