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Any specific REAL math problems you need help with?


ezex

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I've noticed that around the forums all people are concentrated on are the specifics of the IA's and how to do them and what is expected off of them, but I can't help but wonder how come no one ever gets stuck on math problems. For example, I remember back in 9th grade when i was taking Pre-IB Pre-calculus that there were some things that were just really hard to understand. And since IB isn't the most open curriculum out there, I went on to take AP math classes (which are INRCREDIBLY better than any IB math class)

Right now i'm in 11th and i'm taking Calculus IV (differential equations here in Florida), so if you guys have any questions about calculus that maybe you dont need for your class, but simply would like to see the "why", just post here. For example, integration is very limited and watered down in Math SL and HL, and I could more than happily explain other simple methods that could make your life so much easier (like integration by parts or trig substitution)

Edited by ezex
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Guest whatever

Maybe you can help me with this problem about tides

I have a set of data given :

x y

0 7.5

1 10.2

2 11.8

3 12

4 10.9

5 8.9

6 6.3

7 3.6

8 1.6

9 0.9

10 1.8

11 4.0

12 6.9

13 9.7

14 11.6

15 12.3

16 11.6

17 9.9

18 7.3

19 4.5

20 2.1

21 0.7

22 0.8

23 2.4

1) create a function for the given data (cosine/sine function)

2) Modify the function to create a better fit

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Maybe you can help me with this problem about tides

I have a set of data given :

x y

0 7.5

1 10.2

2 11.8

1) create a function for the given data (cosine/sine function)

2) Modify the function to create a better fit

Ya of course, this is the tipical IA that's going around. All right, first off you have to realize that what you're trying to do is get a function that fits the data as good as you can. If you haven't already seen statistics, this might be hard to understand, but it really makes no difference as long as you have a graphing calculator (the theory behind it is just a huge formula called linear regression but it only works for lines so don't worry about it)

Ok what you're supposed to do is plug your data into the table in your GDC and get a good graph so you see what you're working with. I have a TI-84+ sylver edition with me right now and it's the same as an 83 so I'll be using this notation. Since they tell you that it has to be a sine function it makes your life easier, but if they just tell you to find a function, any, you'd have to look at your data's graph and pick what you think is right for it.

Once you have your data, all you do is go to "stat" or wherever your calculator has statistic functions, and press "calc". You should be looking at a screen with all the regression functions. Since this is a sine function (cosine is the same but just moved pi/2 to the left) just press the "SinReg" button. This part is tricky: you have to make sure that your data is as follows: your x values have to be on list1 and y values on list2 or else you're going to have to change them manually which is a pain. If they're like this then all you do is press enter. If not you put this: SinReg L#, L# (# is the list number, the first should be the x and the second the y) Once you press enter you're going to get an answer in the form of a*sin(bx+c)+d.

As you should already know, a is the coefficient of amplitude, b is the inversly proportional coefficient of the period, c is the units to teh left or right the graph moves and d is the units up or down that it moves. So no all you do is put this formula in the "y=" place and you should be getting a pretty accurate graph :P

By the way, to get the graph of the data to show on the graph you have to go to "y=" go up to "plot1" and press on it till it's black. If it's not working its probably because it's off, so just press 2nd (blue button) then y= and the Stat Plot page should pop up. Press enter on plot one and set it on the following: On, Type: the first one, Xlist: L#, Ylist: L#, Mark: whatever.

Give me a few minutes to see if I can get you some pictures or at least accurate results. :P

I got y = 5.713sin(0.506x + 0.196) + 6.587

And the graphs...let me try to take snapshots of my TI, it's not working right now...

Ok I got them but I can't post them here, let me see if i can put them somwhere.

Edited by ezex
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Ugh, I am so sorry...I'm not sure that this counts as a "real math problem"- this is such a stupid question, but I am trying to work on a math review, and I am completely lost on matrices...

I'm trying to solve two basic equations, like 1x+2y=2 and 3x+4y=5, using inverses of matrices... I just chose random numbers... Right now I have it set up like this:

[1 2] = A, A [x] = [2]

[3 4] [y] [5]

I know that I should be able to do this, but for some reason I can't remember anything to do with it, and I am reading my text over and over again, to know effect... I just don't understand the concept... S.O.S!

Sorry, I can't get the two other matrices to allign...

Edited by spaceisland
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Ugh, I am so sorry...I'm not sure that this counts as a "real math problem"- this is such a stupid question, but I am trying to work on a math review, and I am completely lost on matrices...

I'm trying to solve two basic equations, like 1x+2y=2 and 3x+4y=5, using inverses of matrices... I just chose random numbers... Right now I have it set up like this:

[1 2] = A, A [x] = [2]

[3 4] [y] [5]

I know that I should be able to do this, but for some reason I can't remember anything to do with it, and I am reading my text over and over again, to know effect... I just don't understand the concept... S.O.S!

Sorry, I can't get the two other matrices to allign...

All right, now first of...why use matrices for such an easy system of equations >.< I know I know...they probably tell you to do it that way but substitution is soooo much easier :D

Ok the inverse method:

First off you have to find the inverse of the matrix which is inverse-matrix.gif

to get the inverse, you interchange the top left and bottom right entries and change the signs of the top right and bottom left entriess. Finally, the inverse is found by dividing the determinant by the whole matrix, and the determinant is:

top left times bottom right - top right times bottom left (ad-bc in simpler terms)

let me write your 2x2 matrix: top left = 1, top right = 2, bottom left = 3, bottom right = 4

the inverse matrix would then be: (1/(1*2-3*4)) times the matrix, from top to right, 4, -2, -3, 1.

all of this will give you the inverse, from top to right and simplified, -2/3, 1/3, 1/2, -1/6

so now we hvae the inverse and the determinant, though you need mostly the inverse (that 2x2 matrix i wrote on the line above)

now, to get your answer, you multiply your first matrix by the inverse and your end result [2][5] (up and down i just cant write the matrix) by the inverse

Of course when you multiply your ivnerse by the base matrix you get a matrix:

[1 0] [x]

[0 1] [y]

so you get [x]

[y]

and on the other side you have the inverse times the answer, which is...you do the math :P its a 2x2 matrix times a 1x2 , x = 1, y = 1/2

so in the end you get

[x] = [inverse times 2 (2 is your first answer)= 1]

[y] [inverse times 5 (5 is your second answer)= 1/2]

so in reality, to simplyfy, al you're doing, is getting the discriminant to find your inverse function, then setting the matrix x, y equal to the inverse times your inital answer matrix, and once you multiply your inverse times the matrix you should get a 1x2 matrix with 2 numbers, the top is your x and the bottom is your y.

...or you do this:

since 1x + 2y = 2, then x = 2-2y

and since 3x + 4y = 5, then substitue the x from above to get, 3(2-2y) + 4y = 5, y = 1/2, plug this in to the first equation, x + 2(1/2) = 2, so x = 1 and you're back to gettin the same answer as with the inverse functions :P

hope this wasnt too confusing -_- if you need me to explain better just tell me and we'll talk over the shoutbox

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Thank-you, thank-you!!! It's come back to me now. Yeah... I'd do substitution but I have to use matrices for this one. We have the stupidest math program. I'm having a lot of trouble with this review though, so I might need a little bit of further assistance... :S

Ugggh I hate math.

Hey, this is EXACTLY why I started this thread, just for completely random math questions, like right now, I'm trying to finish my math review sheet for the winter break and I can't get this Taylor Series question. When i started this i was thinking of the math packets, so if you have any any any math problem you're stuck on post it here and we can all see it and discuss it.

PS: if anyone knows Taylor Series and can help me out i'd really appreciate it, problem reads:

The Taylor series for sin(x^2)/x^2 centered at x=0 is...

an it gives a lot of answers in general form, I know the general form of sine but with the x^2 as denominator it throws everything off, any ideas?

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  • 2 weeks later...
Eeep! Sorry to bother you again.... This review is going to be the death of me.

Okay, here's what I'm trying to do:

Find the value of p if the product of the roots of the equation:

2px^2 + 3x +7 = 0, is 4...

I'm trying to use the quadratic formula, I keep trying different things, but I am just hitting walls!!!

All right, this seems fairly simple:

What you want to do is, first, recognize that there are going to be 2 roots, x1 and x2. What you do, is get these two roots and multiply them, then equal them to zero. It's hard to explain over the chat, let me add a scan of the work :rolleyes:

scan.pdf

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i don't know what a Taylor series are but isn't that suppose to be something with the idea of limit? so, sin(x^2)/x^2 at x=0 is 1.

lol, almost but not quite. A taylor series is just another way of writing an equation but in polynomial form. For example, if you have sin(x), there is some polynomial Px that will get close to the graph of sin(x). It works for really weird functions in which you need to have a polynomial but there's no way of writing it in such a way. By the way, the limit you were talking about is, in fact, 1 :) It comes from H'lophitals rule: since the limit as x goes to zero is zero on the top and bottom, you can take the derivate of the top over the derivative of the bottom and then plug in. Since the derivative of sin(x^2) is 2xcos(x) and the deriv. of x^2 is 2x, the 2x cancel out and you're left with the lim as x goes to zero of cos(x) which is in fact 1 :lol:

Ugh I have a problem. Im in pre-calc so this might seem a little basic. But the problem goes

81^(3x-13)=xroot(27) solve for x and simplified

Allright, here it goes :)

I worked it out on paper and attached the file but i don't know if you'll be able to download it...

Step-by-step:

1) you have to remember that 81^power is the same as 3^(4 times the power) and that the xroot of 27 is the same as 27^(1/x)

2) if we rewrite the formula, we get 3^(4*(3x-13))= 3^(3/x)

3) drop the bases and set the powers equal to each other: 4(3x-13) = 3/x

4) mulitiply everything by X to get rid of the X on the right side: x[12x-52=3/x]; 12x^2-52x=3

5) minus 3 to both sides: 12x^2-52x-3=0

6) quadratic formula: a=12, b=-52, and c=-3

x=52+/-sqrt((-52)^2-4(12)(-3)) divided by 2(12)

7) solve the inside of the sqrt and simplify the 52/24 to 13/6 and you end up with the two solutions:

x= 13+sqrt(178) divided by 6

x= 13-sqrt(178) divided by 6

Hoped it helped :)

scan.pdf

Edited by ezex
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I'm sorry to interrupt this conversation, but I was curious what topics are going to be on the Math Studies exam? I know that M.S. doesn't cover the same ammount of topics as Mathematics does. Also, how many questions are on the examination? How many marks are necesssary for a 6, 5, 4 etc. Any help is appreciated!

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  • 1 month later...
Can you solve these in terms of complex numbers?

Give the answers in terms of a + bi :

A ) (3 + 2i)/(2 - i)

B )b/7i

That's what I'm here for :D

a little background...

when dividing complex numbers in the form a + bi over c + di, then there are two things you can do. One is use the formula and just plug in all the constants; the other thing you can do is solve the equation, which is fairly simple:

all you do is multiply by the conjugate (conjugate of (a + bi) is (a - bi)) of the denominator and then split the fraction into two to get the final result of (ac + bd) divided by (c squared + d squared) all this plus ((bc - ad)/(c squared plus d squared))*i

so for A), all you do is plug in:

(3*2 + 2*-1)/(2^2 + (-2)^2) + [(2*2 - 3*-1)/(2^2 + (-2)^2)]i

simplified gives: 4/5 + 7/5 i

the second question is pretty easy. Remember that a^-1 is the same as 1/a, so b/7i is the same as -b/7i because 1/i is the same as -i.

hope i helped :P keep posting to get more responses

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