A conical tank has a height that is always 3 times its radius. If water is leaving the tank at the rate of 50 cubic feet per minute, how fast if the water level falling in feet per minute when the water is 3 feet high? Volume of a cone is V=1/3(pi)r^2h
A. 1.000
B. 5.305
C. 15.915
D. .589
E. 1.768
To find the rate at which the water level is falling, we need to determine the rate at which the height of the water is changing with respect to time.
Given that the height of the tank is always 3 times its radius, we can express the radius in terms of the height. Let's assume the radius is r and the height is h. If the height is always 3 times the radius, then we have the equation h = 3r.
The volume of a cone is given by V = (1/3) * π * r^2 * h. Since we want to find the rate at which the water level is falling, we take the derivative of the volume with respect to time, which will give us the rate of change.
dV/dt = (1/3) * π * [(2r * dr/dt * h) + (r^2 * dh/dt)]
We are given that the water is leaving the tank at a rate of 50 cubic feet per minute, which is dV/dt. At the water level of 3 feet, we want to find dh/dt, the rate at which the height is changing.
Substituting the known values into the equation, we have:
50 = (1/3) * π * [(2r * dr/dt * h) + (r^2 * dh/dt)]
Since h = 3r, we can substitute h = 3r into the equation:
50 = (1/3) * π * [(2r * dr/dt * 3r) + (r^2 * dh/dt)]
Now, we can solve for dh/dt by rearranging the equation:
50 = (1/3) * π * [6r * dr/dt + r^2 * dh/dt]
Multiplying everything by 3/π to isolate dh/dt:
dh/dt = 50 * (3/π) * 1 / [6r + r^2 * (3/π)]
At the water level of 3 feet, we can substitute r = h/3 = 3/3 = 1 into the equation:
dh/dt = 50 * (3/π) * 1 / [6 * 1 + 1^2 * (3/π)]
dh/dt = 50 * (3/π) * 1 / (6 + 1 * (3/π))
dh/dt = 50 * (3/π) * 1 / (6 + 3/π)
Let's calculate this value:
dh/dt = 50 * (3/π) * 1 / (6 + 3/π)
dh/dt = 50 * 3 / (π * (6 + 3/π))
dh/dt = 150 / (6π + 3)
dh/dt ≈ 150 / (18.849 + 3)
dh/dt ≈ 150 / 21.849
dh/dt ≈ 6.868 feet per minute
Therefore, the water level is falling at a rate of approximately 6.868 feet per minute when the water is 3 feet high.
The correct answer is not provided in the options given.
To find the rate at which the water level is falling, we need to differentiate the volume of the cone with respect to time (t).
Let's start by writing the formula for the volume of the cone:
V = (1/3)πr^2h
Given that the height (h) is always 3 times the radius (r), we can substitute this relationship into the volume formula:
V = (1/3)πr^2(3r)
Simplifying:
V = πr^3
Now we can differentiate this equation with respect to time (t) using the chain rule:
dV/dt = d(πr^3)/dt
To find dV/dt, we need to differentiate each term:
dV/dt = 3πr^2(dr/dt)
We are given that water is leaving the tank at a rate of 50 cubic feet per minute, which means dV/dt = -50 (negative because the volume of water is decreasing).
Now we can substitute the given information and solve for the rate at which the water level is falling.
-50 = 3πr^2(dr/dt)
Now we need to find the value of r when the water level is 3 feet.
Since the height (h) is always 3 times the radius (r), when the water level is 3 feet, we can substitute this value into the equation:
3 = 3r
Simplifying:
r = 1
Now we can substitute the value of r into the equation:
-50 = 3π(1)^2(dr/dt)
Simplifying further:
-50 = 3π(dr/dt)
Finally, we can solve for dr/dt, which represents the rate at which the water level is falling:
dr/dt = -50/(3π)
Approximating π to 3.14159, we can calculate:
dr/dt ≈ -16.835
Thus, the water level is falling at a rate of approximately 16.835 feet per minute when the water is 3 feet high.
None of the given options (A, B, C, D, E) match this value, so it seems there may be an error in the answer choices provided.