AP Calculus AB Exam Question:

The rate at which water flows into a tank is given by the function R(t) = 2t + 5, where t is the time in minutes and R(t) is the flow rate in gallons per minute. The initial amount of water in the tank is 10 gallons. Find the net change in the amount of water in the tank from t = 0 to t = 5 minutes.

Solution:

To find the net change in the amount of water in the tank, we need to calculate the accumulation of water over the given time interval. The accumulation of a function represents the total change in the function over an interval. In this case, we need to integrate the given flow rate function from t = 0 to t = 5 to find the net change in the amount of water.

The flow rate function is R(t) = 2t + 5, representing the gallons per minute. To find the accumulated amount of water, we need to integrate this function with respect to time:

∫[0,5] (2t + 5) dt

Integrating, we get:

∫[0,5] (2t + 5) dt = ∫[0,5] 2t dt + ∫[0,5] 5 dt

Applying the power rule of integration, we obtain:

[t^2 + 5t] evaluated from 0 to 5 + [5t] evaluated from 0 to 5

Substituting the limits of integration, we have:

[(5)^2 + 5(5)] - [(0)^2 + 5(0)] + [5(5)] - [5(0)]

Simplifying, we get:

[25 + 25] - [0 + 0] + [25] - [0]

= 25 + 25 + 25

= 75 gallons

Therefore, the net change in the amount of water in the tank from t = 0 to t = 5 minutes is 75 gallons.