Corn is poured through a chute at the rate of 10ft^3 / min and falls in a conical pile whose bottom radius is always half the height. How fast is the circumference of the base changing when the pile is 8 feet high?

Please explain

h = 2r, so

v = 1/3 pi r^2 * 2r = 2pi/3 r^3

dv/dt = 2pi r^2 dr/dt
10 = 2pi*4^2 = 32pi dr/dt
dr/dt = 5/16pi

now, c = 2pi*r, so
dc/dt = 2pi dr/dt
= 2pi * 5/16pi = 10/16 = 5/8

Pauline jogs at 5 1/3 miles per hour for 1 5/6 hours. about how many miles has she walked?

To solve this problem, we need to apply related rates. Let's denote the height of the cone as h and the radius of the base as r. We are given that the rate at which corn is poured in is 10 ft³/min.

We are asked to find how fast the circumference of the base is changing (dc/dt) when the height of the pile is 8 feet (dh/dt at h = 8).

We know that the volume of a cone is given by V = (1/3)πr²h. Since the bottom radius is always half the height, we can write r = (1/2)h.

Differentiating the volume equation with respect to time t, we get:

dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)π(r²)(dh/dt)

We are given that dV/dt = 10 ft³/min, and we want to find dc/dt when h = 8 ft. Since the circumference of the base is C = 2πr, we can write dc/dt as d(2πr)/dt.

Now, substituting the values we know into the differentiated volume equation:

10 = (1/3)π(2(8)(dh/dt)) + (1/3)π((1/2)(8)²)(dh/dt)

Simplifying further:

10 = (16/3)π(dh/dt) + (16/3)π(dh/dt)

10 = (32/3)π(dh/dt)

Now, isolate dh/dt:

dh/dt = 10 / [(32/3)π]

Simplifying and calculating:

dh/dt = 30 / (32π)

dh/dt ≈ 0.296 ft/min

Therefore, the height of the pile is increasing at a rate of approximately 0.296 ft/min when the pile is 8 feet high.

To solve this problem, we need to use related rates, which involves finding the rate of change of one quantity with respect to another over time. In this case, we want to find how fast the circumference of the base of the conical pile is changing with respect to time when the pile is 8 feet high.

Let's break down the problem into smaller parts:

1. Let's define the variables:
- h: height of the conical pile
- r: radius of the base of the conical pile
- V: volume of the conical pile

2. We are given that corn is poured through a chute at a rate of 10 ft^3/min. This tells us that the rate of change of the volume of the pile (dV/dt) is constant and equals 10 ft^3/min.

3. The volume of a cone is given by the formula V = (1/3)πr^2h. Since the radius of the base is always half the height (r = h/2), we can rewrite the volume equation as V = (1/12)πh^3.

4. Now, let's differentiate the volume equation with respect to time (t) using the chain rule.
dV/dt = (1/12)π(3h^2)(dh/dt)

5. We know that dV/dt = 10 ft^3/min (the rate at which corn is pouring).

6. We need to find dh/dt when the height (h) of the pile is 8 ft. Substituting these values into the equation:
10 = (1/12)π(3*8^2)(dh/dt)

7. Simplifying the equation:
10 = (1/12)π(3*64)(dh/dt)
10 = 16π(dh/dt)

8. Solving for dh/dt:
dh/dt = 10 / (16π)
dh/dt ≈ 0.198 ft/min

9. Now, we need to find the rate of change of the circumference of the base (C) with respect to time (dC/dt).

10. The circumference of a circle is given by C = 2πr. Since r = h/2, we can rewrite the circumference equation as C = 2π(h/2) = πh.

11. Applying the chain rule, differentiate the circumference equation with respect to time:
dC/dt = π(dh/dt)

12. Substituting the value of dh/dt that we obtained earlier:
dC/dt = π(0.198)
dC/dt ≈ 0.623 ft/min

Therefore, the circumference of the base of the conical pile is changing at a rate of approximately 0.623 feet per minute when the pile is 8 feet high.