An upright cylindrical tank with radius 3 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing?
To find the rate at which the height of the water is increasing, we can use the formula for the volume of a cylinder:
V = πr^2h
where V is the volume, r is the radius, and h is the height.
We are given that the radius of the tank is 3 m, and the tank is being filled with water at a rate of 4 m^3/min. Let's call the rate at which the height is increasing dh/dt.
Since the rate of change of volume with respect to time is given by dV/dt, we can differentiate the volume formula with respect to time:
dV/dt = d(πr^2h)/dt
The derivative of the first term, πr^2, is zero because the radius is constant. So we only need to consider the derivative of the second term, which is h.
dV/dt = πr^2(dh/dt)
Now we can substitute the given values:
4 m^3/min = π(3 m)^2(dh/dt)
Simplifying, we have:
4 m^3/min = 9π(dh/dt)
Finally, we can solve for dh/dt:
dh/dt = (4 m^3/min) / (9π)
Using a calculator, we get:
dh/dt ≈ 0.045 m/min
Therefore, the height of the water is increasing at a rate of approximately 0.045 m/min.
To find out how fast the height of the water is increasing, we need to use the formula for the volume of a cylinder:
V = πr^2h,
where V is the volume, r is the radius, and h is the height.
Given that the radius of the tank is 3 m and the water is being filled at a rate of 4 m^3/min, we can differentiate both sides of the equation with respect to time (t) to solve for dh/dt (how fast the height is increasing):
dV/dt = d(πr^2h)/dt
4 = 2πr^2 (dh/dt).
Substituting the given radius into the equation, we have:
4 = 2π(3^2)(dh/dt)
4 = 18π(dh/dt).
Simplifying the equation:
dh/dt = 4 / (18π).
Calculating the value:
dh/dt ≈ 0.0707 m/min.
Therefore, the height of the water is increasing at a rate of approximately 0.0707 meters per minute.
volume=PI*r^2 *h
dV/dt= PI*3^2 * dh/dt
you know dV/dt, find dh/dt