AP Calc Presentations

7/30/2019 AP Calc Presentations

1/26

AP Calculus Presentations

Table of Contents

AP Calculus Presentations...............................................................................................................1Bradley and Maggie - Riemann Sums and Simpsons Rule........................................................2

What is a Riemann Sum?.........................................................................................................2What is Simpsons Rule?.........................................................................................................3

Gina and Jill - Centre of Mass, Laminas and Centroid Handout.................................................4

Centre of Mass - One Dimension.............................................................................................4Centre of Mass - Two Dimensions..........................................................................................5

Laminas....................................................................................................................................6

Mackenzie - Hyperbolic Functions Handout...............................................................................8

The Basics Hyperbolic Functions and Where to Start..........................................................8

Reciprocal Hyperbolic Functions............................................................................................9Integration with Hyperbolic Functions..................................................................................10

Alex and Ashley - Improper Integrals Handout.........................................................................12

Alastair and Graham - Power Series Handout...........................................................................15

Power Series...........................................................................................................................15The Radius and Interval of Convergence ..............................................................................16

Steps for Solving for Radius & Interval of Convergence .....................................................17

Joyce - Partial Fraction Decomposition Handout......................................................................18Tyler - Newtons Method and Linear Approximations Handout...............................................21

Nicole and Selena - Area of a Surface of Revolution Handout.................................................23

Corey and Paul Arc Length Handout .....................................................................................25


2/26

Bradley and Maggie - Riemann Sums and Simpsons Rule

What is a Riemann Sum?

A Riemann sum is the estimation of the area under a curve (definite integral) by finding

the sum of the areas of rectangles of equal width (x).

The definite integral can be calculated by expressing it as a limit of Riemann sums.

This is how it is expressed as a limit

There are many different types of Riemann sums, the most well-known being Left, Right

and Midpoint sums.

These types of Riemann Sums are all basically the exact same method; the type of

Riemann sum being used only denotes the point at which the height of the rectangle is

being taken (f(x))

Riemann Sums are particularly good for finding the definite integral of functions that are

not as curvy. (IE. A linear function)

The simplified way of solving Riemann Sums

=

1

0

)(n

i

i ttf

How to use the TI Riemann Sums Program

1. Go into Programs2. Choose AREA

3. Enter the Equation

4. Left bound of your interval5. Right bound of your interval

6. Then Number of rectangles you want (N)

7. Choose the type of Riemann Sum(LRAM,

MRAM, RRAM)8. See the graph created by the program

9. Hit enter and see the value of the estimated

area

Example of Riemann sums on the TI-83 plus


3/26

)4(3

rmlhAp

++=

What is Simpsons Rule?

Simpsons Rule allows you to find the area under a function, or the definite integral

Simpsons Rule: Area under a parabolic arc simple formula:

Where h is half the length of the base, l and r are the lengths of the left and right sides,

and m is the altitude at the midpoint of the base

What you will need to be thinking when you are using Simpsons Rule:

)42...424(3123210 nnn yyyyyyy

hS +++++++=

Where the interval is partitioned into an even number of n subintervals of equal length (

3

h); when doing this remember the way we did trapezoidal approximation. Only take the

heights on the end once each and you need two of some heights because you need the

height twice for different intervals.

Simpsons Rule is particularly good at finding the definite integral of functions that are

curvier. (IE. A cubic function)


4/26

Gina and Jill - Centre of Mass, Laminas and Centroid Handout

The centre of mass is the coordinate in a rod or plate where it can be balanced on a point

horizontally. For example: if you are trying to find the centre of mass of a meter stick, the centre

of mass will be at the exact geometric centre because it is of uniform mass. If youre trying tofind the centre of mass of an object of irregular shape, the centre of mass will be nearer to the

heavier end.

Think of a seesaw, the fulcrum is the centre of the seesaw, this is where the centre of mass is

located.

Centre of Mass - One Dimension

m1 and m2 are the measurements of the mass from the centre to the outside point . The distance

between those two points and the fulcrum are d1 and d2.

If we picture the seesaw to be on an x-axis, m1 is the mass from the centre to x1 and m2 is the

same forx2. The fulcrum takes place at the xf (f is for the fulcrum) value between the two points.

This x is the centre of mass of the system as shown in the picture below:

The distance from the fulcrum to the outerx points is the variable d where x1 corresponds to d1

and x2 corresponds to d2 in the equation : m1d1=m2d2These have to be equal to have the centrecoordinate.

d1 = xf x1 and d2 = xf x1. So we can now substitute d1:


5/26

The numbers m1x1 and m2x2 are called the moments of the respective masses. WhereM, the sum

of the individual moments is called the moment of the system, and m, the sum of the masses is

the total mass of the system:

One dimensional cases are rarely used in calculus. This form of finding the centre of mass is onlyuseful in finding the centre of a straight line/rod with masses attached to each end. These

examples are often used in physics when solving for torque. Expect to see examples such as a

rod balanced on a fulcrum, two particles or two masses separated by some distance, or even two

children on a seesaw.

Example Problem

Two reckless children are playingon a seesaw without handle bars. It

is 4 meters long of uniform mass.Where is the centre of mass located

if child A is 40kg and 2.3meters

from the fulcrum and child B is80kg? (note: the image to the left

is not to scale)

Solution

( )( ) ( )( )( )

9.18040

7.1803.240=

++

=x

Centre of Mass - Two Dimensions

The equation above needs to include every single outer point of the shape when we look at two

or three dimensions, so we can convert it to a series that has n points and n moments. In thesecases the same concept applies (sum of the moments divided by sum of the masses) but we must

look at more than just the x axis.

Laminas are used in two dimensions. Laminas are 2-D shapes with a negligible thickness (whichis usually called ). For these cases, you need to take into account the y-axis as well as the x-

axis. You must define the moments of the xs separate from the ys.


6/26

Laminas

Laminas are the area between two curves. Finding the centroid of a lamina uses the following

steps. The equation below gives the mass of the whole area of the lamina. Next step is to find thecentre of that mass. We multiply the thickness () by the area between the two curves.

Next, we find the moments of the lamina: Mx and My

Therefore:In conclusion we know that centre of mass (CM) is:

Our new formula below gives us the x and y coordinates:

Example Problem

Determine the center of mass for the region bounded by and .

Solution

The two curves intersect at and the center of mass marked with a dot.


7/26

Next find the area of the region:

Now find the moments of both x and y:

Now we can find the coordinates of x and y:

The coordinates for x and y are:


8/26

Mackenzie - Hyperbolic Functions Handout

The Basics Hyperbolic Functions and Where to Start

- Hyperbolic functions are analogs of the original trigonometric functions. However, thesefunctions arent defined like the original trig functions, which are circular based and have

trigonometric properties, these functions are hyperbolically based, and parameterize a

hyperbola. However, these functions are similar to the original, though they much morestraightforward. Hyperbolic functions have many uses, and can relate to many real world

properties. For example, hyperbolic cosine represents the curve made when an

unweighted chain is hung between two objects at equal points. This shape is catenary,and we can manipulate hyperbolic cosine to represent it.

- The basic hyperbolic functions are hyperbolic sine and cosine, which derive the

hyperbolic tangent, and are the foundation for all hyperbolic functions.

-2

sinhxx

eex =

-2

coshxx ee

x+=

- From these two definitions, we can derive hyperbolic tangent:

1

1

2

2

cosh

sinhtanh

2

2

+

=+

=+

==

x

x

xx

xx

xx

xx

e

e

ee

ee

ee

ee

x

xx

The graphs of these can be seen below

- As seen above, hyperbolic tangent and sine, both have zeros at (0,0), and hyperboliccosine has a minimum at (0,1)
http://upload.wikimedia.org/wikipedia/commons/7/76/Sinh_cosh_tanh.svg


9/26

Reciprocal Hyperbolic Functions

-From our standard hyperbolic functions, sinh(x), cosh(x) and tanh(x), we can derive ourreciprocal hyperbolic functions: csch(x), sech(x), and coth(x).

- xx eexh

=

2)(csc

-xx

eexh +

= 2)(sec

-1

1

sinh

cosh)coth(

2

2

+==

x

x

e

e

x

xx

Hyperbolic Functions and Their Unique Identities

- Similar to Pythagorean identity of trigonometric functions 1cossin 22 =+ xx , hyperbolicfunctions carry the equation 1sinhcosh 22 = xx

- We can simply prove this: 2222

22sinhcosh

+

=

xxxx eeeexx

factoring out the denominator, and multiplying out:

( ) ( )[ ]

( )

1

44

1

224

1

0

2222

==

=

+++=

e

ee

eeeeeeee

xx

xxxzxxxz

Hyperbolic functions also have other similar identities to trigonometric functions, such asyxyxyx sinsinhcoshcosh)cosh( +=+ , and

xxx coshsinh2)2sinh( =

Derivatives of Hyperbolic Functions

- The derivatives of hyperbolic functions are the same as the trigonometric derivatives,though in hyperbolic form, with the exception that the derivative of hyperbolic cosine is

hyperbolic sine.

-

- However, if one fails to memorize these derivatives, they can always be proven using the

definitions of hyperbolic sine and cosine, and solving using the chain rule or logarithmic

differentiation.


10/26

- For example: solve for the derivative of hyperbolic sine using the chain rule:

xy

eey

eeee

eex

xx

xxxx

xx

cosh'

2'

4

)0)(()(2=y'

2sinh

=

+=

+

=

- Examples of applications using differentiation:

1. Differentiate xy 2sinh2=

[ ]

xxy

xxy

2cosh2sinh4'

)2(2cosh2sinh2'

=

=

2. f(x) = coshx, find f(0)

0

2)0sinh(

2sinh

sinhcosh

00

=

=

=

=

ee

eex

xxdx

dy

xx

Integration with Hyperbolic Functions

- Integration with hyperbolic functions is a bit trickier than differentiation. This is becauseanti-differentiating some identities, require the anti-differentiation of complex fractions.

- For this reason, when integrating hyperbolic functions, if you are not very skilled in all

techniques of integration, memorizing the integral formulas is a necessity.- The formulas are as follows:

+= Cuxuux sinh1

cosh

+= Cuxuux cosh1

sinh

Cuxu

ux += coshln1

tan

Cuxu

ux += )harctan(sin1sec

Cux

uhux += 2tanln

1csc

+= Cuxuux sinhln1

coth

- The integration of hyperbolic functions is pretty straight forward with these formulas, and

can be solved using regular methods of integration


11/26

Examples:

1. 2

3cosh2 x

[ ]

Cxx

xx

+=

=

2

2

3sinh12

)3(63sinh3

2

2.

[ ]

[ ]

04.11

17621.34

)2

(4)2

(4

)0cosh(4)2cosh(4

cosh4sinh4

0022

2

0

2

0

=

+

+=

=

=

eeee

xx


12/26

Alex and Ashley - Improper Integrals Handout

An improper integral is a definite integral that has either one or both limits infinite or an

integrand that approaches infinity at one or more points in the interval of integration. It is often

necessary to use improper integrals in order to find a value for integrals that may not exist due toinfinity bounds or limits. Improper integrals can either converge or diverge. If it is convergent,

the corresponding limit exists. If it is divergent, the corresponding limit does not exist. There are

two types of improper integrals, you can identify each type by the bounds on the integral.

Definition of Type 1 & 2-

Type 1:

a) If

t

adxxf )( exists for every number t a, then

=

t

aat

dxxfdxxf )(lim)(

Provided this limit exists (finite number)

b) If b

t

dxxf )( exists for every number t b, then

=

b

t

b

tdxxfdxxf )(lim)(

Provided this limit exists

c) If both

a

dxxf )( and

b

dxxf )( are convergent, then we define

+=a

a

dxxfdxxfdxxf )()()(

Provided a is a real number.

Example of Type 1-


13/26

12

1

12

10

12

1

)13(3

1lim

)13(3

1lim

)13(3

1

3

1

1

3

1

313

)13(

1lim

)13(

1

1

2

1

2

1

2

=

+=

+

+

=

+

=

+

=

=

=

=+=

+=

+

t

x

x

u

u

dxduxu

dxx

dxx

t

t

t

t

t

convergent

Type 2:

a) If f

is continuous on [a, b) and is discontinuous at b, then

=t

a

b

abt

dxxfdxxf )(lim)(

Provided the limit exist

b) Iffis continuous on (a, b] and is discontinuous at a, then

=b

t

b

aat

dxxfdxxf )(lim)(

Provided the limit exists

c) If fhas discontinuity at c, where a < c < b, and both c

a

dxxf )( and b

c

dxxf )( are

convergent, then we define


14/26

+=b

a

c

a

b

c

dxxfdxxfdxxf )()()(

If one part of the integral is divergent the whole integral is divergent, therefore the limit

does not exist.

Example of Type 2-

( )

32

032

232lim

22lim

2lim

2

22

1

2

5

2

5

2

5

2

5

2

=

=

=

=

=

=

t

x

x

dx

x

dx

xatVAdxx

t

tt

tt

convergent


15/26

Alastair and Graham - Power Series Handout

Arithmetic Geometric

-

= +++++=0 ...3210n nn - ( )n

n

n

arraraara ++++=

=

...)()(1)(2

1

1

Common Ratio is found in geometric series using the formula:

n

n

a

aR 1+=

The Ratio Test

The Ratio Test is used to find whether a series converges and if so, for which values of x it does

so. The Ratio Test does this by finding the limit of the different ratios in a sequence. The Ratio

Test is represented algebraically by the following:

La

aLim

n

n

n=+

1

- If L < 1, then the series is absolutely convergent.

- If L > 1 or L = , then the series is divergent.

- If L = 0, then no conclusion can be made.

Power Series

The formula for a power series is given by the formula:

n

n

n

n

n xcxcxccxc ++++=

=

...22100

Or more generally:

n

n

n

n

naxcaxcaxccaxc )(...)()()(

2

210

0

++++=

=

Where cis an integer coefficient and xis always a variable. The a is also an integercoefficient, and the series is stated as being centred about this point a. A power series is

similar to a geometric series, but there is a distinct difference. Unlike previous series, a powerseries contains a variable x; and depending on this variable x:

A power series can be convergent for some values of x, but divergent for others.


16/26

The Radius and Interval of Convergence

For a given power series

=

0

)(n

n

naxc there are only three possibilities:

i.) The series converges only when x = a.

ii.) The series converges for all x, ().iii.) There is a positive numberRsuch that the series converges if Rax .

The third possibility is the most important and consequently the most difficult to understand,

it can also be written as:

- The series converges if RaxR


17/26

Steps for Solving for Radius & Interval of Convergence

1.) Set up & solve the limit for the ratio test.

2.) To solve for where it is convergent, we make the solution to the Limit < 1, (This comesfrom the details of the ratio test).

3.) Solve for which values of x the series will converge.

4.) Determine the Radius & Interval of Convergence.


18/26

Joyce - Partial Fraction Decomposition Handout

Partial Fraction Decomposition is a type of integration method used to break apart rational

functions.

Rational Functions consist of two polynomials in this form:

)(

)(

xg

xf

Thus partial fraction decomposition is a technique used to find:

dxxgxf

)(

)(

Partial Fraction Decomposition rules

1. The degree of the polynomial in the numerator must be less than that of the denominator. If

this is not the case then you must divide the numerator by the denominator to obtain a

numerator of a smaller degree than the denominator.

ie.2

222

234

+

++++

xx

xxxx=

22 + xx

00030

2

22

2

234

234

+++++

+

++++

x

xxx

xxxx

633

223

2

2

+

++

xx

xx

8x

2

222

234

+

++++

xx

xxxx=

2

83

2

2

+

++

xx

xx

2. When substituting variables, if the denominator is linear then only one variable is needed. If

the denominator is quadratic the formula of a linear function is used.

ie. Linear Denominator:111

12 +

+

= x

B

x

A

x

Quadratic Denominator:

44

523

2

+

++=

+

+

x

CBx

x

A

xx

x

3. If the denominator has the same factor to the thn power, then it should follow this form:

Linear:( ) ( ) n

n

n

kx

A

kx

A

kx

A

kx

x

+++

+=

+

+...

)(

52

212

1A , 2A , nA are unknown constants.

Quadratic:n

nn

ncbxax

BxA

cbxax

BxA

cbxax

BxA

cbxax

x

)(...

)()(

5222

22

2

11

2

2

++

+

++

++

++

+=

++

+


19/26

1A , 2A , nA , 1B , 2B , nB are unknown constants.

Integration by Partial Fractions steps

1. Factor the denominator.

+ +=+ dxxx xdxxx )3)(3( 1591522. Separate the factors of the denominator. Assume and substitute the numerator with constants.

33)3)(3(

15

++

=

++

x

B

x

A

xx

x

3. Multiply each term by the denominator of the factored function so that the numerators are

equal.

)3()3(15

1

)3)(3(

31

)3)(3(

31

)3)(3(

)3)(3(

15

++=+

+

++

+

=

+

++

xBxAx

xx

x

Bxx

x

Axx

xx

x

4. Expand then rearrange into function form in terms ofx.BBxAAxx 3315 ++=+

)33()(15 BAxBAx ++=+

y = mx + b

5. Set the variables (A, B) of the degree ofx equal to the same degree ofx of the function on theleft side. Then solve for the variables.

3

1536

15)1(33

1533

=

=

=

=

A

A

AA

BA

2

31

1

1

=

=

=

=+

B

B

AB

BA

6. Rewrite the integral and integrate.

+

=

+

dxxx

dxx

x

3

2

3

3

9

152

cxx ++= 3ln23ln3


20/26

Shortcut step 4:

4. Sometimes expanding the equation is too much work. To save time, you can substitute

numbers forx to solve for the constants.

ie. )7()8(422 ++= xBxAx

15

941594

)77()87(4)7(2 2

==

++=

A

A

BA

15

12415124

)78()88(4)8(2 2

=

=

++=

B

B

BA

Shortcut Formulas:

1. cnxAdxnx

A+=

ln11

ie. cxdxx

+= 2ln525

2.( )

( ) cnxk

Adx

nx

A kk

++

=

+ 111

1fork >1

ie.( )

( ) cxdxx

+

=

23

22

9

2

9

3.

++=

++ n

x

n

Bnx

Adx

nx

BAx 12222

tanln2

ie.

++=++ 3tan3

59ln

2

3

9

53 122

xxdx

x

x

4.( )

+

+

=+

+ nx

n

B

nxn

AnBxdx

nx

BAx 13222

2

222tan

2)(2

ie.( )

+

+

=+

+ 4tan1287

)16(32

327

16

72 1222

x

x

xdx

x

x


21/26

Tyler - Newtons Method and Linear Approximations Handout

Newtons Method: The use of Newtons method is to compute the zeros of any function byusing tangent line approximations.

Explanation:

To start the approximation we make and initial guess and go up to that point and draw a

tangent line mark this point as

x0

This tangent is defined is defined by the point slope form equation:

y=f(x0)+f'(x0)(x10 Now we set y=0 and by doing some simple algebra we set the equation equal to

x1 and

we are left with the equation

x1=x0

f(x0)

f'(x0)

This will give us the root to the tangent of

x0. Now this will usually not be the zero of

the original function so to get a better approximation we will find the zero of the tangent

at

x1 by plugging its value back in to our equation. This time we will use

x2 to

represent the new zero.

And since we can always get more accurate we can generalize this equation as

xn+1=xn

f(xn)

f'(xn)and we can use this equation until we reach the right level of

precision.

Key equation:

xn+1=xnf(xn)

f'(xn)Given the equation

f(x and

'(x) our new approximation

xn+1

will always be closer to the original

xn

because the slope of the tangent line is always getting

smaller witch mean the zero of that tangent line is always getting close to the zero of the original

equation.


22/26

Linear Approximations: The use of linear approximations is to approximate a value of anyfunction by using tangent lines and the closeness of the tangent line to the desired point on the

function.

Explanation:

As we already learned from Newtons method the point slope form for the equation of a

tangent line is

y=f(a)+f'(a)(x

But we will set

y=L( to represent the linearization. For values of

x closer to

x= , we expect

f(x and

L(x to have similar values.Since

L(x is a linear function we have a linear approximation of function

f.

Key Equation:

L(x)=f(a)+f'(a)(x . This equation will work best when for approximating the values

of the function as long as we stay close to

x= , if we go further away the approximation willget worse due to the slope of the tangent line.


23/26

Nicole and Selena - Area of a Surface of Revolution Handout

You can get the Area of a Surface of Revolution by rotating the curve about the given

axis as you would a volume question like we did previously in the year. When doing this, you

are finding the entire surface area of the rotated curve. The equation is also very similar to that ofthe volume questions, which means that you would solve it basically the exact same way. We

know how to find the surface area of simple 3-d objects such as a cylinder rhSarea 2= or acone rlSarea = , but what about those more complex shapes? Those that arent given a name but

are just curves rotated about a given axis? The surface area of these shapes can be found with the

general equation in terms of x:

( ) dxxfxfSb

a

area += 2)('1)(2 bxa

In Leibniz notation, in terms of x:

dxdx

dyyS

b

a

area

+=

2

12 bxa

When in terms of y, the equation changes to:

dydy

dxxS

d

c

area

+=

2

12 dxc

Example of setting up the integral

xy sec= 4

0

x Y-axis

Find the derivative

xxy tansec'=

Set up the integralYou take your equation, and plug it in for the x or y values depending on the axis of revolution,

then plug in your derivative. When done your equation should look like this:

dxxxxSarea +=

4

0

2)tan(sec12

Example of evaluating the integral


24/26

xy = 94 x

Find the derivative

xy

21'=

Set up the integral as shown before

dxx

xSarea

+=

9

4

2

2

112

Anti-differentiate and then solve by plugging in your boundaries.

dxx +=9

4412

9

4

2

3

4

1

2

32

+= x

( )9

4

2

3

148

1

3

4

+= x

( )171737376

=

14.81


25/26

Corey and Paul Arc Length Handout

What is arc length?

Simply put, arc length is the length of a curve on a given interval.

Deriving the Arc Length Formula

- We start with approximating the arc length by

connecting linear line segments to the curve.

- In order to find the length of these linear linesegments we make a right triangle, making the line

segment the hypotenuse

- The other two sides are y and x

- To calculate the length of the linear line segment we use

Pythagorean Theorem: 22 bac +=

- We apply that to our problem: 22 )()( yxc +=

- Because of the mean value theorem, we know that )(' xfx

y=

somewhere on that interval

- We rearrange the equation for y giving us xxfy = )('

- Now we substitute y into our equation above: 222 )())('()( xxfxc +=

- Next we factor out2)( x to give us: xxfc += 2)]('[1

- So the approximate length of the curve would be all of our line segments added together. Lets

say we have n line segments. Therefore the approximate arc length is: xxfLn

k

+==1

2)]('[1

- To get a more accurate approximation, we need to make more and more line segments. To find

the exact arc length we need to have an infinite number of line segments. We make this into a

limit where n approaches : =

+=

n

kn

xxfL1

2)]('[1lim

Now using the definition of the definite integral this simplifies nicely into our finished arc length

formula: +=b

a dxxfL2

)]('[1

Example Problems


26/26

Find the arc length of 84)( 2 += xxxf from 30 xFirst lets find the derivative of the function: 42)(' += xxf

Now we simply plug this into our arc length equation: +=b

adxxfL 2)]('[1 and solve.

++=3

0

2)42(1 dxxL

++=3

0

2 17164 dxxxL

Now we can just solve on the calculator (see bottom of page)

23.21=L

Now find the arc length of )sin()( xxf = from x0)cos()(' xxf =

+=

0

2 )(cos1 dxxL

Now use your calculator

82.3=L

Why its difficult to solve by hand

You might be asking yourself cant I solve these without a calculator? Of course you can, the

only problem is when using )()( aFbF you need to anti-derive the function. This is difficultbecause the arc length formula has us squaring an equation, and on top of that its under a square

root. So our two examples would look like 21

2)17164( ++ xx and 2

1

2 )](cos1[ x+ , defiantly not

easy things to anti-derive, but dont let us stop you from trying anyway!

Review: Solving Integrals with a TI

Press the Math button, then select 9:fnInt(Your screen will then look like this:

fnInt(

Enter your equation, followed by the variable used (x), then the lower bound, then the upperbound. All separated by commas.

fnInt(equation,variable,lowerbound,upperbound)

Then press enter and you have your answer.

Ill use the problems above as examples:

fnInt )3,0,),17164(( 2 xxx ++

fnInt ),0,),))(cos(1(( 2 xx+

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