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Summer Bridge 2014 of 14

Notes for CETA’s Summer Bridge August 2014 Prof. Lee Townsend

Townsend’s Math Mantra: What’s the pattern?

What’s the rule? Apply the rule. Repeat.

Precalc Crash Course

Classic Errors 1) You can cancel factors but not terms across the division line. Terms – added or subtracted.

Factors – multiplied or divided x2 − 4x2 − 2x

The x2 do not cancel. Instead, factor both numerator and denominator.

x2 − 4x2 − 2x

=x − 2( ) x + 2( )x x − 2( ) The x − 2( ) do cancel as they are factors in both the

numerator and denominator.

x2 − 4x2 − 2x

=x − 2( ) x + 2( )x x − 2( ) = x + 2

x

2) Negative exponents.

The negative exponent means change the sign and put the term on the other side of the divisor line.

y−3 ≠ −y3 y−3 = 1y3


3) Division of an equation by a common factor. In differential equations I saw this error too many times.

If you have had calculus: x d2ydx2

− 2x2 dydx

+ 4x = xex

If you have not had calculus: xA − 2x2B + 4x = xC There is an x in each term. Most of my students divide it out of the first term and the right hand side of the equation but forget the middle terms. The safest way to do the problem is to first factor the left hand side then divide the equation by the common factor.

Factor: x d 2ydx2

− 2x dydx

+ 4⎛⎝⎜

⎞⎠⎟= xex x A − 2xB + 4( ) = xC

Divide: d 2ydx2

− 2x dydx

+ 4 = ex A − 2xB + 4 = C

4) Trig errors The calculator is in the wrong mode – radians are degrees. Inverse trig functions.

Radians Degrees Inverse trig functions 0 ≤θ ≤ 2π 2π = 6.28 0 ≤θ ≤ 360 Graphs of trig functions – One complete cycle:

If you are setting the window as if you are in radian mode, (xmin=0, xmax=2π, you will get essentially a straight line if your TI is in degree mode. 2π =6.28 – not very many degrees.

If you are setting the window as if you are in degree mode, (xmin=0, xmax=360, you will get 360/6.28 ⋍57.3 cycles in your graph if your TI is in radian mode.

Notation: sin cos tan mean nothing if there is no argument. You need something like sin(x), cos(x), tan(x). Notation for powers of trig functions and logs: books and handwritten: sin2(x) calculator entry: sin(x)( )^ 2 = sin(x)( )2

books and handwritten: ln2 (x) calculator entry: ln(x)( )^ 2 = ln(x)( )2


A) Basic Algebra Make your problem look like a rule. Here are some rules. Exponent rules:

Name Exponents Product Rule u v u vb b b += Division Rule

bu

bv = bu−v b−ν =

1bν

Power Rule ( )vu uvb b=

Unity 0 1b = Identity logb uu b=

Product/Factoring rules:

Name Full Product (left to right) or Factors (right to left) Distributive Law a(b ± c) = ab ± ac a + b( ) a − b( ) = a2 − b2 (a ± b)2 = a2 ± 2ab + b2 FOIL a + b( ) c + d( ) = ac + ad + bc + bd General method: a + b + c( ) d + e( ) = a d + e( ) + b d + e( ) + c d + e( ) = ad + ae+ bd + be+ cd + ce

See the handouts for sample exams with solutions.


B) Matrix equations – n equations in n unknowns Solve: 2x + y = 1 5x − 2y = −11

1) You can solve by intersecting y = 1− 2x and y = 5x +11( ) 2 . Type in the equations Press F2/6 (ZoomStd): xmin=-10, xmax=10, xscl=1 ymin=-10, ymax=10, yscl=1 Select F5/5 (Intersection) It ask which curves. You only have Y1 and Y2 so press ENTER ENTER

Lower bound is any x value left of the intersection. You can scroll there or type in an x value (read the tick marks). ENTER

Upper bound is any x value right of the intersection. You can scroll there or type in an x value (read the tick marks). ENTER

2) You can solve by substitution. y = 1− 2x . Plug into the other equation 5x − 2 1− 2x( ) = −11 You need to get the x’s together so distribute the -2. 5x − 2 + 4x = −11

Add 2 to both sides to get all terms with x on the left and all terms without x on the right.

5x + 4x = −9 Combine the x’s 9x = −9 Isolate the x by dividing both sides by 9. x = −1 ✓ Plug this value into the y equation to find the value of y:

y = 1− 2 −1( ) = 3 ✓ 3) Do this math on the TI-89 Let → mean you have pressed

Solve(2x+y=1,y) → y = 1− 2x 5x−2y=-11| then arrow up to y = 1− 2x and press ENTER 5x−2y=-11| y = 1− 2x → 9x−2=-11 then arrow up to 9x−2=-11 and press ENTER ENTER The equation 9x−2=-11 should now be highlighted on the command line. Type +2 . The command line now says ans(1)+2 ans(1)+2 ENTER. → 9x=-9


Now type ÷9. The command line now says ans(1)÷9. ans(1)÷9 ENTER. → x=-1 ✓ CLEAR the command line. Scroll up to y = 1− 2x . ENTER to put it on the command line. Append | scroll up to x=-1 and press ENTER → y=3 ✓

4) Use matrices First line up the equations so the x’s are on top of each other on the left, the y’s are on top of each other on the left and the numbers are on the right. 2x + y = 1

5x − 2y = −11 Make a matric of the coefficients:

2 15 −2

⎡

⎣⎢

⎤

⎦⎥

You can either use the data/matrix app or I find it much easier just to type 2,1;5,−2[ ] ENTER Row elements are separated by commas and the rows themselves are separated by semi-colons. Let's call this matrix a. 2,1;5,−2[ ] STO a The calculator shows 2,1;5,−2[ ] ⇾ a Let’s make a vector of the right hand side and call it b.

1;−11[ ] ⇾ b → 1−11

⎡

⎣⎢

⎤

⎦⎥

Note the semi-colon as there are two rows. We now write this in traditional matrix notation.

2 15 −2

⎡

⎣⎢

⎤

⎦⎥

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1

−11⎡

⎣⎢

⎤

⎦⎥

i.e. Ax = b where both x and b are vectors with x representing xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥

To solve this equation multiply both sides by A−1 Ax = b( ) . But A−1A = I . I is the identity matrix. For the 2D case

I = 1 00 1

⎡

⎣⎢

⎤

⎦⎥ - the matrix equivalent of one.

Therefore x = A−1b Since you have already defined A and b, type


a ^ −1*b → −13

⎡

⎣⎢

⎤

⎦⎥ but that’s x

y⎡

⎣⎢⎢

⎤

⎦⎥⎥

✓

Now look at three equations in three unknowns.

Solve: x + y + z = 2x − z = 1x + y = 1

for x, y, and z.

5) Use Solver F2/1 solve(2x + y = 1 and 5x − 2y = −11 ,{x,y}) → x=-1 and y=3 ✓ i.e. you are solving for the set {x,y}) solve(x+y+z=2 and x–z=1 and x+y=1,{x,y,z})→ x=2 and y=-1 and z=1 ✓


C) Trig in four quadrants Start with basic definitions. They are visual. Just look at the triangle.

The basic trig triangle Basic Definition Definition using the diagram

cosθ = adjhyp

sinθ = opphyp

tanθ = oppadj

cosθ = ac

sinθ = bc

tanθ = ba

The above chart leads to SOHCAHTOA. It is easier to see the meaning if we insert some

spaces: SOH CAH TOA. i.e. sinθ = OH

, cosθ = AH

, tanθ = OA

with O=opp, A=adj, H=hyp.

Now take the same triangle and put a circle around it.

The origin of the circle is at the lower left hand corner of the triangle. The radius of the circle is the length of the hypotenuse. The origin is symbolized by the letter O.

Now what do we do if I change the angle so it is out of quadrant I?

The basic definitions no longer work. What’s the opposite? Where’s the triangle?

θa

bc

θa

bc

O

θO


To figure out what to do, we need to change the notation. We redraw the original triangle in a circle using the nomenclature of the axes.

cosθ = xr

sinθ = yr

tanθ = yx

From here we rethink what these formulas mean. Instead of thinking of x and y as the length of the legs, think of them as the coordinates of the point on the circle. Now let’s look at the point on the circle (x, y) = (−4,3) .

cosθ = −45

= −0.8

sinθ = 35= 0.6

tanθ = 3−4

= −0.75

θx

yr

O

(x,y)

θO

(-4,3)

5


The quadrants are labeled

Signs of coordinates Quadrant x y

I + + II - + III - - IV + -

Signs of trig functions (r is positive so they follow the signs of the coordinates.)

Quadrant cosθ sinθ tanθ

I + + + II - + - III - - + IV + - -

Inverse Trig Functions

We read sinθ = 35= 0.6 as “What is the angle whose sine is 0.6?”

Let’s put it in the calculator to find out the answer. I find that θ = sin−1 0.6 = 36.87 . In radians we get θ = sin−1 0.6 = 0.6435 . Note: 0.6435degrees is less than 1/360th of the way around the circle. If you are looking for degrees and get really small numbers, you’re in the wrong mode. Inverse Trig Functions are not functions. Why? Because there are two answers in a circle. The calculator gives you one of them. Inverse sine and tangent calculator answers are in quadrants I and IV. Inverse cosine answers are in quadrants I and II.

III

III IV


θ = cos−1u θ = sin−1u θ = tan−1u

Some “rules” for finding the second angle that satisfies a trig equation cosθ = u , sinθ = u , or tanθ = u and knowing the calculator angle, θcalculator , along with the signs of the x and y coordinates follow. 1) θ = cos−1u :

if y < 0 then θother = −θcalculator

since cosθother = cosθcalculator =xr

Note that to make 0 ≤θother < 360

, add 360 to θother :

θother = 360 −θcalculator

2) θ = sin−1u :

if x < 0 then θother = 180

−θcalculator since

sinθother = sinθcalculator =yr

θx

yr

O

(x,y)

θother

calculator

(x,-y)

θx

yr

O

(x,y)

θother

calculator

(-x,y)

-|x|

y


3) θ = tan−1u :

if x < 0 then θother = 180 +θcalculator

since

tanθother =−y−x

= tanθcalculator =yx

I try to use θ = tan−1u as the rule if x < 0 then θother =180

+θcalculator since is so easy to remember; i.e. if x is negative, add 180 (or π if you are in radian mode) to the calculator answer. A piece of advice: When solving problems given data, always use the original data instead of calculated values which are rounded off to keep the error in your final answer to a minimum.

θx

yr

O

(x,y)

θother

calculator

(-x,-y)

-x

-y


D) Notes on TI-89 entry TI-89 Note: ♦7 gives you log(. Fill in the rest with log(8,2). When you press ENTER, this becomes log2 8 which is 3. TI-89 Note: ♦9 gives you root(. Fill in the rest with root(8,3). When you press ENTER, this becomes 83 which is 2. TI-89 Note: When you are unsure about how to enter formulas in the calculator, do an example where you know the answers. That is what I always do with the above two notes. I am never quite sure of the order so I guess. If the answer is wrong then I guess with the numbers switched I know: 8 = 23 hence the numbers used above. Also check the Catalog. For example, the integration formula (F3/2). In the Catalog got to A (press the = button) then scroll up to the integral sign. In the Catalog integration looks like

(∫ integrate At the bottom of the screen you will see EXPR,var[,low,high] Translation: EXPR is the integrand – i.e. what you are integrating var is the variable of integration – i.e. the variable x in dx. The brackets indicate the next two arguments are optional. The are the limits.

The integral: x2 dx∫ TI: (∫ x^2,x)

The integral: x2 dx1

2

∫ TI: (∫ x^2,x,1,2)

I recommend that you always do the indefinite integral first to check your input. Then go back and put in the limits to get the numerical value of the definite integral.


E) Logs and Exponents Transform between log and exponent form Problem 1) convert log2 8 = 3 to exponential form. Match the form of the last column 3= log2 8 By pattern match find u, v, and b u = 3 , v = 8 ,b = 2 Put the variables in the right place in column 1 8 = 23

v = bu v b u u = logb v 8 = 23 8 2 3 3= log2 8

Problem 2) solve for v in the following problem: 3log4 x = 9 . Match the form of the last column – u = logb v

First we have to get rid of the coefficient of the log as the form in column 5 has no coefficient. So, divide the equation by 3. log4 x = 3

By pattern match find u, v, and b u = 3 , v = x ,b = 4 Put the variables in the right place in column 1 v = 43

v = bu v b u u = logb v x = 43 = 64 x 4 3 3= log4 x

Problem 3) solve for x in the following problem: 16 = 2 ⋅35x Match the form of the first column – v = bu

First we have to get rid of the coefficient of the b as the form in column 1 has no coefficient. So, divide the equation by 2. 8 = 35x

By pattern match find u, v, and b u = 5x , v = 8 ,b = 3 Put the variables in the right place in column 5 5x = log3 8 Divide by 5 to solve for x

v = bu v b u u = logb v

8 = 35x

8 3

5x

5x = log3 8

x = 15log3 8


Use solver (F2/1) to check your work. Problem 2: solve(3*log(x,4)=9,x) x=64 ✓

Problem 3: solve(16=2*3^(5x),x) x = 3ln(2)5 ln(3)

???

Log conversion rule: logb u =loga uloga b

with a=any base. Pick e. Then our calculated

answer becomes x = 15log3 8 =

15ln8ln3

However 8 = 23 . The power rule of logs says n logb u = logb u

n .

Now our answer is x = 15ln8ln3

= 15ln23

ln 3= 35ln2ln3

✓

Just FYI, there are two more log rules.

Product Rule: logb uv = logb u + logb v

Division Rule: logbuv= logb u − logb v

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