Download AP Calculus AB

AP Calculus AB
Formulas & Justifications
Limits at Infinity:
To find lim f ( x) think
1
2
x
Top Heavy  limit is ±∞
Bottom Heavy  limit is 0
Equal  limit is ratio of coefficients
Limits with Infinity (at vertical asymptotes):
When finding a one-sided limit at a vertical asymptote,
the answer is either ±∞.
Justifying that a function is continuous at a point:
f is continuous at c iff:
1. f (c ) is defined
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2. lim f ( x ) exists
x c
3. f (c ) = lim f ( x )
x c
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Definition of the Derivative:
f ( x  h)  f ( x )
f ( x  x)  f ( x)
f '( x)  lim
 lim
h 0

x

0
h
x
f ( x)  f (a)
(Alternate form for a derivative at a given value.)
xa
Justifying that a derivative exists at a point, c :
Show algebraically that lim f '( x)  lim f '( x) .
f '(a)  lim
x a
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x c
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9
10
x c
Average Rate of Change of f on [a, b]:
f (b)  f (a )
y
a.r.c. =
(algebra slope of
)
ba
x
Instantaneous Rate of Change of f at a :
f '(a)
(derivative at the given value)
Power Rule:
d n
 x   nx n 1
dx
Common Derivatives to Remember:
d  1  1

dx  x  x 2
Trig Function Derivatives:
d
sin x   cos x
dx
d
 tan x   sec2 x
dx
d
sec x   sec x tan x
dx
AB Calculus Formulas & Justifications
(slope of secant line)
(slope of tangent line)
d  
1
x 

dx
2 x
d
 cos x    sin x
dx
d
cot x    csc2 x
dx
d
csc x    csc x cot x
dx
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Derivatives of Inverse Trig Functions:
d
1
[arcsin x] 
dx
1  x2
d
1
[arctan x] 
dx
1  x2
d
1
[arc sec x] 
dx
x x2 1
d
1
[arccos x] 
dx
1  x2
d
1
[arc cot x] 
dx
1  x2
d
1
[arc csc x] 
dx
x x2 1
Derivatives of Exponential and Logarithmic Functions:
d
1
d
1
[ln x]  , x  0
[log a x] 
dx
x
dx
x ln a
d x
d x
[e ]  e x
[a ]  a x ln a
dx
dx
Justifications for horizontal tangent lines:
dy
 0.
f ( x ) has horizontal tangents when
dx
Chain Rule:
dy dy du
d


 f ( g ( x))  f '( g ( x))  g '( x)
dx du dx
dx
Product Rule:
d
 f ( x) g ( x)  f ( x) g '( x)  g ( x) f '( x)
dx
Quotient Rule:
d  f ( x)  g ( x) f '( x)  f ( x) g '( x)

2
dx  g ( x) 
 g ( x) 
Derivatives of Inverse Functions:
The derivative of an inverse function is the reciprocal of the derivative of the
original function at the “matching” point.
1
If (a, b) is on f ( x) , then (b, a) is on f 1 ( x) and ( f 1 ) '(b) 
.
f '(a)
Justifications for vertical tangent lines:
dy
f ( x ) has vertical tangents when
is undefined.
dx
Justifications for Particle Motion:
Particle is moving right/up because v(t )  0 (positive).
Particle is moving left/down because v(t )  0 (negative).
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Particle is speeding up (|velocity| is getting bigger) because v (t ) and a(t ) have
same sign.
Particle is slowing down (|velocity| is getting smaller) because v (t ) and a(t )
have different signs.
AB Calculus Formulas & Justifications
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Intermediate Value Theorem:
If f is continuous on [a, b] and k is any number between f (a ) and f (b ) , then
there is at least one number c between a and b such that f (c )  k .
Extreme Value Theorem:
If f is continuous on the closed interval [a, b], then f has both a minimum
and a maximum on the closed interval [a, b].
Justification for an Absolute Extrema.
1. Find critical numbers.
2. Identify endpoints.
3. Find f (critical numbers) and f (endpoints) .
4. Determine absolute max/min values by comparing the y-values. State in a
sentence.
Mean Value Theorem:
If f is continuous on [a, b] and differentiable on (a, b) then there exists a
f (b)  f (a )
number c on (a, b) such that f '(c) 
.
ba
(Calculus slope = Algebra Slope)
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Rolle’s Theorem:
If f is continuous on [a, b] and differentiable on (a, b) and if f (a )  f (b) ,
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Justification for a Critical Number:
x  c is a critical number because f '( x)  0 or f '( x) is undefined.
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then there exists a number c on (a, b) such that f '(c)  0 .
Justification for Increasing/Decreasing Intervals:
Inc: f ( x) is increasing on [____, ____] b/c f '( x)  0 .
Dec: f ( x) is decreasing on [____, ____] b/c f '( x)  0 .
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Justification for a Relative Max/Min Using 1st Derivative Test:
Local Max: f '( x) changes from + to -.
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Justification for Relative Max/Min Using 2nd Derivative Test:
Local Max: f '(c)  0 (or und) and f ''( x)  0 .
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Justification for a Point of Inflection:
Using 2nd derivative: f ''( x)  0 (or dne) AND f ''( x) changes sign.
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Justification for Concave Up/Concave Down:
Concave Up: f ( x) is concave up on (____, ____) because f ''( x)  0 .
Local Min: f '( x) changes from - to +.
Local Min:
f '(c)  0 (or und) and f ''( x)  0 .
Using 1st derivative: f ''( x)  0 (or dne) AND slope of f '( x) changes sign.
Concave Down: f ( x) is concave down on (____, ____) because f ''( x)  0 .
AB Calculus Formulas & Justifications
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Justifications for linear approximation estimates:
A linear approximation (tangent line) is an overestimate if the curve is
concave down.
A linear approximation (tangent line) is an underestimate if the curve is
concave up.
Integration Rules:
1 n1
x n dx 
x C
 cos xdx  sin x  C
n 1
1
 cos  kx  dx  k sin  kx   C
 sin xdx   cos x  C
1
2
 sin  kx  dx   k cos  kx   C
 sec xdx  tan x  C
2
 csc xdx   cot x  C
 sec x tan xdx  sec x  C

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1
 x dx  ln x  C
 e dx  e  C
 csc x cot xdx   csc x  C
 tan xdx   ln cos x  C
e
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kx
dx 
x
1 kx
e C
k
x
2
x
1
1
x
dx  arctan    C
2
a
a
a
Justifications for Reimann Sums:
Left-Riemann Sums:
The sum is an overestimate if the curve is decreasing.
The sum is an underestimate if the curve is increasing.
Right-Riemann Sums:
The sum is an overestimate if the curve is increasing.
The sum is an underestimate if the curve is decreasing.
First Fundamental Theorem of Calculus:

b
a
f '( x)dx  f (b)  f (a)
(Finds the signed area between a curve and the x-axis)
Properties of Integrals:


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


b
f ( x)  g ( x)dx 
a
b
f ( x)  g ( x)dx 
a
b
cf ( x)dx  c
a
b
f ( x)dx  
a
a




b
f ( x)dx 
a
b
f ( x)dx 
a


b
g ( x)dx
a
b
g ( x)dx
a
b
f ( x)dx
a
a
f ( x)dx
b
f ( x)dx  0
a
AB Calculus Formulas & Justifications
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Average Value of a Function:
b
1
f avg 
f ( x)dx
ba a
Second Fundamental Theorem of Calculus:
d x
d g ( x)
f
(
t
)
dt

f
(
x
)
f (t )dt  f ( g ( x))  g '( x)
dx  a
dx  a
“Net Change” Theorem:

b
f ( x)dx represents the “net change” in the function f from time a to b.
a
Finding Total Amount:
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f (b)  f (a) 

b
f '( x)dx
(want = have + integral)
a
Steps for Solving Differential Equations:
“Find a solution (or solve) the separable differentiable equation…”
1. Separate the variables
2. Integrate each side
3. Make sure to put C on side with independent variable (normally x)
4. Plug in initial condition and solve for C (if given)
5. Solve for the dependent variable (normally y)
Exponential Growth and Decay:
“The rate of change of a quantity is directly proportional to that quantity”
dy
Gives the differential equation:
 ky
dt
Which can be solved to yield: y  Cekt
Particle Motion Formulas:
Velocity:
v(t )  s '(t )
Speed:
speed  v(t )
Acceleration:
a(t )  v '(t )  s ''(t )
s (b)  s (a )
ba
b
1
v(t )dt
(given v (t ) )
ba a
v (b)  v (a )
Average Acceleration:
(given v (t ) )
ba
b
1
a (t )dt
(given a(t ) )

ba a
Average Velocity: (given s (t ) )
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b
Displacement:

Total Distance:

Position at b:
s (b)  s (a)   v(t )dt
AB Calculus Formulas & Justifications
a
b
a
v(t )dt
v(t ) dt
b
a
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Areas in a Plane:
Perpendicular to x-axis:
  f ( x)  g ( x) dx
b
a
f ( x ) is top curve, g ( x ) is bottom curve, a and b are x-coordinates of
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point of intersection
Perpendicular to y-axis:
  f ( y)  g ( y) dy
b
a
f ( y ) is right curve, g ( y ) is left curve, a and b are y-coordinates of
point of intersection
Steps to Finding Volume:
Volume =
 Area
1. decide on whether it’s a dx or dy
2. find a formula for the area in terms of x or y
3. find the limits (making sure they match x or y)
4. integrate and evaluate
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Volumes Around a Horizontal Axis of Rotation or Perpendicular to x-axis:
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b
Disc:
V    r 2 dx
a and b are x-coordinates
Washer:
V    R 2   r 2  dx
a
b
a and b are x-coordinates
a
Slab (Cross Section):
b
V   A( x)dx
a
A( x ) is the area formula for the
cross section
Volumes Around a Vertical Axis of Rotation or Perpendicular to y-axis:
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b
Disc:
V    r 2 dy
a and b are y-coordinates
Washer:
V    R 2   r 2  dy
a
b
a and b are y-coordinates
a
Slab (Cross Section):
b
V   A( y )dy
a
A( y ) is the area formula for the
cross section
Volume=
AB Calculus Formulas & Justifications
Surface Area= 6𝑠 2 ,
2𝜋𝑟ℎ + 2𝜋𝑟 2 ,
𝐷𝑜𝑛′ 𝑡 𝑛𝑒𝑒𝑑 4𝜋𝑟 2
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