AP Advantage: AP Calculus

AP Advantage: AP Calculus AB
Classroom Matters
Instructor: Shashank Patil
May 6-7, 2017
Exam Day:
Tuesday, May 9 @ 8:00AM
Exam Structure:
Time: 3 hours 15 minutes
Section I: Multiple Choice | 45 Questions | 1 hour and 45 minutes | 50% of Final Exam Score
Part A — 30 questions | 60 minutes (calculator not permitted)
Part B — 15 questions | 45 minutes (graphing calculator required)
Section II: Free-Response | 6 Questions | 1 hour and 30 minutes | 50% of Final Exam Score
Part A — 2 problems | 30 minutes (graphing calculator required)
Part B — 4 problems | 60 minutes (calculator not permitted)
Note: You may not take both the Calculus AB and Calculus BC exams within the same year.
Content Review
Limits
• Properties of Limits:
o lim
𝑥→𝑐
(𝑓(𝑥) + 𝑔(𝑥)) = lim
𝑥→𝑎
𝑓(𝑥) + lim
𝑥→𝑎
𝑔(𝑥)
o lim
𝑥→𝑐
(𝑓(𝑥) − 𝑔(𝑥)) = lim
𝑥→𝑎
𝑓(𝑥) − lim
𝑥→𝑎
𝑔(𝑥)
o lim
𝑥→𝑐
(𝑓(𝑥)𝑔(𝑥)) = lim
𝑥→𝑎
𝑓(𝑥) × lim
𝑥→𝑎
𝑔(𝑥)
o lim
𝑥→𝑎
[𝑐𝑓(𝑥)] = c lim
𝑥→𝑎
𝑓(𝑥)
o lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
lim
𝑥→𝑎
𝑓(𝑥)
lim
𝑥→𝑎
𝑔(𝑥)
Finding Limits of Equations
- General: If lim
𝑥→𝑎+
𝑓(𝑥) = L and lim
𝑥→𝑎−
𝑓(𝑥) = L, then the limit lim
𝑥→𝑎
𝑓(𝑥) exists.
o If not, the limit does not exist.
- Limits by direct substitution
o F is continuous at x= a if lim
𝑥→𝑎
𝑓(𝑥) = f(a)
- Finding limits by factoring
o Find limit as x approaches 2 of f(x) = (x2
+ x – 6)/(x-2)
▪ Factor numerator and simplify to f(x) = x + 3
▪ Solution: 5
- If k and n are constants, |x| > 1, and n >0, then lim
𝑥→ 
𝑘
𝑥 𝑛
= 0, and lim
𝑥→ −
𝑘
𝑥 𝑛
= 0

- Rational function points of discontinuity
o Strategy
▪ Functions are undefined for inputs that make the denominator equal to
zero
▪ Inputs where the function is defined and the numerator is equal to zero are
zeros of the function. These are simply the x-intercepts of the graph of
the function.
▪ Simplify the function expression by canceling out common factors. Any
undefined input that no longer makes the denominator equal to zero is a
removable discontinuity. The remaining undefined inputs are vertical
asymptotes.
o Rationalizing functions with square roots
▪ Multiply by conjugate if you get the indeterminate form 0/0 when you
plug in the value x approaches into the function
• Ex. Conjugate of √4𝑥 + 4 + 4 is √4𝑥 + 4 − 4
o Polynomial Limits
▪ If the highest power of x in a rational expression is in the numerator, then
the limit as x approaches infinity is infinity.
▪ If the highest power of x in a rational expression is in the denominator,
then the limit as x approaches infinity is zero.
o Limits of Trigonometric Functions
▪ There are four standard limits you can memorize – with these, you can
evaluate all of the trigonometric limits that appear on the test:
1. lim
𝑥→0
sin(𝑥)
𝑥
= lim
𝑥→0
𝑥
sin(𝑥)
= 1 (where x is in radians)
a. Remember that both sin(x) and x have approximately the same
slope near the origin as x gets closer to zero
2. lim
𝑥→0
cos(𝑥)−1
𝑥
= 0
3. lim
𝑥→0
sin(𝑎𝑥)
𝑥
= 𝑎
4. lim
𝑥→0
sin(𝑎𝑥)
sin(𝑏𝑥)
=
𝑎
𝑏
Continuity
A function f is continuous at x = a if and only if:
1. f(a) exists
2. lim
𝑥→𝑎
𝑓(𝑥) exists
3. lim
𝑥→𝑎
𝑓(𝑥) = 𝑓(𝑎)
Types of Discontinuity (Draw an example of each of these!)
• Jump Discontinuity
o Occurs when the curve “breaks” at a particular place and starts somewhere else.
In other words, lim
𝑥→𝑎+
𝑓(𝑥) ≠ lim
𝑥→𝑎−
𝑓(𝑥).

• Point Discontinuity
o Occurs when the curve has a “hole” in it from a missing point because the
function has a value at that point that is “off the curve.” In other words,
lim
𝑥→𝑎
𝑓(𝑥) ≠ 𝑓(𝑎) .
• Removable Discontinuity
o Occurs when you have a rational expression with common factors in the
numerator and denominator.
▪ Since the factors can be canceled, the discontinuity is “removable.”
• Essential Discontinuity (Vertical Asymptotes)
o Occurs when the curve has a vertical asymptote.
Derivatives: Definition & Formulas
• Derivatives = slope of the tangent line to the graph of f(x) at some point
o Tangent line: straight line that touches a function at only one point = represents
instantaneous rate of change of a function at one point
o Secant line: straight line joining two points on a function, which gives us the
average rate of change (= slope between the two points)
o Limit definition of a derivative
▪ “Use slope of secant line between two nearby points to approximate slope
of tangent line at a certain point”
▪ f’(x) = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
• Power Rule
o If y = 𝑦 = 𝑥 𝑛
, 𝑡ℎ𝑒𝑛
𝑑𝑦
𝑑𝑥
= 𝑛𝑥 𝑛−1
a. If y = x, then
𝑑𝑦
𝑑𝑥
= 1
b. If y = kx, then
𝑑𝑦
𝑑𝑥
= 𝑘 (where k is a constant)
c. If y = k, then
𝑑𝑦
𝑑𝑥
= 0 (where k is a constant)
• Addition Rule
o If 𝑦 = 𝑎𝑥 𝑛
+ 𝑏𝑥 𝑛
, where a and b are constants, then
𝑑𝑦
𝑑𝑥
= 𝑎(𝑛𝑥 𝑛−1) +
𝑏(𝑛𝑥 𝑛−1
).
• Product Rule
o If f(x) = uv, then 𝑢𝑣′ + 𝑣𝑢′
• Quotient Rule
o If f(x) = u/v, then 𝑓′(𝑥) =
𝑣𝑢′− 𝑢𝑣′
𝑣2
• How to Remember: “Low De-High minus High De-Low, Cross the line
and Square the Below”
• Chain Rule
o If y = f(g(x)), then y’ = g’(x) • f’(g(x))
• How to Remember: “Derivative of the inside times the derivative of the
outside of the inside”
• Trig Functions:
o
𝑑
𝑑𝑥
sin( 𝑥) = cos( 𝑥)
o
𝑑
𝑑𝑥
cos(𝑥) = − sin(𝑥)

o
𝑑
𝑑𝑥
tan(𝑥) = sec2
(𝑥)
o
𝑑
𝑑𝑥
sec(𝑥) = sec(x)tan(x)
o
𝑑
𝑑𝑥
csc(𝑥) = -csc(x)cot(x)
o
𝑑
𝑑𝑥
cot(𝑥) = -csc2
(x)
Exponential and Logarithmic Derivatives
o If y = ln(x), then
𝑑𝑦
𝑑𝑥
=
1
𝑥
• Similarly, if y = ln(u), then
𝑑𝑦
𝑑𝑥
=
1
𝑢
(
𝑑𝑢
𝑑𝑥
)
o If y = ex
, then
𝑑𝑦
𝑑𝑥
= 𝑒 𝑥
• Similarly, if y = eu
, then
𝑑𝑦
𝑑𝑥
= 𝑒 𝑢
(
𝑑𝑢
𝑑𝑥
)
o If y= log 𝑎 𝑥, then
𝑑𝑦
𝑑𝑥
=
1
𝑥𝑙𝑛(𝑎)
• Similarly, if y = log 𝑎 𝑢, then
𝑑𝑦
𝑑𝑥
=
1
𝑢𝑙𝑛(𝑎)
(
𝑑𝑢
𝑑𝑥
)
o If y = ax
, then
𝑑𝑦
𝑑𝑥
= e(xlna)
(ln a) = (ax
)(ln a)
• Similarly, if y = au
, then
𝑑𝑦
𝑑𝑥
= au
(ln a) (
𝑑𝑢
𝑑𝑥
)
Implicit Differentiation
o Use when you cannot isolate y in terms of x, so you derive in terms of y AND x
o Ex. Find
𝑑𝑦
𝑑𝑥
if y3
– 4y2
= x5
+ 3x4
• 3𝑦2
(
𝑑𝑦
𝑑𝑥
) − 8𝑦 (
𝑑𝑦
𝑑𝑥
) = 5𝑥4
(
𝑑𝑥
𝑑𝑥
) + 12𝑥3
(
𝑑𝑥
𝑑𝑥
)
•
𝑑𝑥
𝑑𝑥
= 1, 𝑠𝑜
𝑑𝑦
𝑑𝑥
(3𝑦2
− 8𝑦) = 5𝑥4
+ 12𝑥3
• Thus:
𝑑𝑦
𝑑𝑥
=
5𝑥4+12𝑥3
3𝑦2−8𝑦
Derivative of an Inverse Function
o Strategy
• 1. Solve the equation with respect to x (e.g. x = something)
• 2. Switch the x and y. You now have the inverse function
• 3. Take derivative of the function.
• 4. Plug in point value into the equation and solve.
• Remember: You will be plugging in the given y value for x
(Because this is the inverse function!)
Derivative Applications
o Finding the equations for the tangent and normal line to the curve f(x) = y at a
point (x1, y1) as
• Finding the slope: Take the derivative of the function and then plug in the
value of x into the derivative function. The resulting value is the slope
(m). To find slope of the normal (perpendicular) line, take the negative
reciprocal of m.
• Finally, plug in values of m, x1, and y1 into the point-slope equation:
• y- y1 = m(x-x1)

o Mean Value Theorem for Derivatives
• If y = f(x) is continuous on the interval [a,b] and is differentiable
everywhere on the interval (a,b), then there is at least one number c
between a and b such that,
• 𝑓′(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
• “There’s some point in the interval where the slope of the tangent
line equals the slope of the secant line that connects the endpoints
of the interval”
• Rolle’s Theorem (Special Case of MVT)
• Same as Mean Value Theorem but in this case f(a) = f(b) = 0, so
the formula simplifies to finding the value of c where f’(c) = 0
o Maxima and Minima
• A maximum or minimum of a function occurs at a point where the
derivative of the function is zero, or where the derivative fails to exist.
• Relative (local) max/min: means that the curve has a horizontal
line at that point, but it is not the highest or lowest value that the
function attains.
• Absolute (global) max/min: occurs at an end point or an x-value
where there is a vertical asymptote
• Second Derivative Test: If a function has a critical value at x = c, then that
value is relative maximum if f’’(x) < 0 and it is a relative minimum if
f’’(x) > 0
• Intuition: Let’s look at the slope of the graph of f’(x)
• Application: Knowing the maxima or minima will help us to optimize
functions
o Curve Sketching
• Find Intercepts
• Set f(x) = 0, then solve for x. This tells you the function’s x-
intercepts (or roots)
• Set x= 0 to find y-intercept(s)
• Find Asymptotes
• Horizontal Asymptotes: Find limits of f(x) as x approaches + and
-
o If they give you an interval, evaluate f(x) at the endpoints
• Vertical Asymptotes: Find values of x that make f(x) undefined
• Test the First Derivative
• Find where f’(x) = 0. This tells you the critical points
(maxima/minima).
• When f’(x) >0, the curve is rising; when f’(x) < 0, the curve is
falling.
• Test the Second Derivative
• Find where f’’(x) = 0. This tells you the inflection points.
• When f’’(x) > 0, the curve is concave up; when f’’(x) < 0, the
curve is concave down

o Motion
• Related rates
• You’ll be given an equation relating two or more variables. These
variables will change with respect to time, and you’ll use
derivatives to determine how the rates of change are related.
• Position, Velocity, & Acceleration (Rectilinear Motion)
• Position Function x(t)
• Velocity Function v(t) = x’(t)
• Acceleration Function a(t) = v’(t) = x’’(t)
• When the velocity is negative, the particle is moving to the left.
o When the velocity is positive, the particle is moving to the
right.
• When the velocity and acceleration of the particle have the same
signs, the particle’s speed is increasing.
o When the velocity and acceleration of a particle have
opposite signs, the particle’s speed is decreasing (or
slowing down).
o When the velocity is zero and the acceleration is not zero,
the particle is momentarily stopped and changing direction.
o L’Hopital’s Rule
• Used to find the limit of an expression if it results in an indeterminate
form (
0
0
𝑜𝑟
∞
∞
)
• If f(c) = g(c)= 0, and if f’(c) and g’(c) exist, and if g’(c) ≠ 0, then
• lim
𝑥→𝑐
𝑓(𝑥)
𝑔(𝑥)
=
𝑓′(𝑥)
𝑔′(𝑥)
• Similarly, if f(c) = g(c)= ∞, and if f’(c) and g’(c) exist, and if g’(c) ≠ 0,
then:
• lim
𝑥→𝑐
𝑓(𝑥)
𝑔(𝑥)
=
𝑓′(𝑥)
𝑔′(𝑥)
• Translation: If the limit of an expression is undefined, take the derivative
of the top and bottom until you get a determinate expression. Then, take
the limit.
o Differentials/Linearization
o Differential = very small quantity that represents a change in a number (∆x)
o Used to approximate the value of a function
o Take the definition of a derivative, replace h with ∆x, and get rid of the
limit:
▪ 𝑓′(𝑥) ≈
𝑓(𝑥+∆𝑥) – 𝑓(𝑥)
∆𝑥
▪ Rearrange to get:
• 𝒇(𝒙 + ∆𝒙) ≈ 𝒇(𝒙) + 𝒇′(𝒙)∆𝒙
• This equation only works with radians (No Degrees!)
o A similar function used to estimate the error or change in a
measurement:
▪ dy = f’(x)dx

______________________________________________________________________________
Integrals (Antiderivative)
• Used to help find area under curve or evaluate volume enclosed by a function
• Power Rule: If f(x) = xn
, then ∫ 𝑓(𝑥)𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
+ 𝐶 (except when n = -1)
o It follows that:
▪ ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥
▪ ∫[𝑓(𝑥) + 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥
▪ ∫ 𝑘𝑑𝑥 = 𝑘𝑥 + 𝐶
• Integrals of Trig Functions
o ∫ sin(ax)𝑑𝑥 = −
cos(ax)
a
+ C
o ∫ cos(ax)𝑑𝑥 =
sin(ax)
a
+ C
o ∫ sec(ax) tan(𝑎𝑥) 𝑑𝑥 =
sec(ax)
a
+ C
o ∫ sec2
𝑎𝑥 𝑑𝑥 =
tan(ax)
a
+ C
o ∫ csc(ax) cot(𝑎𝑥) 𝑑𝑥 = −
csc(ax)
a
+ C
o ∫ csc2
𝑎𝑥 𝑑𝑥 = −
cot(ax)
a
+ C
o ∫
du
u
= ln |u| + C
o ∫ tan(x)𝑑𝑥 = −ln|cos 𝑥| + C
o ∫ cot(x)𝑑𝑥 = ln|sin 𝑥| + C
o ∫ sec(𝑥) = ln|sec(x) + tan(𝑥)| + C
o ∫ csc(x)𝑑𝑥 = −ln|csc(x) + cot(𝑥)| + C
o ∫ ex
dx =
1
k
ekx
+ C
• Integration technique: U-substitution
o ∫ un
du =
un+1
n+1
+ C
o Pick one expression to equal u, derive u, and then plug in values of u and du into
original equation.
Evaluating Integrals using Geometry
o Using Riemann Sums to estimate the integral of a function
▪ Translation: Finding area under a curve by adding up areas of rectangles
▪ Formulas (Don’t memorize these!):
• Left-Handed Sum
o
𝑏−𝑎
𝑛
(y0+ y1 + y2 + y3 + y4 + …+yn-1)
• Right-Handed Sum
o
𝑏−𝑎
𝑛
(y1 + y2 + y3 + y4 +y5 …+yn)
• Midpoint Sums
o
𝑏−𝑎
𝑛
(y1/2 + y3/2 + y5/2 + y7/2 +y9/2 …+y[2n-1]/2)
o Note: Fractional subscript means to evaluate the function at
the number half-way between each integral pair of n-values

• Where a = left endpoint of the interval, b= right endpoint of the
interval, and n= number of triangles we’re using
o Trapezoid Rule = Better at estimating area than Riemann Sum method
▪ Finding area under a curve by adding up areas of trapezoids
▪ Formula
• Using formula for area of a trapezoid A= ½ (b1+ b2)(h) you can
derive the formula:
• (
1
2
) (
𝑏−𝑎
𝑛
)(y0+2y1+2y2+2y3+…+2yn-2+2yn-1+yn)
The First Fundamental Theorem of Calculus
o ∫ f(x)dx = F(b) − F(a); where F(x)is the antiderivative of f(x)
b
a
The Second Fundamental Theorem of Calculus
o If f(x) is continuous on [a,b], then the derivative of the function F(x) = ∫ 𝑓(𝑡)𝑑𝑡
𝑥
𝑎
is:
▪
dF
dx
=
𝑑
𝑑𝑥
∫ 𝑓(𝑡)𝑑𝑡 = 𝑓(𝑥)
𝑥
𝑎
• All we are concerned with is swapping the upper variable term (in
this case x) with the variable in the function (in this case t)
o The bottom term could be any constant value
• If the upper variable term is a function of x (e.g. x2
), we multiply
the answer by the derivative of that term (e.g. 2x)
Mean Value Theorem for Integrals
o Enables you to find the average value of a function
o If f(x) is continuous on a closed interval [a,b], then at some point c in the interval
[a,b] the following is true:
▪ ∫ 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑐)(𝑏 − 𝑎)
𝑏
𝑎
• Translation: The area under the curve of f(x) on the interval [a,b] is
equal to the value of the function at some value c (between a and
b) times the length of the interval.
o This equation can be rearranged to find the average value of a function:
▪ 𝑓(𝑐) =
1
𝑏−𝑎
∫ 𝑓(𝑥)𝑑𝑥
𝑏
𝑎
Area Between Two Curves
o Vertical Slices
o If a region is bounded by f(x) above and g(x) below at all points of the interval
[a,b], then the area of the region is given by:
▪ ∫ [ 𝑓( 𝑥) − 𝑔( 𝑥)] 𝑑𝑥
𝑏
𝑎
o Horizontal Slices
o If a region is bounded by f(y) on the right and g(y) on the left at all points of the
interval [c,d], then the area of the region is given by
▪ ∫ [𝑓(𝑦) − 𝑔(𝑦)]𝑑𝑦
𝑑
𝑐

Volume of a Solid of Revolution
o Washers (Disk) Method: Finding the volume of a complex shape by adding up volumes
of many thin discs
o Think: “You’re adding up the volume of each thin pancake in a stack!”
o Disk: If you have a region whose area is bounded by the curve y = f(x) and the x-
axis on the interval [a,b], each disk has a radius of f(x), and the area of each disk
will be π[f(x)]2
(Just like the area of a circle of a circle is A = πr2
!)
▪ To find the volume, evaluate the integral:
• π ∫ [𝑓(𝑥)]2
𝑑𝑥
𝑏
𝑎
(where dx is some small depth along the x-axis)
o Washer: If you have a region whose area is bounded above the curve y = f(x) and
below by the curve y = g(x), on the interval [a,b], then each washer will have an
area of π[f(x)2
– g(x)2
] (We’re subtracting the area of one circle from another!)
▪ To find the volume, evaluate the integral: (rotated about the x-axis)
• π ∫ [𝑓( 𝑥)2
− 𝑔( 𝑥)2
]
𝑏
𝑎
dx
▪ Finding volume for washers when the region is rotated about y –axis:
• π ∫ [𝑓(𝑦)2
− 𝑔(𝑦)2
]
𝑑
𝑐
dy
o Remember: dy and dx just stand for really (infinitesimally) small depths along the
y or x axis respectively!
o Cylindrical Shells Method
o Think: “Finding the volume of each layer of an onion” (Shrek!)
o If you have a region whose area is bounded above by the curve y= f(x) and below
by the curve y = g(x), on the interval [a,b], then each cylinder will have a height
f(x) – g(x), a radius of x, and an area of 2 πx[f(x)-g(x)].
▪ Volume when the region is rotated around the y-axis:
• 2π∫ 𝑥[𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑏
𝑎
o If you have a region whose area is bounded on the right by the curve x =f(y) and
on the left by the curve x =g(y), on the interval [c,d], then each cylinder will have
a height of f(y) – g(y), a radius of y, and an area of 2 πy[f(y)-g(y)].
▪ Volume when the region is rotated around the x-axis:
• 2𝜋 ∫ 𝑦[𝑓(𝑦) − 𝑔(𝑦)]
𝑑
𝑐
dy
o Volumes of Solids with Known Cross-Sections
o If you’re given an object where you know 1) the shape of the base and 2) that the
perpendicular cross-sections are all the same regular, planar geometric shape
▪  You can integrate using the area of that shape!
▪ Strategy: Find the area of the regular (=all sides are equal) shape, multiply
by it some small depth dy or dx (depending on how the object is oriented),
and then integrate from the endpoints of an interval
Differential Equations
o Equations that relate a function with one or more of its derivatives
o How to solve differential problems:
o Separate the variables
o Integrate both sides
o Solve for the constant

o Applications:
o Position, Velocity, and Acceleration functions
▪ Derivative of Position Function  Velocity Function
▪ Derivative of Velocity Function  Acceleration Function
▪ We can go backwards as well:
• Integral of Acceleration function  Velocity function
• Integral of Velocity function  Position function
o Exponential Growths and Decay
Slope Fields (Direction Fields)
o Making a graphical representation of the slope of a function at various points in the plane.
o You will be given
𝑑𝑦
𝑑𝑥
. Plug-in different values for x and y and graph the slopes at the
respective coordinates.
Test-Taking Strategies
- Show all of your work (Free Response)
o Remember that the grader is more interested in seeing if you know HOW to solve
a problem rather than just checking to see if your answer is correct/incorrect
- Do not round partial answers
o Store them in your calculator so that you can use them unrounded in further
calculations
▪ Be sure you’re in the correct mode  Most always radian mode
o Unless otherwise specified, your final answers should be accurate to three places
after the decimal point
- Use Process of Elimination and Bubble in for all problems
o There is No Guessing Penalty
- Pro-Tip: During the second timed portion of the free-response section (Part B), you are
permitted to continue work on problems in Part A, but you are not permitted to use a
calculator during this time.
- Breathe. You’ll do great! ☺

AP Advantage: AP Calculus

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AP Advantage: AP Calculus