Water is draining from a swimming pool in such a way that the remaining volume of water after t minutes is V = 200(50 - t)^2 cubic meters. Find : (a) the average rate at which the water leaves the pool in the first 5 minutes
To find the average rate at which the water leaves the pool in the first 5 minutes, we need to calculate the rate of change of the volume with respect to time within that time interval.
The given function for the volume of water remaining in the pool is V(t) = 200(50 - t)^2, where V represents the volume of water in cubic meters and t represents time in minutes.
To find the rate at which the water is leaving the pool, we need to find the derivative of the volume function with respect to time, which gives us the rate of change of the volume.
First, let's find the derivative of V(t) with respect to t:
dV/dt = d/dt[200(50 - t)^2]
To differentiate the function, we can use the chain rule. The derivative of the outer function (200) with respect to t is zero since it is a constant. The derivative of the inner function (50 - t)^2 can be found using the chain rule.
Using the chain rule, the derivative of (50 - t)^2 is:
2(50 - t) * (-1)
Simplifying this expression, we get:
dV/dt = 2(50 - t) * (-1) * 200
Now, let's simplify further:
dV/dt = -400(50 - t)
The rate at which water leaves the pool can be determined by substituting t = 0 into the derivative expression, since we want to find the average rate within the first 5 minutes:
dV/dt at t = 0 = -400(50 - 0)
Simplifying further, we get:
dV/dt at t = 0 = -400(50)
Finally, calculating the value:
dV/dt at t = 0 = -20000
Therefore, the average rate at which the water leaves the pool in the first 5 minutes is 20000 cubic meters per minute.
V(0) = 500,000
V(5) = 405,000
(V(5)-V(0))/(5-0) = -19,000 ft^3/min