AP Calculus BC:
Study Guide
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AP Calculus BC:
Study Guide
AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.
Key Exam Details
The AP®^ Calculus BC exam is a 3-hour 15-minute, end-of-course test comprised of 45 multiple- choice questions (50% of the exam) and 6 free-response questions (50% of the exam).
The exam covers the following course content categories:
- Limits and Continuity: 4–7% of test questions
- Differentiation: Definition and Fundamental Properties: 4 – 7 % of test questions
- Differentiation: Composite, Implicit, and Inverse Functions 4 – 7 % of test questions
- Contextual Applications of Differentiation: 6 – 9 % of test questions
- Analytical Applications of Differentiation: 8 – 11 % of test questions
- Integration and Accumulation of Change: 17–20% of test questions
- Differential Equations: 6– 9 % of test questions
- Applications of Integration: 6 – 9 % of test questions
- Parametric Equations, Polar Coordinates, and Vector-Valued Functions: 11–12% of test questions
- Infinite Sequences and Series: 17–18% of test questions
This guide offers an overview of the main tested subjects, along with sample AP multiple-choice questions that look like the questions you’ll see on test day.
Limits and Continuity
About 4–7% of the questions on your exam will cover Limits and Continuity.
Limits
The limit of a function f as x approaches c is L if the value of f can be made arbitrarily close to L x sufficiently close to c (but not equal to c ). If such a value exists, this is denoted
. If no such value exists, we say that the limit does not exist, abbreviated DNE.
Limits can be found using tables, graphs, and algebra.
Important algebraic techniques for finding limits include factoring and rationalizing radical expressions. Other helpful tools are given by the following properties.
, and a is any real number.
by taking lim x → c f ( ) x = L
Suppose lim x → c f ( ) x = L , lim x → c g x ( ) = M ,lim x → L h x ( ) = N
Example
The function shown has the following limits:
- x l →− im 2 − f ( ) x = − 1
- x li →− m 2 + f ( ) x = 1
- li x →− m 2 f ( x ) DNE
- l x im → 1 − f ( ) x = 4
- l x im → 1 + f ( ) x = 4
- l x im → 1 f ( ) x = 4
Note that f (1) = 3, but this is irrelevant to the value of the limit.
Infinite Limits, Limits at Infinity, and Asymptotes
When a function has a vertical asymptote at x = c , the behavior of the function can be described using infinite limits. If the function values increase as they approach the asymptote, we say the limit is ∞, whereas if the values decrease as they approach the asymptote, the limit is - ∞. It is important to realize that these
limits do not exist in the same sense that we described earlier; rather, saying that a limit is is simply a convenient way to describe the behavior of the function approaching the point.
We can also extend limits by considering how the function behaves as x → . If such a limit
exists, it means that the function
x lim → f^ ( ) x^ = L
approaches a horizontal line as x increases y = L
or decreases without bound. In other words, if , then f has a horizontal asymptote. It is possible for
a function to have two horizontal asymptotes since it can have different limits as x^ → and x → −.
The Squeeze Theorem
The Squeeze Theorem states that if the graph of a function lies between the graphs of two other functions, and if the two other functions share a limit at a certain point, then the function in
between also shares that same limit. More formally, if f ( x ) g x ( ) h ( x )for all x in some
interval containing c , and if lim x → c f ( ) x = lim x → c h x ( ) = L , then lim x → c g x ( ) = L as well.
Example
The sine function satisfies − 1 sin x 1 for all real numbers x , so 1 sin 1 1 x
^
− is also true
for all real numbers x. Multiplying this inequality by x^2 , we obtain^2 x^2 sin^1 x^2
x
− x ^
. Now
the functions on the left and right of the inequality, x^2 and − x^2 , both have limits of 0 as x → 0.
Therefore, we can conclude that lim x 0 x^2 sin 1 0 → x
(^) also.
Continuity
The function f is said to be continuous at the point x = c if it meets the following criteria:
1. f c ( )exists
- lim x → c f ( x )exists
- lim x → c f ( ) x = f c ( )
In other words, the function must have a limit at c , and the limit must be the actual value of the function.
Each of the above criteria can fail, resulting in a discontinuity at x = c. Consider the following three graphs:
Suggested Reading
- Hughes-Hallett, et al. Calculus: Single Variable. 7 th edition. Chapter 1. New York, NY: Wiley.
- Larson & Edwards. Calculus of a Single Variable: Early Transcendental Functions. 7 th^ edition. Chapter 2. Boston, MA: Cengage Learning.
- Stewart, et al. Single Variable Calculus. 9 th^ edition. Chapter 2. Boston, MA: Cengage Learning.
- Rogawski, et al. Calculus: Early Transcendentals Single Variable. 4th^ edition. Chapter 2. New York, NY: Macmillan.
- Sullivan & Miranda. Calculus: Early Transcendentals. 2 nd Edition. Chapter 1. New York, NY: W.H. Freeman.
Sample Limits and Continuity Questions
Consider the following graphs of f and g :
Compute x lim → 2 + 3 f ( ) x − g x ( ), provided the limit exists.
A. Does not exist B. 4 C. 6 D. 7
Explanation: The correct answer is D. First, use linearity to write:
x^ lim → 2 +^ ^3^ f^ ( ) x^^ −^ g x ( )^^ =^ 3 lim x → 2 +^ f^ ( ) x^^ − x lim→ 2 +^ g x ( )^.
. Substituting these values above yields .
Now, observe that x lim → 2 + f ( ) x = 3 and (^) x lim→ 2 + g x ( ) = 2
x^ lim → 2 +^ ^3^ f^ ( ) x^^ −^ g x ( )^ ^ =^ 3 lim x → 2 +^ f^ ( ) x^^ −^ x lim→ 2 +^ g x ( )^ =^ 3(3)^ −^2 =^7
Suppose that f ( ) x =2 cos( 3 x )and the graph of g ( x ) is given by
Compute (^) x lim → − 3 f ( ) x g x ( ), provided it exists.
A. – 6 B. 4 C. 6 D. Does not exist
Explanation: The correct answer is C. Use the fact that the limit of a product is the product of the limits, provided they both exist independently, to compute:
( ) ( ) ( (^ )) (^ ) ( (^ )) (^ )
3 3 3 3
lim ( ) ( ) lim ( ) lim ( )
2 cos ( 3) 3 2 cos 3 2 3 6
x f^ x^ g x^ x f^ x^ x g x
→ − ^ =^ → − → −
Which of the following limits does not exist?
Differentiation: Definition and
Fundamental Properties
About 4–7% of the questions on your AP exam will cover Differentiation: Definition and Fundamental Properties.
Definition of the Derivative
The average rate of change of a function
x = a + h
f over the interval from x = a x = a + h f ( ) x f a ( ) x a
to is f a ( h ) f a ( ) h
- −. Alternatively, if , this can be written. When h is made
smaller, so that it approaches 0, the limit that results is called the instantaneous rate of change of f at x = a
0 ( ) lim^ (^ )^ ( ) h a f a^ h^ f a h
f = →^ +^ −
, or the derivative of f at x = a , and is denoted f ( ) a
( ) lim ( )^ ( ) x a a f^ x^ f a x a
f = (^) →^ − −
f ( ) a
That is, , or equivalently,.
If this limit exists, f is said to be differentiable at a. Graphically, represents the slope of the
line tangent f^ ( ) x
y − f ( a ) = f ( a )( x − a )
to the graph of at the point where x = a. Therefore, the line tangent to f ( x ) at x = a is.
If the function y = f ( ) x is differentiable at all points in some interval, we can define a new
function on that interval by finding the derivative dy dx x = a
at every point. This
0 ( ) lim^ (^ )^ ( ) h x f^ x^ h^ f^ x h
f = →^ +^ −
f ( ) a
x a
dy dx (^) =
new function, called the
derivative of f , can be denoted f^ ( ) x , y , or , and is defined by.
The value of the derivative at a particular point can then be denoted or
If f is differentiable at x = a , then it also must be continuous at x = a. In other words, if a function fails to be continuous at a point, it cannot possibly be differentiable at that point. Another way that differentiability can fail is via the presence of sharp turns or cusps in a graph.
Free Response Tip
When specific function values are given, the derivative at a point can be approximated by finding the average rate of change between surrounding points. For example, if you are given values of a function at x = 3, 4, and 5, then the derivative at 4 can be approximated by the average rate of change between 3 and 5.
Basic Derivatives and Rules
There are several rules that can be used to find derivatives. Assume f and g are differentiable functions, and c is a real number.
- The constant rule:^ d c 0 dx
- The power rule: d xn^ n^1 x
nx d
=^ −, for any real number n
( )^ ( )^ ( )^ ( )
d (^) f x g x f x g x dx
( )^ ( )^ ( )^ ( )
d (^) f x g x f x g x dx
d (^) cf ( ) x (^) cf ( ) x dx
( )^ ( )^ ( )^ ( )^ ( )^ ( )
d (^) f x g x f x g x f x g x dx
^2
d f x x g x f x g x dx g x (^) g x
(^) = f −
- The sum rule:
- The difference rule:
- The constant multiple rule:
- The product rule:
- The quotient rule:
d (^) cx c dx
= , and^ d x 1 dx
As special cases of the power rule, note that =.
In addition to these rules, the derivatives of some common functions are as follows:
f ( x ) f^ ( ) x
ex ex
ln x 1 x sin x cos^ x cos x −sin x
tan x sec^2 x
sec x sec x tan x csc x −csc x cot x
cot x − csc^2 x
An object moves along the curve y =^1 x
, starting at x =^1 10
. As it passes through the point (1,1) its
x - coordinate increases at a rate of 2 inches per second. How fast is the distance between the object and the origin changing at this instance in time?
A. 0 inches per second B. 4 2 2
inches per second C. inches per second D. – 2 inches per second
Explanation:
The correct answer is A. The position of a point on this curve is of the form ( x y , ) =( x , 1 x ). So,
the distance D between it and the origin is
D = ( x − 0 ) 2 + ( y − 0 ) 2 = ( x − 0 ) 2 + ( 1 x − 0 )^2 = x^2 + x −^2.
While we could differentiate both sides with respect to t directly, the D^2 = x^2^ + x −^2
presence of the radical makes this inconvenient. So, we square both sides first to get and now differentiate both sides with respect to t. Doing so yields: 3
3
2 D dD^ 2 x dx^ 2 x dx dt dt dt D dD^ x dx^ x dx dt dt dt
−
−
Note that when the object is at the point (1,1), we know that x = 1, D^^ =^12 +^1 −^2 =^2 , and dx 2 dt
2^ dD 1 2 1 2 0 dt
inches per second. Substituting this information into the above equation yields
So,^ dD 0 dt
For how many values of x in the interval 0, 2 is the tangent line to the curve
g x ( ) = sec x +csc x parallel to the line x = y?
A. 0 B. 2 C. 3 D. 4
Explanation:
The correct answer is B. First, use the graphing calculator to graph g( x ) on 0, 2 :
Now, reasoning using the continuity of the graph and the vertical asymptotes reveals that there
must be tangent lines to g( x ) with slope 1 at one x - value in each of the following intervals:( a ,^ 2 )
and (^) ( , c ).
Note that differentiating (^) xy required an application of the product rule, and that every time an
expression in terms of y was differentiated, the derivative was multiplied by dy dx
. Now all of the
terms with dy dx
can be gathered on one side of the equation, and dy dx
can be solved for:
( )
2 2
2 2
2 2
y dy^ x dy y x dx dx dy (^) y x y x dx dy y x dx y x
= −^ −
This technique can also be applied to find the derivatives of inverse functions. Consider an invertible function f , with inverse f −^1. By definition this means that f (^) ( f −^1 ( ) x (^) )= x. Now,
differentiating both sides with respect to x , we get f ( f −^1 ( x )) ( f −^1 )( x ) = 1. Solving for
( f^ −^1 )( ) x^ ,^ we have^ ( ). ( )
1 1
f x f f x
− = (^) −
Applying this rule to the inverse trigonometric functions, we can find the following derivatives:
f f^ arcsin x 2
1 − x arccos x 2
1 x
arctan x 2
1 + x arccot x 2
1 x
arcsec x 2
x x − 1 arccsc x 2
x x 1
Higher Order Derivatives
The derivative f of a function f is itself a function that may be differentiable. If it is, then its
derivative is f , called the second derivative of f. The relationship of f and f is identical to
the relationship between f and f . Similarly, the derivative of f is f , the third derivative of
f. This process can continue indefinitely, as long as the functions obtained continue to be
differentiable. After three, the notation changes, so that the 4th^ derivative of f is denoted^ f (4),
and the n th^ derivative is f (^ n ).
If y^ = f^ ( ) x , then higher order derivatives are also denoted^ y^ ,^ y ^ ,^ y^ (4)^ ,,^ y (^ n )^ ,, or
2 3 2 ,^3 ,^ ,^ ,
n n
d y d y d y dx dx dx
Suggested Reading
- Hughes-Hallett, et al. Calculus: Single Variable. 7 th edition. Chapter 3. New York, NY: Wiley.
- Larson & Edwards. Calculus of a Single Variable: Early Transcendental Functions. 7 th^ edition. Chapter 3. Boston, MA: Cengage Learning.
- Stewart, et al. Single Variable Calculus. 9 th^ edition. Chapter 3. Boston, MA: Cengage Learning.
- Rogawski, et al. Calculus: Early Transcendentals Single Variable. 4th^ edition. Chapter 3. New York, NY: Macmillan.
- Sullivan & Miranda. Calculus: Early Transcendentals. 2 nd Edition. Chapter 3. New York, NY: W.H. Freeman.
(^14) 1 1 3 3 4 4 4 2
f
This is the slope of the tangent line at x = 14. To get the y - value of the point of tangency,
compute f ( 14 ):
f ( ) 41 = sin− (^1) ( 14 ) = sin−^1 ( 21 )=^ 6
So, using the point-slope form for the equation of a line, the equation of the tangent line is y − 6 = (^2 33) ( x −^14 ).
What is the slope of a line perpendicular to the tangent line to the curve defined implicitly by x^2 y^2 – xy = 42 at the point (2, – 3)?
A.
B.
C.
D.
3 26 − (^23) − (^263) 3 2
Explanation: The correct answer is B. This is the correct answer. Implicitly differentiate both sides with respect to x :
( )
^ + − + =
2 2 2 2 2 2 2 2
x y y y x xy y x y y xy xy y y x y x y xy
y y^ xy x y x y y^ xy x xy
y y x So, the slope of the tangent line at the point (2, – 3) is− (^ −^3 )^ =^3 2 2
. Therefore, any line perpendicular
to the tangent line at this point would have a slope equal to − 23.
Contextual Applications of Differentiation
Around 6–9% of the questions on your AP exam will cover the topic Contextual Applications of Differentiation.
In any context, the derivative of a function can be interpreted as the instantaneous rate of change
of the independent variable with respect to the dependent variable. If y = f ( ) x , then the units of
the derivative are the units of y divided by the units of x.
Straight-Line Motion
Rectilinear (straight-line) motion is described by a function and its derivatives.
If the function s t ( )represents the position along a line of a particle at time t , then the velocity is
given by v ( t )= s ( t ). When the velocity is positive, the particle is moving to the right; when it is
negative, the particle is moving to the left. The speed of the v t ( )
particle does not take direction into account, so it is the absolute value of the velocity, or.
The acceleration of the particle is a (^^ t^ )^ =^ v^ ( )^ t =^ s ( t ). The velocity is increasing when a t ( )is
positive and decreasing when a t ( )is negative. The speed, however, is only increasing when
v t ( ) and a t ( )have the same sign (positive or negative). When v t ( )and a t ( )have different
signs, the particle’s speed is decreasing.
Related Rates
Related rates problems involve multiple quantities that are changing in relation to each other. Derivatives, and especially the chain rule, are used to solve these problems. Though the problems vary widely with context, there are a few steps that usually lead to a solution.
- Draw a picture and label relevant quantities with variables.
- Express any rates of change given in the problem as derivatives.
- Express the rate of change you need to solve for as a derivative.
- Relate the variables involved in the rates of change to each other with an equation.
- Differentiate both sides of the equation with respect to time. This may involve applying many derivative rules but will always involve the chain rule.
- Substitute all of the given information into the resulting equation.
- Solve for the unknown rate.
Example
The length of the horizontal leg of a right triangle is increasing at a rate of 3 ft/sec, and the length of the vertical leg is decreasing at a rate of 2 ft/sec. At the instant when the horizontal leg is 7 ft and the vertical leg is 1 ft, at what rate is the length of the hypotenuse changing? Is it increasing or decreasing?