graphs of sin and cos

i dont understand what a phase shift is, i know it can be defined through the equation by -c/b but how is it represented ON the graph?

For Those of Us in a Mathematical Science Class

Chances are that if you were/are in an AP science class at payton that you have used a lot of different constants to solve equations. Therefore I find it handy to store these constants as different variables on my TI-89, and recall them for later use simply by pressing "x" or "y." Now let's say that you're in some crazy Payton math class where you have to use fancy calculators, where you may overwrite the variables you stored. So, let's create a program to restore these variables to what they originally were.


To create a program you have to go under the "Apps" menu and choose "Program Editor." From here choose "New," and you'll be prompted with the familiar dialogue box where you must give the program a name. Let's call our progam "set."









Now when you press enter, you should be presented with a screen that looks like the one below.








set() refers to the name of the program, and it is what you put between Prgm and EndPrgm. You can pretty much enter any normal function like you can on the home screen, so to store variables we just use the "STO" key. Let's set avagadros's number to variable "a," which we all know to be 6.022e23. So in between Prgm and EndPrgm, you would enter "a->6.022e23." If you wanted to set more variables just place them somewhere between Prgm and EndPrgm. To make a new line, press enter.








Since everything on the TI-89 is nice and saves for you, let's quit back to the Home screen. To run the program all you have to do is type "set()," and it will set the variable "a" to avagadro's number. Remember that you have basically an umlimited number of variables of things you can set, because you can use xy, xz, ab, etc.

Pythagorean Style Solving

When solving the types of questions asking you what the cos(x) if the sin(x) = 7/8 (or some similar problem. Set an equation equal to cos^2(t) =1 - sin^2 (x). Substitute the value: 1- (7/8)^2 and simplify.

64/64 - 49/64
15/64=cos^2(x)
square root of 15/64= root 15/8

Integer notation

If you have a problem where the solution is x+2*pi*n when n is an integer, is there a way that you can write this. I think Mr. Karafiol showed us something but I don't remember now.

6.3.48

The problem asks to simplify (sin t - cos t)^2 and I don't understand how you would simplify that, if sin = y/r and cos = x/r?

Response to 6.3.51

For this problem, we could take both sqrt(sin^3 t * cos t) and sqrt(cos t) and combine them to get sqrt(sin^3 t * cos^2 t) because it is ok to multiply square root values.
Next, we can actually calulate the square root of this eq., and the answer comes to be
sqrt(sin^2 t * sin t * cos^2 t) = abs(sin t * cos t) * sqrt(sin t)
We need absolute value signs b/c the values of sqrt's can't be negative. As far as geting to the answer, it is a matter of breaking up the multiples.

# 57 on 6.3

In 57, I tried finding a common denominator and when I simlified, I did not get cos t for an answer, but I got
(1-sin^2 t)/cos t. Could someone help me with this problem?

6.3.51

I also am having trouble understanding this one....can someone please help?

6.3.51

Can someone please explain this one to me

#28 6.4

How many solutions does the equation have between o and 2 pi. Sin t = k , where k is a nonzero constant such that -1<1. So you have the domain of all possible sines. When you imagine this on the unit circle, imagine it as the y-value. For any one y-value, how many points are there on the unit circle. There are two. The points are reflected over the y-axis.

6.4 #38

can anyone explain how to answer number 38 in section 6.4 by comparing graphs rather than using algebra?

homework

I dont understand how to do number 64 in 6.3 and #28 in 6.4. thanks!
melissa

HW #39 Section 6.3

Can anyone explain the proof in 36a?

Even v. Odd Functions

I got to number 64 in section 6.3 and was having trouble figuring out which functions were even versus odd. Can anyone post a quick refresher? Thanks

graph assignment

can someone please explain to me what criteria he is looking for in this effort graph assignment? wat are we supposed to use for the basis of our measurment?

notation

can someone remind me what it means when you see something like f'(4)?

Help me!

Can anyone please scan and send me there final review questions, the one we got today. Please send it to goojiu@gmail.com thx!

# 2a

For #2, i'm not quite sure what would be the best way to think about this. is it helpful to think about it as 31=3.1e1? or does that lead me the wrong way? and how do the properties of logs work in here?
~monica~

homework

Which # homework are we suppose to do?
I did # 39 and I dont understand how to do number 64 in 6.3 and #28 in 6.4
thanks
melissa

#6 on Review

I have completely drawn a blank on how to do this. Any help?

#12 and 17d help

for 12 you could subtract g(x) from f(x) to get h(x) using polynomial subtraction

By the way, I do not necessarily remember how to calculate the IRC without a calculator in 17d. Could someone please help.

#1

I found an answer for the inverse function of x as e^((x-a)/2) but it doesnt match up to any of the multiple choice answers how can I change this answer to get one of the multiple choice answers.

#1

I found an answer for the inverse function of x as e^((x-a)/2) but it doesnt match up to any of the multiple choice answers how can I change this answer to get one of the multiple choice answers.

#12/8

I also need what Donavan asked for (tips for #8) and I was wondering if anyone had any tips forr number 12.

complex factoring

can anyone suggest a method for factoring x^8-1 over the complex numbers?

# 2 on review and log properties

I was going over perhaps a nice way of remembering how to find logs with the table for those of us who would rather use numbers (first and foremost, this is an easier way for me to think about it). When there is a log of a number like 2.9, and we are asked to give the log of 290, think of 290 as 2.9*10^2, becuase the answer is the log of 2.9 plus that number 2. It works for negative powers too, in which case you would subtract. Then, when you are given a nicer log like the log of 3 and the problem asks for the log of 27, you would think of log 27 as log 3^3 and take the initial value for log 3 and multiply by the power, 3. Again, it works for negative powers too. Here's the example
log 3 = .477
log 27 = log 3^3 = 3 log 3 = .477*3 = 1.43
log 1/27 - log 3^-3 = .477*-3 = -1.43

Review sheet #6

I'm also having trouble with #6. But I did it a different way. First I plugged in -4 for x and I got 16b + d=52, then I plugged in 2 + 5i for x and I got -80i - 21b + 20ib + d=148. Then I subtracted the first equation from the second to get rid of d. But after that I'm stuck.

Questions on Final Exam Review

What is the difference between "factor and find all root over R" and "factor and find all roots over C"?
Since i don't know what R and C stand for, i can't get #7 and 8 on final exam review. can someone help me? thanks.

#6 on final review sheet

I'm having trouble with number 6 on the review sheet. I tried taking the zeroes and dividing P(x) by the terms (x-r1)*(x-r2) to get a remainder but it got messy. Any other suggestions on how to do this problem?

1 and 3

1) For 1 I used a system of equations, 4^2+4k+m=-6 and (-2)^2-2k+m=8, and solved for k and m. Next, i found the sum of k and m, which is -9.
3) The difference quotient (f(x+h)-f(x))/h, and I defined f(x)=(2x+1)^2. Next, I plugged it into my 89 and got 8x+4h+4.

#s 7 and 8

7) This graph is simply the graph of abs(x)+abs(y)=2 translated 2 up and 3 to the left. In conclusion, the equation for this graph would look like abs(x+3)+abs(y-2)=2.
8) This graph is the graph of abs(x)+abs(y)=2 but it is scaled horizontally by 1/2. The final equation is abs(x)+abs(y/2)=2

9 and 18

For number nine, i wrote that the statement is true. If there is a linear function, it is in the form of y=mx+b. To find the inverse, simply replace x and y with each other to get x=my+b. Solve for y and you get y=(x-b)/m which is still a linear equation.

For number 18, this is what i did:
logx=a logy=b
log(10sqrt(x)/y^2) can be rewritten as log10+log(sqrt(x))-logy^2 which can be rewritten as
1+.5*logx-2logy
which is 1+.5a-2y

35 and 36 on review

35:
The form for a geometric sequence is a_n=a_0*r^n.
Since a_2=18 and a_5=-2/3 we know that over three steps a_n has been multiplied by -1/27 we know that over one step it was multiplied by (–1/27)^(1/3)=-1/3.
So r=-1/3.
Here I take a short cut and turn a_2 into a_0. Then I have to turn a_100 into a_98.
18(-1/3)^98

36:
First, I set the parabola on a coordinate graph with the vertex at (0,16).
Then I wrote the equation in vertex form: y-16=a(x-0)^2
Then I inserted the x and y values of one of the points I had, (5,0), the right base of the arch: 0-16=5^2a.
Then I solved for a: a=-4/5
Then I inserted 3 for the x value in my equation: y=-4/5*3^2+16=44/5

#'s 3 and 8

3. Find and simplify the Df(x) when f(x)=(2x+1)^2. First find f(x+h)-f(x), which is (2(x+h)+1)^2-(2x+1)^2= (4x^2+8hx+4x+4h^2+1)-(4x^2+4x+1). The 4x^2, 4x, and 1 cancel out so your left with 8xh+4h^2+4h. Then you take that all over h and the h's cancel out so you're left with 8x+4h+4.

8. First look at the original graph and see how x changes. x changes from 2 to 3 and from -2 to -3 which means x was scaled by 1.5, but you have to write it as the reciprocal, so part of the equation becomes abs(3/2x). Then you see how y changed. y went from 2 to 1 and from -2 to -1 so y was scaled by 1/2. So the rest of the equation is abs(2y), so the entire equation is abs(2/3x)+abs(2y)=2. You can always chack it by putting in values for x and y.

number 15, 17

for number 15, i just thought of it like this, x^n+1 if n is odd will have (x+1) as a factor, it will be (x+1)(y{whatever y may be}) is divisible by (x+1).

number 17, x^3=4x+4, which equals to x^3-4x-5, so i graphed it and set my x min to 0 and max to 5, then i used the zero key to get around 2.383

#20 & #22

#20
the correct answer for the domain f(x)=ln((x-1)/x)) is b (-∞,0) union (1,+∞) because when x is within the first set (two negative numbers) the result is positive and within the domain of natural log, and the equation is also within the domain for natural logs when x is in the second set.

#22
e^(-ln(x))=1^(-x) [because the e and ln cancel each other out]=1/x

5&6

5
Since you scale before you shift, scale the x-values by 1/2, so you have
sqr rt (2x) Move it to the left 6, sqr rt (2(x+6))=2x+12

6
Graph the equation, and calculate the minimum, using 0 as the lower bound. This happens at one hour, with a 1/3 percent oxygen. This is the lowest possible amount of oxygen in the stream.
(O(4)-O(2))/2 gives you the average rate of change over the interval [2,4]
6.52685 is the answer. This is how much, on average, the percent oxygen in the stream increases in an hour in between the second and fourth hour.

solution to #25 and #29

25.
a function is odd if f(-x) = -f(x).
g(-x)=e^-2x- 1/(e^-2x), which simplifies to 1/(e^2x)-e^2x, which = -g(x). Therefore, the function is odd.

For 29, the graph is of y=0, which makes sense because log(x^4-1)-log(x^2-1)-log(x^2+1) = log((x^4-1)/(x^2-1)/(x^2+1)). (x^4-1)/(x^2-1)=x^2+1, and (x^2+1)/(x^2+1)=1, so the equation is = log(1), which equals 0.

If anyone could explain how to prove #15, that would be great.

# 10 and 21

For problem 10, we know that the descriminant must be equivalent to 0 in order for there to be one solution because a number + or - zero will only yield one answer. There for we set the descriminat of the problem to zero and solve for k.

b^2 - 4ac = 0
20^2 -4*3*k = 0
400- 12k =0
-12k = -400
k = 100/3

For problem 21 we are trying to find the inverse of the function. In order to do this we must call f(x) a variable y and re write the equation switching the x and the y variable. Then solve for the y variable, which will give you the inverse equation.

f(x) = log(2x - 1)
y = log(2x - 1)
x = log(2y - 1)
10^x = 2y - 1
(10^x +1)/2 = y

#1 and 12 question on 16

For number one I created two equations one for the point (-2,8) and another for the point (4,-6). Then I added these two equations together and this canceled out K, leaving me with the value for m: -14/3. Then I plugged this back into one of the original equations to find the value for K: -3.333. So k+m is -3.333 + -14/3 which is -8.

For number 12, the equation of a parabola is ax^2+bx+c And so we can find a b and c by plugging in points from the graph into equations and then solving for them. To find C i plugged in 0,-2 which took A and B out of the equation leaving C which when solved for is -2. Then after that I plugged -2 into the equation again along with the point -2,-3. There I solved for B in terms of A to get B= 2a+.5 Finally I plugged that in for B along with the C value and point 1,0 ans solved for A and then used that value of .5 to solve for B giving me a final equation of .5x^2+1.5x+(-2)

Can someone post another solution to 16

#21

#21
f(x)=log(2x-1) and to do the reverse f(x) changes to y so then y=log(2x-1). then switch x and y=>x=log(2y-1) and solve for y which is y=(10^x+1)/2 or f^-1(x)=(10^x+1)/2 and it is B.
Melissa

#24

#24
A. P(t)=700/(1+e^(-.4t)), if t goes to infinant that means that the exponent of e goes smaller thus it becomes insignificant=>700/(1+0) so then if t goes to infinant it would equal 700. This means that no matter how many classes you take at the Test Logistics your score will not get more than 700.
B. For the inverse of P(t) then change the equation to y=700/(1+e^(.4x)) then switch the x and y so it becomes x=700/(1+e^(.4y)) and solve for y in terms of x so then y=2.5*-In((x-700)/x). so then the P^-1(t)=2.5*-In((t-700)/t).
Melissa

Solutions to # 4 and 5

4) the difference quotient of ln(x)=(ln(x+h)-ln(x))/h. Then simplify the numerator by writting, (ln{(x+h)/x})/h. The X's on the numerator and denomenator will then cancel making the final equation=1/h*ln(1+h/x).

5) The first thing u do in this problem is factor the 2 out of sqrt(2x+12), so the equation would equal sqrt(2*(x+6)). Doing this makes it an easier equation. Basically, it would Scale horizontally by 1/2 make the original equal to sqrt(2x). Then you would translate the graph left by 6, making the new equation equal sqrt(2(x+6).

Answers to 14 and 34

#14
In order to find forumlas for the graph, we would need to find teh zeroes or the factors of the polynomial.The factors are (a+3)(a-1)(a-5) now by looking at the graph we would have to determine the sign changes. The signs chaneg between the iintervals except for (-infinity,-3) so that factor would be raised to a even power to prevent the sign change. So one equation for the graph would be y=2/5((a+3)^2(a-1)(a-5)). To find another equation for the graph, i just raised teh factor (a+3) to the 4th power. So a second equation would be y=2/225((a+3)^4(a-1)(a-5)).

#34
To find the remainder of the division of the the two polynomials, simply use the remainder theorem. Find the zero of the the divisor (x-1) and substitute that value into the dividend for the x value. So (1-1)=0---> (1)^27+(1)^24+(1)^21...(1)^3+1=10 So the reminder of the quotient would be 10 or 10/(x-1).

solution to # 38

The reason there are two equations is because they are parametric (both x and y are given as a function of t). I'm still a little hazy about when parametric graphing is helpful, but I think an example we used in class was the height and width of a function changing at different rates during time t. (? Maybe?) Don't trust me on that one...

But in terms of graphing it, it isn't too hard. You press the MODE button on your calculator and switch from "function" to "parametric" in the row labeled "graph." Then when you go to the y= screen, you will see xt1, yt1, xt2, yt2, etc. instead of y1, y2, etc. For #38, enter 16t-42 in the xt1 row and t^3-12t^2 in the yt1 row.

When I graph these, I'm always worried that I'm missing part of the graph, because sometimes a parametric curve is a tangled mess. My viewing window was -50 < x < 100 and -50 < y < 10, with 0 < t < 5 and tstep=.1. With these dimensions, I found at least one x-intercept (x=-24) and at least one y-intercept (y=-23.625). I'm wary that I'm missing something though, since Mr. Karafiol asked us to find "all x- and y- intercepts."

solution to #2 (re: brian)

If we are told that g is an odd function, this means that we can replace both x with -x and y with -y and we will get the same function. This means that if (2,-5) is a point on the graph, replacing 2 with -2 and -5 with 5 is valid. And this means that (-2,5) is a point on the graph and that g(-2) is 5. I think this is what Mr. Karafiol was hoping we'd conclude.

#26 + 27

#26. Solve for x: log(2x)+log(x+5)=2

log(2x*(x+5))=2 -----------------------using formula: loga+logb=log(a*b).

2x^2+10x=100-------------------------using theory: loga=b, 10^b=a.

x=-10 and 5.

check: x cannot be -10, so x=5.-----------------using theory: loga=b only if a > 0 .

#27. ab^t=ae^rt

b^t=e^rt-----------------------divide a on both sides.

ln(b^t)=ln(e^rt)----------------take ln on both sides.

ln(b^t)=rt----------------------using theory: ln(e^a)=a.

t ln b=rt------------------------ln (a^b)=b ln a.

lnb=r---------------------------divide t on both sides.

#38 question

what does he mean by a curve? and why are there two equations?

Question on #2

can someone please explain what he wants you to conclude?

Assn 36-37

we have to do both of these by tues.

assn 36 - sec 6.1 #'s 1, 6(there are 16 angles in the diagram), 7, 11, 13, 15, 17, 19, 25, 29, 31, 35, 43, 47, 49, 51, 53, 59, 61, 69, 73, 74
talks about angular radians
assn 37 - sec 6.1 #'s 21, 23, 41, 48, 68, 71, 75, 77, 79; sec 6.2 No Calc. #'s 1-9 odd, 15-29 odd (learn the values, no calc?), 30, 31, 35, 37, 39, 41, 47, 49, 50a, 51, 57
assn 37 will be reviewed mon. when we get back, if i'm not mistaken.

Assignment 36?

can someone tell me what it is? i think i lost my assignment sheet
thnx

Solution to 1 & 16 and request for 13

For number 1, we can take the given equation and plug in the point, say, (-2,8), and use this to solve the equation for k in terms of m...

y = x^2 + kx + m , such that (x,y) = (-2,8)

8 = (-2)^2 + k(-2) + m

k = -2 + 1/2*m

Then plug in the solution for k to the original equation, this time using (4,-6) as x and y, in order to solve for m...

y = x^2 + kx + m , such that (x,y) = (4,-6)

-6 = 4^2 + (-2 + 1/2*m)*4 + m

m = -14/3

Now plug in the solution for m into the k equation, and finally, find k + m...

k = -2 + 1/2* (-14/3)

k = -13/3

k + m = -13/3 + -14/3 = -27/3

= -9

For number 16, we can take the given equation and assume that using the point P (1,2), there is a point on the graph y = 8 - x^2 that is (x,2), and plug the y value into the equation...

y = 8 - x^2

2 = 8 - x^2

x = +/- sqrt(6)

Then solve for the 2 cases when x = sqrt(6) and when x = -sqrt(6), and we get y values of 2 and 14, respectively, giving us (sqrt(6),2) and (-sqrt(6),14) for points on the graph.

I was also wondering if someone had a sol'n for number 13. I always have problems w/ these.