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

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

contact me if you have any questions you are welcome.

to complete this square you have to rotate about the origin 270 degrees clockwise.