Final answer:
The function r(t) is not continuous at t=5 as the left-hand and right-hand limits are not equal. The average rate of draining is found by integrating r(t) from 0 to 8 and dividing by the time period. The value of r'(3) corresponds to the acceleration of the water flow, and to find the time when the tank has 9000 liters left, we need to solve an integral for the volume drained.
Step-by-step explanation:
To determine if r(t) is continuous at t=5, we need to check if the limit of r(t) as t approaches 5 from the left is equal to the limit as t approaches 5 from the right and if both are equal to r(5). So, we have:
Left-hand limit:
Limit as t approaches 5 from the left of t+3 = 5+3 = 8.
Right-hand limit:
Limit as t approaches 5 from the right of 600t = 600*5 = 3000.
Since the left-hand and right-hand limits are not equal, r(t) is not continuous at t=5.
To find the average rate of water draining from the tank between t=0 and t=8, we integrate r(t) from 0 to 8 and divide by the time interval, which is 8 hours:
Average rate = \(\frac{1}{8-0} \int_{0}^{8} r(t) dt\).
To find r'(3), we differentiate r(t) for the interval 0≤5 where r(t)=t+3 and evaluate at t=3:
r'(t) = 1, hence r'(3) = 1.
The value of r'(3) represents the instantaneous rate of change of the volume of water flowing out of the tank at time t=3 hours; in this context, it would be 1 liter per hour per hour, or the acceleration of the water flow rate.
To find the time A when the tank has 9000 liters of water, we need to solve an integral that represents the total volume of water drained from the tank:
\(12000 - \int_{0}^{A} r(t) dt = 9000\).
This integral equation involves finding the value of A that makes the tank's volume equal to 9000 liters after some water has drained out.