Sand is falling into a conical pile so that the radius of the base of the pile is always equal to one half its altitude. If the sand is falling at the rate of 10 cubic feet per minute, how fast is the altitude of the pile increasing when the pile is 5 feet deep?
volumepile=1/3 (PI r^2)h
but r=h/2, so
volume=1/12 PI h^3
dv/dt=10 ft^3/min
but dv/dt=1/12 PI 3h^2 dh/dt
solve for dh/dt
This assumes you mean by "altitude" the height. If you mean altitude as slant height, you have to adjust the fromula
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To solve this problem, we can use related rates and the formula for the volume of a cone.
Let's denote the altitude of the pile as h and the radius of the base as r.
From the given information, we know that the radius is always equal to one-half the altitude:
r = (1/2)h
The volume of a cone is given by the formula:
V = (1/3)πr^2h
We need to find dh/dt, the rate at which the altitude is changing, when h = 5 feet.
Step 1: Find an expression for the volume of the cone in terms of h only.
Substitute the expression for r in terms of h into the formula for the volume of a cone:
V = (1/3)π[(1/2)h]^2h
V = (1/3)π(1/4)h^3
V = (1/12)πh^3
Step 2: Find an expression for the rate of change of the volume with respect to time, dV/dt.
We are given that the sand is falling at a rate of 10 cubic feet per minute, so the rate of change of the volume is constant:
dV/dt = 10 ft^3/min
Step 3: Differentiate the volume equation with respect to time, t.
dV/dt = (1/12)π * 3h^2 * dh/dt
10 = (1/4)πh^2 * dh/dt
Step 4: Solve for dh/dt.
We want to find dh/dt when h = 5 feet. Substitute h = 5 into the equation:
10 = (1/4)π(5)^2 * dh/dt
10 = (1/4)π(25) * dh/dt
10 = (25/4)π * dh/dt
dh/dt = 10 * 4/(25π)
Simplifying the expression, we find:
dh/dt = 16/(5π) ft/min
Therefore, when the pile is 5 feet deep, the altitude of the pile is increasing at a rate of 16/(5π) ft/min.
To solve this problem, we can use related rates. We need to find how fast the altitude of the pile is changing with respect to time when the pile is 5 feet deep.
Let's denote the radius of the base of the pile as r and the altitude (height) of the pile as h. According to the problem, the radius is always half of the altitude, so we have the equation:
r = 1/2h
We are given that sand is falling at a rate of 10 cubic feet per minute. This means that the volume V of the pile is increasing at a rate of 10 cubic feet per minute. The volume of a cone is given by:
V = (1/3)πr^2h
Since r = 1/2h, we can substitute it into the volume equation:
V = (1/3)π(1/2h)^2h
V = (1/12)πh^3
Differentiating both sides of the equation with respect to time t gives:
dV/dt = (1/4)πh^2 dh/dt
We know that dV/dt = 10 cubic feet per minute, so we have:
10 = (1/4)πh^2 dh/dt
Now we need to find dh/dt, which represents the rate at which the altitude h is changing with respect to time t. We can solve for dh/dt:
dh/dt = 40/(πh^2)
To find the rate at which the altitude is changing when the pile is 5 feet deep, we substitute h = 5 into the equation:
dh/dt = 40/(π(5)^2)
dh/dt = 40/(25π)
dh/dt = 8/π
Therefore, the altitude of the pile is increasing at a rate of 8/π feet per minute when the pile is 5 feet deep.