AP Calculus AB Cha 2 A Determine Average Rates of Change Using Difference Quotients Questions

Practise AP Calculus AB 2.1 questions by calculating average rates of change from intervals, tables, graphs and applied data.

Exam points

calculate average rate of change from endpoint values, tables or applied measurements
compare interval rates to identify the greatest average change over a domain

[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.

The student response accurately includes both of the criteria below.

□

Answer
□

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

All question bank results loaded

AP Calculus AB Cha 2 A Determine Average Rates of Change Using Difference Quotients Questions question 1