[table]
As air is pumped into an exercise ball, the surface area of the ball expands. The table shows the surface area of the ball at 2 seconds intervals for 30 seconds. Determine the average rate of change of the surface area of the ball for the interval between 20 seconds and 30 seconds. |
[/table]

Question

BlackTom AI · Accepted Answer

To start, identify the two time points for the interval: 20 seconds and 30 seconds, and read off the corresponding surface areas from the table.
At t = 20 s, A = 1266.0 cm^2.
At t = 30 s, A = 2836.0 cm^2.
Next, compute the change in surface area over the interval: ΔA = A(30) − A(20) = 2836.0 − 1266.0 = 1570.0 cm^2.
Then determine the time span: Δt = 30 − 20 = 10 seconds.
The average rate of change is the ratio of the change in surface area to the change in time: (ΔA)/(Δt) = 1570.0 / 10 = 157.0 cm^2 per second.
A quick check: since the surface area increases as time progresses, a positive rate is expected, and 157.0 cm^2/s is consistent with the provided data.
The value 130.620 cm^2/s would imply a smaller total area increase over 10 seconds (1306.2 cm^2 total), which does not match the actual difference between 1266.0 cm^2 and 2836.0 cm^2, so that alternate result is not supported by the table.

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