An inverted conical tank is being filled with water, but it is discovered that it is also leaking water at the same time. The tank is 6 meters high and its diameter at the top is 4 meters. The water is being added to the tank at a constant rate. Some of this water is found to be leaking at a rate of 8,500 cm3/min. Suppose that the water level in the tank is found to be rising at a rate of 20 cm/min when the height of the water is 2 meters. Find the rate at which water is being added to the tank.
when the water has height y, its surface has diameter 4/6 y.
So,
v = 1/3 pi y^2 (2y/3) = 2/9 pi r^3
dv/dt = 2/3 pi r^2 dr/dt
Now just plug in your numbers.
167.5516
To solve this problem, we'll use the concept of related rates. We have the following given information:
- The height of the tank is 6 meters.
- The diameter at the top of the tank is 4 meters.
- The water level is rising at a rate of 20 cm/min when the height is 2 meters.
- Water is leaking from the tank at a rate of 8,500 cm^3/min.
Let's start by finding the rate at which the water level is rising when the height is 2 meters. We can use similar triangles to relate the rates of change in height and radius.
The radius of the tank at the water level is given by the proportion:
(r1 / h1) = (r2 / h2),
where r1 is the initial radius (4/2 = 2 meters) and h1 is the initial height (2 meters). r2 is the radius at the unknown height and h2 is the unknown height (which we're looking for).
Rearranging the equation, we get:
r1 / h1 = r2 / h2,
r2 = (r1 / h1) * h2.
Substituting the known values, we have:
r2 = (2 / 2) * 6 = 6 meters.
Now we have the height and radius of the water level, and we can find the volume of the water in the tank. The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h,
where V is the volume, π is approximately 3.14159, r is the radius, and h is the height.
Substituting the known values, we have:
V = (1/3) * π * (6^2) * 2 = 24π meters^3.
Now we can differentiate the volume with respect to time to find the rate at which the water level is rising.
dV/dt = (1/3) * π * [2r * dr/dt * h + r^2 * dh/dt].
We're given that dV/dt is equal to 20 cm^3/min, r is 6 meters, dr/dt is what we're trying to find, and dh/dt is 20 cm/min.
Substituting the known values, we have:
20 = (1/3) * π * (2 * 6 * dr/dt * 2 + 6^2 * 20).
Simplifying the equation, we have:
20 = (4/3) * π * (12 * dr/dt + 720).
Now, we can solve for dr/dt:
4/3 * 12 * π * dr/dt = 1/3 * (20 - 480 * π),
dr/dt = (20 - 480 * π) / (16 * π) ≈ -29.56 m/min (negative value implies the radius is decreasing).
Finally, we have found that the rate at which the radius is decreasing is approximately -29.56 m/min. Since the water is being added to the tank at a constant rate, the rate at which water is being added to the tank is equal to the rate at which the water is leaking from the tank, which is 8,500 cm^3/min (or 8.5 L/min).