AP Calculus AB Unit 2: Differentiation Basics

Explore AP Calculus Unit 2 questions on average and instantaneous rates, derivative notation, tangent lines, continuity, and basic derivative rules.

Covered syllabus points

2.1 Defining Average and Instantaneous Rates of Change at a Point 2.2 Defining the Derivative of a Function and Using Derivative Notation 2.3 Estimating Derivatives of a Function at a Point 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist 2.5 Applying the Power Rule 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple 2.7 Derivatives of cos x, sin x, eˣ, and ln x 2.8 The Product Rule 2.9 The Quotient Rule 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

[Maximum number: 4]

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 $g(x)=\int_{0}^{x} f(t) d t$ .

(a)

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.

[ 2 ]

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.

(b)

The radius of the circular cross section at height h above the base of the bottle can also be modeled by the function f,

where h and f(h) are measured in centimeters. Let $f(h)=3 \sqrt{2-1.25 \cos \left(\frac{\pi}{100}(h-20)^{2}\right)}$ for $0 \leq h \leq 20$ . Based

on the model, at a height of h=12 centimeters, is the radius increasing or decreasing as h increases? Give a reason

for your answer.

What is the absolute minimum value of g on the closed interval [-1, 3]? Justify your answer.

[ 2 ]

To earn the second point, a response does not need to provide a value for $f^{\prime}(12)$ ; the sign of $f^{\prime}(12)$ is sufficient.

Any presented value of $f^{\prime}(12)$ must be correct for the digits presented, up to three decimal places.

If a response presents an incorrect sign or value for $f^{\prime}(12)$ , the response does not earn the second point.

Degree mode: A response that presents answers obtained by using a calculator in degree mode does not earn the first point

it would have otherwise earned. The response is eligible for all subsequent points (unless no answer is possible in degree

mode or the question is made simpler by using degree mode). In degree mode, $f^{\prime}(12)=-0.0006660598$ . If a

response presents this value, it earns the first point but is not eligible to earn the second point.

2022 Form I FRQ

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

1

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

Considers $f^{\prime}(12)$
\item[□]

Answer with reason

Solution:

$f^{\prime}(12)=-0.535902$

At a height of 12 centimeters, the radius is decreasing as h increases because $f^{\prime}(12)<0$ .

Part D

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

[Maximum number: 1]

$\lim _{h \rightarrow 0} \frac{e^{2+h}-e^{2}}{h}$ is

$e^{2}$

$2 e^{2}$

nonexistent

[Maximum number: 2]

A student starts reading a book at time t=0 minutes and continues reading for the next 10

minutes. The rate at which the student reads is modeled by the differentiable function R, where

R(t) is measured in words per minute. Selected values of R(t) are given in the table shown.

Table for Question Unit 2: Differentiation: Definition and Fundamental Properties question 3 — AP Calculus AB

Approximate $R^{\prime}(1)$ using the average rate of change of R over the interval $0 \leq t \leq 2$ . Show

the work that leads to your answer. Indicate units of measure.

A student starts reading a book at time t=0 minutes and continues reading for the next 10 minutes. The rate

at which the student reads is modeled by the differentiable function R, where R(t) is measured in words per

minute. Selected values of R(t) are given in the table shown.

t (minutes)	0	2	8	10
R(t) (words per minute)	90	100	150	162

Model Solution

Scoring

A Approximate $R^{\prime}(1)$ using the average rate of change of R over the interval $0 \leq t \leq 2$ . Show the work

that leads to your answer. Indicate units of measure.

$\begin{aligned} & R^{\prime}(1) \approx \frac{R(2)-R(0)}{2-0} \\ & =\frac{100-90}{2}=\frac{10}{2}=5 \text { words per minute per minute } \end{aligned}$

Answer with setup	Point 1 (P1)
Units	Point 2 (P2)

Scoring Notes for Part A

- To earn P1, a response must present the answer along with the supporting work of a difference and a

quotient using values from the table.

○ $\frac{100-90}{2-0}, \frac{10}{2-0}, \frac{100-90}{2}$ , or $\frac{R(2)-R(0)}{2-0}=5$ is sufficient to earn P1.

○ $\frac{R(2)-R(0)}{2-0}$ by itself is not sufficient to earn P1.

- P2 is earned for correct units, whether or not they are attached to a numerical value for the average

rate of change.

- P2 is also earned for the units "words/minute ${ }^{2}$ ."

[Maximum number: 1]

If $f(x)=\left\{\begin{array}{ll}x^{2} & \text { for } x \leq 1 \\ 2 x-1 & \text { for } x>1\end{array}\right.$ , then

f(x) is not continuous at x=1

f(x) is continuous at x=1 but $f^{\prime}(1)$ does not exist

$f^{\prime}(1)=2$

$\lim _{x \rightarrow 1} f(x)$ does not exist

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Covered syllabus points

Unit 2: Differentiation: Definition and Fundamental Properties question 1

Question (a)

Question (b)

Unit 2: Differentiation: Definition and Fundamental Properties question 2

Unit 2: Differentiation: Definition and Fundamental Properties question 3

Unit 2: Differentiation: Definition and Fundamental Properties question 4