Need Math Help?

[PsychoIncarnate]

Some perpendicular, some parallel. That specific example was perpendicular. We went through A LOT of chapters before this quiz and that was the one I was getting stuck on.

I wasted a day studying something else too, and when I finally got it I realized it wasn't something we were suppose to be studying which made me mad

 

[terr13]

Just to reiterate the point that Hot_ArmS said: None of the values inherently have a negative or positive attribute. This is the reason why I think you don't quite understand the concept. You do not need to consider if x and y are positive or negative, you just need to plug in the number into the formula, and it will give you the answer. So saying if y is negative, you need to swap values is a weird way to put it, although it is correct ( if -y = 3, then y = -3).

For example, given x = -1, y = -1, slope = 2, we can use the formula to solve for b
y=mx +b
-1 = 2(-1)+b
-1=-2+b
b=1

We don't need to care whether y is negative or x is negative at all.

 

[ajp_anton]

just remember that just because it says y=mx + b doesnt mean that B has to be positive, B could be negative and would make it y=mx-b

Well, if it's +b, then it's always +b and never -b.
You ADD b, whether b is positive or negative.

If b=-5, adding -5 is the same as subtracting 5, but then you're not subtracting b because b isn't 5, it's -5.

I'm sure you know this, but I just think it was poorly explained so I'm trying to make it more clear without shortcuts that may lead to a wrong way of thinking.

You either add a b that is positive, or you add a b that is negative, but in either case you're ADDING b, and b itself will decide if the numeric calculation will be addition or subtraction.
There are never* any if's and then's in math, everything will work itself out if you just use the same rules for everything. Adding a b that happens to be -5 will just mean you will subtract 5, but until you know that b is -5, you're NOT subtracting anything, you're adding a black box called b.

*not at this level at least

 

[PsychoIncarnate]

I don't think I can learn math the way this course is taught. The only reason I know any math is because I memorize things like lectures when I am drawing. I just drew through every lecture course I ever took and memorized everything, occasionally writing notes if there was something I needed. I can only memorize through a mixture of listening and drawing.

The math classes at this college are self teaching courses where the teacher never lectures. He's basically a baby sitter as we sit on computers doing assignments.

That might even be fine if the computer teaching us just gave the formula on how to solve the equations, but it has to draw out it's lesson into segments that attempt to explain every little detail. When it drags on for page after page I start loosing the connection and wish they would just give me a formula and why I am doing it that way

 

[Kal]

Anyone can memorize a formula. Learning the actual content, to a point of understanding, falls down to more than a formula, usually. Your teacher has the right idea behind this class, though I can't comment on the execution since I haven't observed anything.

 

[PsychoIncarnate]

Actually I talked with my teacher and he said that 90% if students that take this class after taking higher level math classes have trouble remembering the exact same thing as me.

It's been 10 years since I took the math class that taught that.

As we get to higher level algebra I'm remembering more. I decided to check out the future things we are learning.

Strangely, there were people in the class that were having trouble with todays lesson. Which was also review. It's multiplying exponents, which is really simple.

Strange how people could do the things I had trouble with but have trouble with things so simple

 

[Hot_ArmS]

have you tried using pauls online notes

 

[sakuraZaKi]

Ahh... I just started taking Differential Equations as an undergrad, and I'm a bit confused with the intuition behind Exact Equations...

I'm pretty much stumped and dunno what I should be doing to solve them.

 

[Kal]

Been a while since I've studied Differential Equations, but if you post the problem you're on, I think we can work through it.

Anyway, the general idea is, given a differential equation:

I(x,y)dx + J(x, y)dy = 0.

you look for a function P(x,y) that satisfies:

P_x = I
P_y = J

then you know that d/dx P(x,y) = P_x + P_y dy/dx (by the Multivariable Chain Rule). From this, you know that P(x, y) = c is an implicit formula for y.

This is a pretty good read on the topic.

 

[sakuraZaKi]

Thanks, I am going to read that right now, but I'll post this up if you need it. One of my math problems:

"Use the "mixed partials" check to see if the following differential equation is exact.
If it is exact find a function F(x,y) whose differential, dF(x,y) is the left hand side of the differential equation. That is, level curves F(x,y)=C are solutions to the differential equation:

(1x^4+1y)dx+(3x+2y^3)dy=0

Find: M_y(x,y) and N_x(x,y)

If the equation is not exact, enter not exact, otherwise, what is F(x,y) "

 

[Kal]

In this case, you want to see if there is a function P(x,y) which satisfies:

P_x = x^4 + y
P_y = 3x + 2y^3

However, we know that, if there is such a function, it will satisfy:

P_xy = P_yx

so you take the partial with respect to y of x^4 + y, and the partial with respect to x of 3x + 2y^3, and see if they match. If they match, you've got an exact differential equation. If they don't, then the differential equation is not exact.

 

[sakuraZaKi]

After doing some reading, apparently, this is multivariable calculus, which I never took the class for (not required to)... No wonder I am so confused with partial derivatives. So through many attempts of trying to get the answer, I'm still not doing so hot finding M_y and N_x, ugh.

From what I've been looking up (and from what Kal said), in:

(1x^4+1y)dx+(3x+2y^3)dy=0,

x^4+y would be my M while 3x+2y^3 would be my N, right? Then M_y(x,y) would be the derivative of M with respect to y? And treat x like a constant?

EDIT: Ohhhh, M is 1 and N is 3. I forgot that I was still taking the derivative with respect to y or x, so I ended up with M = y and N = . Took me long enough. It really helped me to write it like this: M_y(x,y) = d/dy (C^4 + y) = 0+1

Oh, new problem. In a case where M = (-2xy^2+y), if I take the derivative with respect to y, and with x treated as a constant, where would I go with this?:

d/dy (-2C[y^2]+y) = ?

 

[Kal]

If M = -2xy^2 + y, then M_y = -2x(d/dy y^2) + (d/dy y), since (-2x) is a constant you can just factor outside of the derivative. So the question boils down to differentiating y^2 and y with respect to y.

 

[sakuraZaKi]

Woo! Thanks Kal, was able to finish my homework with little error. Seems like I finally cracked a bit of exact equations.

Now to learn exact equations with integration factors.

 

[Kal]

I'm craving some math; let's get some discussion going! Is anyone taking any interesting classes or studying anything interesting at the moment? I'm reading "Codes and Cryptography" by Dominic Welsh, and this stuff is breaking my balls.

 

[Teczer0]

I'm taking a cool class in statistical estimation/detection theory.

Its more of an engineering course and the math isn't very complex but its still fun.

Just a lot of gaussian functions, Q functions, and linear algebra.

 

[sakuraZaKi]

I got sick and missed a class so i had a friend of mine tell me what we went over.

Linear Equations with Multiple Orders,

I'm like... what?

 

[Kal]

Can you post an example problem or something?

 

[Yonder]

I have to learn quadratic factoring by tomorrow...huzzah.

 

[Kal]

Factoring quadratics isn't too bad. For a monomial: x^2 + Bx + C, you want to write it as:

(x - a)(x - b)

which expands to x^2 - (a + b)x +ab. So basically, you need two things which add up to -B and multiply to C. For example, factoring x^2 + 5x + 6, you need two things that add up to -5, but multiply out to 6. The obvious choices are -2 and -3, which means it factors as

(x + 2)(x + 3)

For one which is not a monomial, it's a little harder. This is a pretty good guide. Feel free to ask questions if you need any help.

One nice thing about factoring is you can always check your answers. So, if you ever have a test or quiz about factoring, you can expand (FOIL) your answer to see if it matches the quadratic.

 

[Yonder]

Thanks for the site and assistance. Tomorrow's test is going to be mainly factoring, long factoring, and foiling. I'm doing ok right now with a 90, hope I can keep that grade. I like learning those quick memory tricks that help keep the content in my head.

Math is my worst subject, so I usually need a lot of help.

 

[PsychoIncarnate]

Can you say that Square roots are basically negative exponents?

 

[Kal]

Not quite. When you start to define exponents more generally, you realize that they satisfy really only one rule:

a^x * a^y = a^(x+y)

from which everything else really follows (though I'm sure more care needs to be taken to make sure this stays true; I'm being rather lazy). So, if we take sqrt(x), we see that sqrt(x)*sqrt(x) = x. If sqrt(x) = x^a, then this means x^a * x^a = x^{2a} = x, so a = 1/2. In other words, sqrt(x) = x^{1/2}.

More generally, a root is always a fractional power. An nth root is just the power 1/n.

Now, to acquire a similar result for negative exponents, we first need to define x^0. We see that, if x^0 is defined, then x^0 * x^0 = x^{0+0} = x^0, hence it's only natural to define x^0 = 1. From this, defining a negative exponent is pretty easy:

x^{-a} * x^a = x^{-a + a} = x^0 = 1
x^{-a} = 1/x^a

So, to summarize: a root is a fractional exponent. A negative exponent is a multiplicative inverse.

 

[PsychoIncarnate]

Oh I see.

a-^ = 1/a^ while, (Sqrt)a= a 1/^ (^=2 since it's a square)

Put in terms dealing with what I'm doing now

 

[Life]

x^(a/b) = (b)root(x^a)

So x to the power of 1/3 equals the third root of x; x to the power of 2/3 equals the third root of x^2, etc.

x^(-a) = 1/x^a

If you see a negative exponent, you move it to the other part of the fraction (if it's not already a fraction, make it one).

This is the first year in my life since I was in like kindergarten where I didn't have a math class, and I'm having withdrawal. <.<

 

[ajp_anton]

Think of the infinite series
___n= ..., -3 , -2 , -1 , 0 , 1 , 2 , 3 , ...
2^n= ..., 1/8,1/4,1/2, 1 , 2 , 4 , 8 , ...

For 2^n, for each step you increase/decrease n, you multiply/divide the answer by 2. In the "middle" where n=0 (the additive identity, negatives are the inverses of positives), 2^n = 1 (the multiplicative identity, <1 are the inverses of >1).
The n tells you how many 2's you have. For positive n, you already know. For n=0, you have no 2's and you're left with a 1. If you take away even more 2's (n is negative), you get fractions. Note that with addition, "nothing" is 0, but in multiplication, "nothing" is 1.

Basically what the others already said, but from another point of view. There are many different ways to think of it, pick one that suits you (or all of them =)).

 

[PsychoIncarnate]

With functions, do you have to just guess and check?

like with f(x)=(sqrt)-5x+4.5

I was doing well until they had graphs again

 

[Kal]

What are you trying to do with this function?

 

[PsychoIncarnate]

Well, the answer was (-infinity, 0.9] but TBH I'm not really sure

My teacher didn't seem to know, because he gave me the wrong answer

Edit: Actually, I think I get it now.

 

[Elyssa Xey Hexen]

Okay, I'm given a central force which is represented as F(r) = -c*exp(-br)/r^2. Where r is the distance to a particle from an origin. c and b are positive constants. Originally, I am asked to find the potential which is simply the integral of the F(r) with respect to dr.
Now, from the lagrangian, I have to take ∂U(r)/∂r, which should be

∂{ int(-c*exp(-br) /r^2)dr }/∂r I am unsure what to do about this expression. From the fundamental theorem of calculus this would be nothing more than -c*exp(-br) /r^2, however, I am uncertain about dealing with the limits of integration. For a typical 1/r^2 force, it is defined from 0 to infinite with 0 force at infinity. Given this information is this a reasonable course of action?

 

[Teczer0]

Is potential an integral for all r?

I thought it was defined as the integral from r to inf, not 0 to inf. Which would pretty much be the potential of that charge with respect to potential at inf (which is 0).

At first glance off the top of my head, your reasoning seems fine.

 

[Elyssa Xey Hexen]

Well, if there was a spherical mass that the force came from, then I think that would be true since the inside of the sphere would have no net force and the only force would start at the radius r of the sphere. Although, this is point masses I from my understanding of this "kepler problem", so I expected zero to infinity since.

 

[PsychoIncarnate]

For some bizarre reason I thought that i = -1, when i^2 = -1. LOL, that caused me problems for a bit.

i = (Sqrt)-1

 

[ajp_anton]

You should write the parenthesis
i = Sqrt(-1)

 

[sakuraZaKi]

Smeone help me out here.

Solve the diff. Equation:

y'=(y^2-2x^2)/xy

Leave answer in implicit form.

What the heck is implicit form?

 

[Kal]

Implicit form just means that you won't have the answer in the form y = f(x), but rather f(x, y) = c.

 

[THuGz]

I need to show that the integral closure of an affine domain A (a polynomial ring over a field K, modded out by a prime ideal P) in a finite extension of Frac(A) is also an affine domain.

Help!

 

[Doser]

Let A be an affine domain over k with field of fractions K, denote F a finite field extension of K, and let B be the integral closure of A in F.

By Noether normalization, A is integral over a subring A_0=k[x_1,x_2,...x_n] generated by algebraically independent elements. From the transivitivity of integrals, B is the integral closure of A_0 in F. Therefore we can let A_0=A.

If we assume B is a finiteley generated A-module, WLOG we can further assume F/K is a normal extension.

Let E be the subfield of elements of F that are purely inseparable over K.

If char(K)=0 then you can apply the fact that if A is an integrally closed domain with field of fractions K and F a finite separable field extension of K, then the integral closure B of A in F is contained in a finitely generated A-module. (Hence if A is Noetherian then B is as well, and if A is affine over a field k, then B is affine over k)

Otherwise we only need to prove that the integral closure of A in E is what we previously claimed.

Since E=k(z_1, z_2,...z_m) is a finite extension of k(x_1,x_2,...x_n) there exists an exponent q=p^(d) s.t. z^q is an element of k(x_1,x_2,....x_n) for all z in E.

Let k' be the finite extension of k obtained by adjoining to k the qth roots of all the coefficients that occur when z^q_i is written as a rational function of the x_j.

Thus E is contained in E', and E'=k_0(x^(1/q)_1,....x^(1/q)_n). But the ring B'=k_q[x^(1/q)_1,....x^(1/q)_n] is an integral closure of A in E', and B must also be contained in it.

QED

 

[THuGz]

KINDA EZ THo

 

[PsychoIncarnate]

My most common mistake is math is reading my own handwriting and the fact too many of the characters I write look identical. Such as s looking exactly like my 5 and a like d, etc.