AP Calculus BC 4.1: Meaning of the Derivative

$$\begin{aligned} g^{\prime}(140) \approx-0.004908 g^{\prime}(140) =-0.005(\text { or }-0.004) \end{aligned}$$$g^{\prime}(140)$

1 point

The rate at which gasoline is flowing into the tank is decreasing at
a rate of 0.005 (or 0.004 ) gallon per second per second at time
t=140 seconds.

Interpretation

1 point

Scoring notes:

- The exact value of $g^{\prime}(140)$ is $\frac{1}{500} \cos \left(\frac{49}{36}\right)-\frac{49}{9000} \sin \left(\frac{49}{36}\right)$.

- The value of $g^{\prime}(140)$ may appear only in the interpretation.

- To be eligible for the second point a response must present some numerical value for $g^{\prime}(140)$.

- To earn the second point the interpretation must include "the rate of flow of gasoline is changing

at a rate of [the declared value of $g^{\prime}(140)$ ]" and "at t=140 " (or equivalent).

- An interpretation of "decreasing at a rate of -0.005" or "increasing at a rate of 0.005" does not

earn the second point.

- Degree mode: In degree mode, $g^{\prime}(140)=0.001997$ or 0.00187.

Total for part (d) for question 19 points

Part A (BC): Graphing calculator required