Practice AP Calculus Unit 4.1 questions on interpreting a derivative in context as an instantaneous rate of change with units.
A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function f, where f(t) is measured in gallons per second and t is measured in seconds since pumping began. Selected values of f(t) are given in the table.
Using the model g defined in part (c), find the value of g′(140). Interpret the meaning of your answer in the context of the problem.
Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.
g′(140) -0.004908
g′(140)=-0.005( or -0.004)
g′(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′(140) is 1/500 (49/36)-49/9000 (49/36).
- The value of g′(140) may appear only in the interpretation.
- To be eligible for the second point a response must present some numerical value for g′(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′(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′(140)=0.001997 or 0.00187.