AP Calculus AB 2.1 Defining Average and Instantaneous Rates of Change Question Bank

Practise AP Calculus AB 2.1 questions by connecting interval averages, difference quotients and instantaneous rates in context.

Exam points

calculate average rates from functions, tables and real-world measurements over intervals
connect difference-quotient limits with instantaneous rate of change at a point

Covered syllabus points

CHA-2.A—Determine average rates of change using difference quotients CHA-2.B—Represent the derivative of a function as the limit of a difference quotient

[Maximum number: 2]

Juice is sold in 20-centimeter-tall bottles with horizontal cross sections parallel to the base that are circles, as shown in the

figure. The radius of the circular cross section at height h above the base of the bottle is given by a differentiable function

r, where h and r(h) are measured in centimeters. Selected values of r(h) are given in the table shown.

Let f be the continuous function defined on the closed interval [-1,3] whose graph, consisting of three line segments, is

shown. Let g be the function given by $g(x)=\int_{0}^{x} f(t) d t$ .

Approximate $r^{\prime}(3.1)$ using the average rate of change of r over the interval $0 \leq h \leq 6.2$ . Show the computations that

lead to your answer. Indicate units of measure.

On what intervals, if any, is g decreasing? Give a reason for your answer.

To earn the first point a response must provide a difference and a quotient. A response of only $\frac{2.7-2.5}{6.2}$ or $\frac{0.2}{6.2-0}$ earns the

first point.

To earn the second point a response must have $\frac{\mathrm{cm}}{\mathrm{cm}}$ (or the equivalent) attached to a numerical value. These units do not

reduce to 1 because the numerator, cm, refers to radius of the bottle and the denominator, cm, refers to height of the

bottle.

\begin{tabular}[t]{|l|l|l|}
\hline 0

1

The student response accurately includes both of the criteria below.
\begin{itemize}
\item[□]

Answer
\item[□]

Units

Solution:

$r^{\prime}(3.1) \approx \frac{r(6.2)-r(0)}{6.2-0}=\frac{2.7-2.5}{6.2}$

0.032 centimeter per centimeter

Part B

Select a point value to view scoring criteria, solutions, and/or examples to score the response.

[Maximum number: 1]

If $f(x)=\ln x$ , then $\lim _{x \rightarrow 3} \frac{f(x)-f(3)}{x-3}$ is

$\frac{1}{3}$

$e^{3}$

$\ln 3$

nonexistent

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Covered syllabus points

2.1 Defining Average and Instantaneous Rates of Change at a Point question 1

2.1 Defining Average and Instantaneous Rates of Change at a Point question 2