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

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

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4