A cylindrical tank with a circular base of 8m in diamante is being filled at 4m3/min. How fast is the level rising when it is half full?
Known:
d=8m
2r=d
r=d/2
d(v)/dt=4.0m3/min
Confusion: How to apply the "half full" concept.
it is not clear to me the orientation of the cylinder. Vertical, or Horizontal?
It is not specified in the question. It is a tank, so perhaps up right, vertical?
if it is upright, then the fluid level rises at a constant rate.
The cross-section is 16π m^2, so
dh/dt = dV/dt / 16π = 4.0/16π = 1/(4π) m/min
To apply the "half full" concept, we need to determine the volume of the cylindrical tank when it is half full.
The formula for the volume of a cylinder is V = πr^2h, where V is the volume, r is the radius of the base, and h is the height.
Since the tank is cylindrical and has a circular base with a diameter of 8m, the radius is half of the diameter, which is 4m.
Let the height when the tank is half full be h. Since we know the radius is 4m, we can use this information to find the value of h.
When the tank is half full, the volume is half the total volume of the tank. The total volume of the tank is given by V = π(4m)^2h = 16πh.
So, when the tank is half full, the volume is (1/2) * 16πh = 8πh.
Now we have the volume, and we know the rate at which the tank is being filled, which is given as dV/dt = 4m^3/min.
To find how fast the level is rising (dh/dt) when the tank is half full, we can differentiate both sides of the volume equation with respect to time:
d/dt (8πh) = d/dt (V)
8π(dh/dt) = 4
Now we can solve for dh/dt:
dh/dt = 4/(8π) = 0.1591 m/min
Therefore, the level of the tank is rising at a rate of approximately 0.1591 meters per minute when it is half full.