Consider a rectangular bathtub whose base is 12 ft^2. At what rate is water pouring into the tub if the water level rises at a rate of 0.9 ft/min?
Please show steps. (I want to know how)
area of tub x depth of water
Therefore a water level rise of 0.9 ft /min
Then, 12 x 0.9 = 10.8 cubic ft/min
To find the rate at which water is pouring into the rectangular bathtub, we need to use the formula:
Volume = Base Area × Height
Given that the base area is 12 ft², we need to determine the rate at which the height (water level) is increasing.
Let's denote the height of the tub as h (in feet) and the time as t (in minutes). We are given that the rate of increase in height is 0.9 ft/min. Therefore, the rate of change of height with respect to time can be represented as:
dh/dt = 0.9 ft/min
We need to find the rate at which water is pouring into the tub, which is the rate of change of volume with respect to time. This can be represented as:
dV/dt = ?
To relate the volume to the height, we multiply the base area (12 ft²) by the height (h):
V = 12h
Now, we need to differentiate the volume equation with respect to time to find the rate of change of volume with respect to time:
dV/dt = d(12h)/dt
Using the product rule of differentiation, the equation becomes:
dV/dt = 12 dh/dt
Substituting the given value, dh/dt = 0.9 ft/min, we can solve for dV/dt:
dV/dt = 12 × 0.9 ft²/min
dV/dt = 10.8 ft³/min
Therefore, the rate at which water is pouring into the tub is 10.8 ft³/min.