A tank has the shape of an inverted circular cone with a base radius of 5 meters
and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters
per minute, find the rate at which the water level is rising when the water is 7
meters deep
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To find the rate at which the water level is rising, we need to use related rates and the volume of a cone formula.
The formula for the volume of a cone is:
V = (1/3) * π * r^2 * h
Where:
V = volume of the cone
π = pi (approximately 3.14159)
r = base radius of the cone
h = height of the cone
Given:
r = 5 meters
h = 20 meters
We want to find the rate at which the water level is rising when the water is 7 meters deep. Let's call this rate dh/dt, where h is the height of the water.
To solve this problem, we need to differentiate the volume equation with respect to time (t):
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
The value of dV/dt is given as 2 cubic meters per minute, so dV/dt = 2.
Substituting the given values into the equation, we have:
2 = (1/3) * π * (2 * 5 * dr/dt * 20 + 5^2 * dh/dt)
Simplifying, we have:
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
Now we plug in the value of h = 7 meters and solve for dh/dt:
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)
2 = (10/3) * π * (100 * dr/dt + 25 * dh/dt)+
To find the rate at which the water level is rising, we need to calculate the derivative of the height of water with respect to time.
Let's consider the volume of water in the tank at a certain height 'h'. The volume of a cone can be calculated using the formula V = (1/3) * π * r^2 * h, where 'V' is the volume, 'r' is the radius of the base, and 'h' is the height.
Given that the base radius is 5 meters, the volume of water when the tank is filled to height 'h' is V = (1/3) * π * 5^2 * h.
Now, we can differentiate both sides with respect to time 't' to find the rate at which the volume of water changing.
dV/dt = (1/3) * π * 5^2 * dh/dt
Here, 'dh/dt' represents the rate at which the height is changing. Since water is being pumped into the tank at a constant rate of 2 cubic meters per minute, we have dh/dt = 2.
Now we can substitute the given values into the equation:
dV/dt = (1/3) * π * 5^2 * 2
= (1/3) * π * 25 * 2
= (2/3) * π * 25
Therefore, the rate at which the water level is rising when the water is 7 meters deep is (2/3) * π * 25 cubic meters per minute.