A swimming pool is 40 feet long, 20 feet wide, 8 feet deep at the deep end, and 3 feet deep at the shallow end; the bottom is rectangular. If the pool is filled by pumping water into it at the rate of 40 cubic feet per minute, how fast is the water level rising when it is 3 feet deep at the deep end?
To find out how fast the water level is rising when it is 3 feet deep at the deep end, we can use related rates.
Let's denote the water level as y, measured in feet, and the time as t, measured in minutes.
We need to determine dy/dt, the rate at which y is changing with respect to t.
Given:
Length of the pool (l) = 40 feet
Width of the pool (w) = 20 feet
Depth at the deep end (d1) = 8 feet
Depth at the shallow end (d2) = 3 feet
Rate of filling the pool (V) = 40 cubic feet per minute
We know that the volume of a rectangular prism is given by the formula V = l * w * h, where h is the height or depth.
Let's express the changing volume of water as a function of y, the water level:
V(y) = l * w * h(y)
Since the pool is rectangular, the depth increases linearly from 3 feet to 8 feet over its length of 40 feet. So, we can express the depth as a linear function of y:
h(y) = m * y + c
Here, m represents the slope of the line and c is the y-intercept.
We can find m and c by using the given information:
When y = 0, h(0) = d2 = 3 feet
When y = 20, h(20) = d1 = 8 feet
Using the slope equation: m = (d1 - d2) / (20 - 0)
Substituting the values:
m = (8 - 3) / (20 - 0) = 5 / 20 = 1/4
Now, we can write the equation for the depth as a function of y:
h(y) = (1/4) * y + 3
Substituting this equation into the volume equation:
V(y) = l * w * [(1/4) * y + 3]
V(y) = 40 * 20 * [(1/4) * y + 3]
V(y) = 200 * (1/4) * y + 2400
V(y) = 50 * y + 2400
Now, we can differentiate both sides of the equation with respect to t:
dV/dt = d(50y + 2400) / dt
Since the rate of filling the pool is known to be 40 cubic feet per minute, the left-hand side can be expressed as:
dV/dt = 40
On the right-hand side, we have:
d(50y + 2400) / dt = 40
Applying the chain rule of differentiation, we get:
50 * dy/dt = 40
Rearranging the equation to solve for dy/dt:
dy/dt = 40 / 50 = 4/5
Therefore, the water level is rising at a rate of 4/5 feet per minute when it is 3 feet deep at the deep end.
To find the rate at which the water level is rising, we need to use the information given about the dimensions of the pool and the rate at which water is being pumped in.
Let's first determine the volume of water in the pool. Since the bottom is rectangular, we can find the average depth by adding the deep and shallow depths and dividing by 2:
Average Depth = (8 ft + 3 ft) / 2 = 5.5 ft
Next, we can find the area of the bottom of the pool by multiplying the length and width:
Area = 40 ft * 20 ft = 800 ft^2
Now, we can find the volume of water in the pool by multiplying the area of the bottom by the average depth:
Volume = 800 ft^2 * 5.5 ft = 4400 ft^3
Since we know the rate at which water is being pumped in is 40 cubic feet per minute, the rate at which the water level is rising can be found by taking the derivative of the volume with respect to time:
dV/dt = 40 ft^3/min
Finally, we can substitute the volume value to find the rate at which the water level is rising when it is 3 feet deep at the deep end:
40 ft^3/min = dV/dt = d/dt (800 ft^2 * 5.5 ft) = dA/dt * 5.5 ft
dA/dt = 40 ft^3/min / 5.5 ft = 7.27 ft^2/min
Therefore, the water level is rising at a rate of approximately 7.27 ft^2/min when it is 3 feet deep at the deep end.