Water is leaking out of an inverted conical tank at a rate of 500 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 26 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
i know V=1/3pi r^2 h
i get: 413.5121 and it's not right, neither is 4135.121 or 41351.2 i don't know what im doing wrong.
No, that is not the volume of the tank. r varies with h in a cone. relate r to h, then make the Volume formula. From the height being 10 m, and r at the top is 4.5 m, then it appears that r= 4.5(h/10)
V=1/3pi [4.5(h/10)]^2 h
Take the derivative of that.
Then you know that dV/dt= ratein-rate out.
change the 500cm^2/min to m^3/min, so your dimensions are consistent.
To solve this problem, let's first find the relationship between the radius and the height of the tank. We know that the diameter at the top of the tank is 4.5 meters and the height is 10 meters.
The relationship between the radius and the height can be determined by similar triangles. Since the top diameter is 4.5 meters, the top radius is half of that, which is 2.25 meters. And since the height is 10 meters, the radius at that height can be calculated as follows:
radius = 2.25 * (height / 10)
Now that we have the relationship between the radius and the height, we can find the volume of the tank and its rate of change.
The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h
Substituting the relationship we found earlier for the radius, this formula becomes:
V = (1/3) * π * (2.25 * (h/10))^2 * h
Next, take the derivative of the volume formula with respect to time to find the rate of change of the volume. This will give us an equation:
dV/dt = (1/3) * π * (2.25 * (h/10))^2 * dh/dt
Now, let's find the rate of change of the volume, which is equal to the rate at which water is being pumped into the tank minus the rate at which water is leaking out.
Given in the problem, the rate at which water is leaking out of the tank is 500 cubic centimeters per minute. Convert this to cubic meters per minute to match the units of the volume:
leak rate = 500 cm^3/min * (1 m/100 cm)^3 = 0.005 m^3/min
Since the problem asks for the rate at which water is being pumped into the tank, we'll denote it as ratein.
Now we can set up the equation:
ratein - leak rate = dV/dt
Substituting the formula for dV/dt that we derived earlier, we get:
ratein - 0.005 = (1/3) * π * (2.25 * (h/10))^2 * dh/dt
We are also given in the problem that the water level is rising at a rate of 26 centimeters per minute when the height of the water is 1.0 meters. This means that:
dh/dt = 26 cm/min * (1 m/100 cm) = 0.26 m/min
Substituting this value into the equation, along with the values for leak rate and other constants, we can solve for ratein:
ratein - 0.005 = (1/3) * π * (2.25 * (1/10))^2 * 0.26
ratein - 0.005 = (1/3) * π * 0.05625 * 0.26
ratein - 0.005 = 0.012 * π
ratein = 0.012 * π + 0.005
Evaluating this expression, we get:
ratein ≈ 0.012 * 3.14159 + 0.005 ≈ 0.0377 + 0.005 ≈ 0.0427 m^3/min
Finally, to convert this rate to cubic centimeters per minute, remember that 1 cubic meter is equal to 1,000,000 cubic centimeters. So, we just need to multiply the ratein by 1,000,000:
ratein ≈ 0.0427 m^3/min * 1,000,000 cm^3/m^3 ≈ 42,700 cm^3/min
Therefore, the rate at which water is being pumped into the tank is approximately 42,700 cubic centimeters per minute.