AP Calculus AB Exam Question:

Let a function $f(t)$ represent the net change in a quantity over time. The table below shows the values of $f(t)$ at various time intervals:

a) Use the values in the table to find the total accumulation of the quantity from $t = 1$ to $t = 4$.

b) On which time interval does the accumulation of the quantity increase the most? Justify your answer.

Please answer both parts (a) and (b) below.

Answer:

a) To find the total accumulation, we need to sum up the net changes from each time interval. Let's calculate it step by step:

The net change from $t = 1$ to $t = 2$ is -1. (2 - 3 = -1)

The net change from $t = 2$ to $t = 3$ is 4. (-3 + 1 = 4)

The net change from $t = 3$ to $t = 4$ is -3. (1 - 2 = -1)

Therefore, the total accumulation of the quantity from $t = 1$ to $t = 4$ is -1 + 4 - 1 = 2.

Hence, the total accumulation of the quantity from $t = 1$ to $t = 4$ is 2.

b) To determine the time interval where the accumulation of the quantity increases the most, we need to calculate the differences between consecutive net changes.

The difference between the net changes from $t = 1$ to $t = 2$ and from $t = 2$ to $t = 3$ is 4 - (-1) = 5.

The difference between the net changes from $t = 2$ to $t = 3$ and from $t = 3$ to $t = 4$ is -3 - 4 = -7.

Therefore, the accumulation of the quantity increases the most on the interval from $t = 1$ to $t = 2$ because the difference is 5.

Thus, the accumulation of the quantity increases the most on the interval from $t = 1$ to $t = 2$.

Note: The net change represents the rate of change of the quantity at each time interval. The total accumulation is found by summing up the net changes, while the change in accumulation is determined by calculating the differences between consecutive net changes.