A tank of water has a base a circle of radius 2 meters and vertical sides. If water leaves the tank at a rate of 6 liters per minute, how fast is the water level falling in centimeters per hour? [1 liter is 1000 cubic centimeters]
the cross-section has area π*200^2 cm^2
One liter = 1000 cm^3, so each minute a cylinder of height
6000/(40000π) = 3/(20π) cm
drains out.
Or, using calculus,
v = πr^2h = 40000πh
dv/dt = 40000π dh/dt
-6000 = 40000π dh/dt
dh/dt = -3/(20π) cm/min = -9 cm/hr
Well, it seems like we're in deep water with this question! But fear not, I'm here to help! Let's break it down.
So we have a tank of water with a circular base of radius 2 meters. The volume of water in the tank is constantly decreasing at a rate of 6 liters per minute.
Now, let's find the volume of the tank using the formula for the volume of a cylinder: V = πr²h, where V is the volume, r is the radius, and h is the height.
Since the base of the tank is a circle with a radius of 2 meters, we know that the area of the base is A = πr² = π(2)² = 4π square meters.
Now, let's convert the rate of water leaving the tank from liters per minute to cubic centimeters per minute. Since 1 liter is equal to 1000 cubic centimeters, we have 6 liters per minute * 1000 cubic centimeters per liter = 6000 cubic centimeters per minute.
To find how fast the water level is falling, we'll need to find the rate of change of the height (h).
Let's call the height of the water level in the tank h(t), where t is the time. The volume of water in the tank can be expressed as V(t) = A * h(t), where A is the area of the base.
Differentiating both sides of the equation with respect to time (t), we get dV(t)/dt = A * dh(t)/dt.
Since we know dV(t)/dt is equal to -6000 cubic centimeters per minute (since the volume is decreasing), and A is equal to 4π square meters, we have -6000 = 4π * dh(t)/dt.
Simplifying the equation, we have dh(t)/dt = -6000 / (4π) = -1500/π cubic centimeters per minute.
To find the rate in centimeters per hour, we need to multiply by 60 (since there are 60 minutes in an hour) and divide by 100 (since there are 100 centimeters in a meter). Therefore, the rate of the water level falling is approximately -9000/π centimeters per hour.
So, the water level is falling at a speed of approximately -9000/π centimeters per hour. And remember, the negative sign indicates that the height of the water is decreasing.
I hope that helps! Just remember, water may be going down, but your spirits can always stay up!
To find the rate at which the water level is falling, we need to calculate the change in the height of the water level with respect to time.
Given:
Radius of the circular base (r) = 2 meters
Rate of water leaving the tank (V) = 6 liters per minute = 6000 cubic centimeters per minute (since 1 liter = 1000 cubic centimeters)
Let's assume the height of the water level is h at a certain time.
The volume of water in the tank can be calculated using the formula for the volume of a cylinder:
V = πr^2h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = π(2r)(dh/dt)
Since the radius of the base is constant, we have:
dV/dt = 4π(dh/dt)
Now, we substitute the given value for dV/dt:
6000 = 4π(dh/dt)
To find the rate at which the water level is falling (dh/dt), we solve for it:
dh/dt = 6000 / (4π)
Simplifying further:
dh/dt = 1500 / π
Now, to convert the units to centimeters per hour, we can multiply by the appropriate conversion factors:
dh/dt = (1500 / π) * (100 cm / 1 m) * (60 minutes / 1 hour)
Simplifying:
dh/dt = (9000000 / π) cm/hr
Therefore, the water level is falling at a rate of approximately 2857745.854 cm/hr (rounded to the nearest centimeter).
To find the rate at which the water level is falling, we need to determine how quickly the volume of water in the tank is decreasing with respect to time.
Let's denote the height of the water in the tank as h (in meters) and the radius of the circle at the base as r (in meters).
Given that the radius of the circle is 2 meters, we know that the area of the circular base is A = πr^2.
We need to determine how the height of the water changes with respect to time. Let's denote the time as t (in minutes) and the height as h(t).
We know that the volume of the water in the tank can be expressed as V = A * h.
Differentiating both sides of this equation with respect to time, we can find the rate at which the volume is changing:
dV/dt = d(A * h)/dt
Using the product rule of differentiation, we can expand this equation:
dV/dt = A * dh/dt + h * dA/dt
The first term, A * dh/dt, represents the rate at which the water level is changing (which is what we need to find).
The second term, h * dA/dt, represents the rate at which the area of the base is changing (since the radius is constant).
Since the tank is a closed system, the area of the base doesn't change, so dA/dt is zero.
Therefore, the equation simplifies to:
dV/dt = A * dh/dt
Now, let's plug in the given values:
A = π * (2^2) = 4π square meters
dV/dt = 6 liters/minute = 6000 cubic centimeters/minute
Converting minutes to hours:
6000 cubic centimeters/minute * (60 minutes / 1 hour) = 360,000 cubic centimeters/hour
Now, we need to convert the units of volume to meters:
1 liter = 1000 cubic centimeters
So, 360,000 cubic centimeters/hour = 360 liters/hour
Now, let's substitute the known values into the equation:
360 liters/hour = 4π square meters * dh/dt
Simplifying the equation, we have:
90π = π * dh/dt
Thus, dh/dt = 90.
Therefore, the water level is falling at a rate of 90 centimeters per hour.