Water is flowing into a vertical cylindrical tank of diameter 4 m at the rate of 2 m3/min. Find the rate at which the depth of the water is rising. (Round your answer to three decimal places.)
volumetank=PI*2^2*h
dv/dt=4PI* dh/dt
dh/dt=1/2PI m^3/min
the area of the water's surface is 4π m^2
since volume = area * thickness, the thickness is changing by
2/(4π) = 1/(2π) m/min
We can solve this problem using related rates. Let's denote the depth of the water in the tank as h (in meters) and the time as t (in minutes).
Given:
- Diameter of the cylindrical tank = 4 m
- Rate of water flowing into the tank = 2 m^3/min
We need to find the rate at which the depth of the water is rising, which can be represented by the derivative dh/dt.
The volume V of a cylindrical tank is given by the formula V = πr^2h, where r is the radius of the tank. In this case, the diameter is 4 m, so the radius r is half of the diameter, r = 4/2 = 2 m.
Differentiating both sides of the volume equation with respect to time (using the chain rule), we get:
dV/dt = dπr^2h/dt
Since the tank is vertical, the area of the tank's base (πr^2) remains constant, so we can rewrite the equation as:
dV/dt = π(2r)(dh/dt)
Now, substituting the given values:
dV/dt = 2 m^3/min
r = 2 m
We can solve for dh/dt by rearranging the equation:
dh/dt = (dV/dt) / π(2r)
Substituting the given values:
dh/dt = 2 / π(2)(2)
Simplifying:
dh/dt ≈ 0.159
Therefore, the rate at which the depth of the water is rising is approximately 0.159 m/min.
To find the rate at which the depth of the water is rising, we need to use the concept of related rates.
Let's denote the depth of the water in the tank as "h" (in meters) and the radius of the tank as "r" (which is half the diameter).
Given that the diameter of the tank is 4 meters, the radius would be r = 4/2 = 2 meters.
We are also given that the water is flowing into the tank at a rate of 2 m³/min. This is the rate at which the volume of water in the tank is increasing.
The volume of a cylinder can be calculated using the formula V = πr²h, 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 = π(2r)(dh/dt) + πr² (dh/dt)
The first term represents the rate at which the area of the base is changing, and the second term represents the rate at which the height is changing.
Since the flow rate is given as 2 m³/min, we have dV/dt = 2.
We can substitute the values into the equation:
2 = π(2(2))(dh/dt) + π(2)²(dh/dt)
Simplifying further:
2 = 8π(dh/dt) + 4π(dh/dt)
Combining like terms:
2 = 12π(dh/dt)
Now, we can solve for (dh/dt), which represents the rate at which the depth of the water is rising:
(dh/dt) = 2 / (12π)
(dh/dt) = 1 / (6π)
(dh/dt) ≈ 0.053 m/min
Therefore, the rate at which the depth of the water is rising is approximately 0.053 meters per minute.