Sand is falling off a conveyor belt and onto a conical pile at a rate of 12 cubic feet per minute. The diameter of the base of the cone is approximately 4 times the height of the pile. A) At what rate is the height of the pile changing when the pile is 10 feet high? B) At what rate is the area of the base fo the pile changing when the pile is 10 feet high?
If the pile has height h, then its base has radius 2h, and its volume is
v = 1/3 πr^2 h = 1/3 π (2h)^2 h = 4/3 πh^3
dv/dt = 4πh^2 dh/dt
Now just plug in your numbers
To solve this problem, we can use related rates, which involve differentiating with respect to time.
Let's denote the height of the pile as h (in feet) and the radius of the base of the cone as r (also in feet).
Given:
- The rate of change of sand falling off the conveyor belt onto the conical pile is 12 cubic feet per minute.
We need to find:
A) The rate at which the height of the pile is changing when the pile is 10 feet high.
B) The rate at which the area of the base of the pile is changing when the pile is 10 feet high.
To solve both parts, we'll need two equations relating the variables:
1) The relationship between the height and the radius of the cone:
Since the diameter of the base of the cone is approximately 4 times the height of the pile, we can write:
2r = 4h
Simplifying, we get:
r = 2h
2) The formula for the volume of a cone:
The volume (V) of a cone is given by:
V = (1/3)πr^2h
Now, let's find the rates of change using these equations:
A) Rate at which the height of the pile is changing:
We're given dV/dt (rate at which the sand is falling off the conveyor belt) = 12 cubic feet per minute.
Differentiating the volume equation with respect to time (t), we get:
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr^2(dh/dt)
12 = (1/3)π(2h)(dh/dt) + (1/3)π(2h)^2(dh/dt)
12 = (4/3)πh(dh/dt) + (4/3)π(4h^2)(dh/dt)
12 = (4/3)π(dh/dt)(h + 16h^2)
dh/dt = 3/(4π(h(1 + 16h^2) )
We can plug in h = 10 into this equation to find the rate at which the height is changing when the pile is 10 feet high:
dh/dt = 3/(4π(10(1 + 16(10^2))
B) Rate at which the area of the base is changing:
The area of the base of the cone (A) is given by:
A = πr^2
Differentiating the base area equation with respect to time (t), we get:
dA/dt = 2πr(dr/dt)
Plugging in the value of r using the relation r = 2h, we get:
dA/dt = 2π(2h)(dh/dt)
dA/dt = 4πh(dh/dt)
Now, we can plug in h = 10 into this equation to find the rate at which the area of the base is changing when the pile is 10 feet high:
dA/dt = 4π(10)(dh/dt)
Note: To find the exact values, you'll need to evaluate the expressions numerically using a calculator.
To solve this problem, we can use related rates, which is a technique in calculus used to find the rate at which one quantity is changing with respect to another. In this case, we need to find the rates at which the height and the area of the base of the pile are changing when the pile is 10 feet high.
Let's denote the height of the pile as h and the radius of the base as r. Since the diameter of the base is approximately 4 times the height of the pile, we can write r = 4h.
Part A: Finding the rate at which the height of the pile is changing.
We are given that sand is falling off the conveyor belt onto the pile at a rate of 12 cubic feet per minute. This means that the volume of the pile is increasing at a rate of 12 cubic feet per minute. The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
Differentiating both sides of the equation with respect to time (t), we get dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt)), where dV/dt is the rate of change of volume, r is the radius, h is the height, and dr/dt and dh/dt are the rates of change of radius and height, respectively.
Since we know that dV/dt = 12, r = 4h, and we need to find dh/dt when h = 10, we can rewrite the equation as 12 = (1/3)π(2(4h)(4dh/dt) + (4h)^2(dh/dt)).
Simplifying the equation, we get 12 = (8/3)π(4h^2(dh/dt) + 16h(dh/dt)).
Now, substitute h = 10 into the equation to solve for dh/dt.
12 = (8/3)π(4(10)^2(dh/dt) + 16(10)(dh/dt)).
Simplifying further, we get 12 = (8/3)π(400(dh/dt) + 1600(dh/dt)).
Dividing both sides of the equation by (8/3)π(400 + 1600), we get dh/dt = 12 / ((8/3)π(2000)).
Calculating the value, we find that dh/dt ≈ 0.000239 ft/min (rounded to 6 decimal places).
Therefore, the height of the pile is changing at a rate of approximately 0.000239 feet per minute when the pile is 10 feet high.
Part B: Finding the rate at which the area of the base of the pile is changing.
The area of the base of a cone can be calculated using the formula A = πr^2, where A is the area and r is the radius.
Differentiating both sides of the equation with respect to time (t), we get dA/dt = 2πr(dr/dt), where dA/dt is the rate of change of the area, r is the radius, and dr/dt is the rate of change of the radius.
As we know that r = 4h and we need to find dA/dt when h = 10, we can rewrite the equation as dA/dt = 2π(4h)(dr/dt).
Substituting h = 10 into the equation, we have dA/dt = 2π(4(10))(dr/dt) = 2π(40)(dr/dt).
Since we don't have the value for dr/dt, we cannot find the exact rate at which the area of the base is changing without additional information.