Water is being filled into a water tank at 22 liters/second. The base of the tank has dimensions w = 1 meter and L = 2 meter . What is the rate of change of the height of water in the tank? (express your answer in cm/sec).
assuming parallel sides, when
v = 2x
dv/dt = 2 dx/dt
22 liters/sec = 22/1000 m^3/s
.022 = 2 dx/dt
dx/dt = 0.011 m/s = 1.1 cm/s
owh, i see.. thank you Steve
To find the rate of change of the height of water in the tank, we can use the formula for the volume of a rectangular prism.
Volume = Base Area × Height
Given that the dimensions of the base are w = 1 meter and L = 2 meters, the base area is:
Base Area = w × L = 1 × 2 = 2 square meters
Since the rate of water being filled into the tank is 22 liters/second, we need to convert it to cubic meters per second because the base area is in square meters.
1 liter = 0.001 cubic meters (since 1 liter = 0.001 cubic meters)
So, 22 liters/second = 22 × 0.001 cubic meters/second = 0.022 cubic meters/second
Now, let's differentiate the volume formula with respect to time to find the rate of change of the height of water in the tank:
d(Volume)/dt = d(Base Area × Height)/dt
Since the base area is constant, its derivative is zero:
d(Base Area)/dt = 0
d(Volume)/dt = d(Base Area × Height) / dt = Base Area × d(Height)/dt
Now, substituting the known values into the equation:
0.022 = 2 × d(Height)/dt
To find the rate of change of the height of water (d(Height)/dt), we can rearrange the equation:
d(Height)/dt = 0.022 / 2 = 0.011 meters/second
To express the answer in centimeters per second, we can convert it:
1 meter = 100 centimeters
So, the rate of change of the height of water in the tank is:
0.011 meters/second = 0.011 × 100 centimeters / second = 1.1 centimeters/second
Therefore, the rate of change of the height of water in the tank is 1.1 cm/sec.