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FragileReveries β I love maths
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Published: 2007-01-02 16:26:13 +0000 UTC; Views: 4318; Favourites: 165; Downloads: 127
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Description
Oh wow... my first moving picture-type thing
My first try too... not bad, considering.
Hmm... feel free to use it. Support the wonders of actually liking maths!!!!
(maths rocks... but the lessons can suck sometimes... well most the time)
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Comments: 173
Screw the flamers, maths rocks.
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Im just not good at math
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no problem! i can't like maths because i'm too stupid and have dysgraphia so I just envy you! XD
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You're not stupid!
If you don't like maths, you don't like maths!
I just happen to love it
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No, srsly I love some part of math (like Geometry, especially exercises where I can use intercept theorem or Pythagorean theorem) but when it comes to some harder things.. I often make stupid mistakes (like copy wrong number, write wrog aswer even thought my result is correct or shit like that XD yeah I'm stupid) and that piss me off, but however I enjoy math sometimes (when it comes to drawing figures lol XD)
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Ah, I do that all the time. It's like your hand isn't listening to your brain xD
I think the thing about maths I like is, like, the challenge of figuring out problems... well, when we GET an interesting problem.
But when it comes down to statistics and charts and shit, I just get bored out of my mind
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I LOVE MATHS!! I'm like an alien in class XD
You are so cool for doing this!
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Hell yeah! I love math enough to want to teach it! <3
Thanks for making this!
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awesooome
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Me too... I love maths!
Maybe the fact taht my father is a maths theacher and both my mother and him studied maths infludes...
(but I think I can't put it in my journal cause I'm not a deviantart suscriber... u_u)
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xD
awesome.
my parents both studied maths too... maybe your right...
awww... well at leaset you showed your support
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Mathphobia, the learned problem
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It's "Math" cuz that's the first four letters in mathematics. Not the first four letters and then the last like the UKers do
Math is grand fun. Except topology.
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ahh, but I'm from the UK.
so i'm gonna use "maths"
thanks for the fave though.
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im not sure what these stamp things are, but i sure do support this one's opinion
maths is awesome
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Yay! {{Maths Lovers} β© {Deviant Members}} β Γ!! High school maths is not a happy thing though. In fact it is most decidedly unimpressive. School maths gets decent once one starts A-Level further maths. ^_^ I'll never understand how people don't love maths.
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Haha.
Hell yeah!
Ahhh. Maths sounds like it does get interesting at that time. I'm still only in year 9
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Ugh. I suggest you remove yourself to a sensible year (like upper Sixth) immediately, in order that you might study Vector Geometry, Differential Geometry, Groups, Markov Chains, Multivariable Calculus, Matrices, Complex Numbers and Logic. Eg:
5) Partial Differentiation
To find the rate of change of the dependant variable with respect to another in an E3 space, the obvious thing to do would be to take a cross-section of the surface, giving you a z-x graph for a particular y (or a z-y graph for a value of x), then derivate with respect to x for that particular y (i.e. put y into the function f(x, y), giving z = g(x), then take gβ(x)). This gives the x gradient for one particular y, but what about the others?
As y has been treated as a constant for that one case, why not treat it as a constant for all cases? Take z = x2y -3y + 3x + xy + 3xy2 as an example. Treating y as a constant, and partially differentiating with respect to x yields βz/βx = 2xy + 3 + y + 3y2. Doing the same for y gives βz/βy = x2 - 3 + x +6xy. Letβs just look again at the first term, just to be sure of whatβs happening: β/βx {x2y}(that is, partially differentiating whatever is in {} with respect to x) = 2yx; y is being treated as the constant a from equation (*), 2 is the power, so it comes down, and the power is reduced by one, giving x1, which is of course just x. But β/βy {x2y} = x2. This is because x is being treated as a constant, and the power has gone down to zero, and so y becomes one.
So the derivatives of multivariable functions can also be multivariabled, and have the following f notation: βz/βx = fx(x, y) and βz/βx = fy(x, y).
Turning points
In the plane, there were three types of turning point, where the gradient was zero, but on a surface there are lots of different ways in which the function could behave about them; in the simplest cases they could just be a mountain-top, but they could also be a Pringle or a seat etcetera. Equate both derivatives to zero, then solve them as simultaneous equations to find the x and y values, feed them into f and obtain the z ordinate. To determine the nature of the stationary point without having to take secondary partial derivates (they can be mixed, like β2z/βyβx (in which order were these derivatives taken?), or not, likeβ2z/βx2), look at the behaviour of points local to the point on the surface, such as f(x + Ξ΄x, y + Ξ΄y), where Ξ΄x and Ξ΄y are small; if it goes down on all sides, then you have a mountain top; if it goes up on two but down on the other two alternately, then you have a Pringle.
Tangent Planes
As one can take a tangent line of a curve, so one can take a tangent plane of a surface. Teaching you about plane geometry would be too large a deviation from the main topic (that can be your next email!), but suffice it to say that the elements of a three column vector (a, b, c) (canβt do column vectors on this: convention: bold brackets) corresponding to x, y and z respectively, give the coefficients of x, y and z in the equation of a plane ax + by + cz +d = 0.
The vector product of two vectors gives a vector perpendicular to both, and if the partial derivatives are expressed in vector form they will both be co-planar at the point at which you wish to take the tangent plane, meaning that their vector product will be normal to it, hence when evaluated will yield the three vector (a. b, c) to give the coefficients of the plane, then the ordinates (x, y and z) can be fed into this to find d.
To express the derivatives in vector form, find them and evaluate them at the point in question. This might give for some point βz/βx = p and βz/βy = q, then for every one you go along x, z changes by q. Hence it has the vector (1, 0, q) and (0, 1, p) for y. The vector cross product is found by the determinant of a three by three matrix (oh the joy of matrices!) whose rows are composed thus: i, j, k (the unit vectors in the x, y and z directions respectively) on the top, 1, 0, q in the middle and 0, 1, p on the bottom, and happens to be (p, q, -1).
This is fed into the vector equation of the plane, (r - a)Β·d, and is the right hand component of the scalar product (i.e. d). The left hand side is the vector (x - x1, y - y1, z - z1), and this comes out to be z = p(x - x1) + q(y - y1) + z1 (can you see an alternative notation?). This can then be rearranged into the form ax + by + cz +d = 0.
Directional Derivatives
What if you need to know the rate of change in a particular direction? That is, at an angle to the flat x and y axes (the z sticks up out of the table if x and y are the edges). fx(x, y) and fy(x, y), only give you the gradients along the y and x axes respectively.
Consider taking the gradient of z = x2y -3y + 3x + xy + 3xy2 in the direction of 18Β° from the x axis at the point (2, 3, 69). The derivatives evaluated at that point are, form the equations for them in the first section, are 45 and 39, and these gradients need the weightings of the angle applied to them. More understandably, that means multiplying them by the factors of cos 18 and sin 18 respectively, as cos 18 is the factor by which it has been βstretched' in the x direction, and sin 18 is that of y.
Using the equation Ξ΄z = (βz/βx)Ξ΄x + (βz/βy)Ξ΄y, and given that Ξ΄x = cos 18 and Ξ΄y is sin 18, feed in the values 45 and 39 to get Ξ΄z. This means that a horizontal shift of one at that angle , 18Β° (not on the surface, along the x-y plane for the point (2, 3, 69); remember that it is horizontal), produces a vertical shift of [whatever that comes out to be] bringing you back up to the surface. You can then use arc-tan ([whatever that comes out to be]) (can you se why?) to calculate the angle of the surface to the horizontal along 18Β° from the x axis.
More generally, to take the directional derivative at angle q,
Ξ΄z = (βz/βx, βz/βy)Β·(cos q, sin q) = û·grad f
Note: I havenβt done multivariable integration yet, but the notation is
β«β« f(x, y) dx dy for integrating the function f(x, y) with respect to x then y (I think).
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I just exploded.
>_____<
I'm doing A Level maths, and it's just entirely in Greek!
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I miss partial derivatives... that was back when math wasn't far-too-complicated...
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Ahhh, trying to read all that makes my brain twirl.
Now, I would love to actually learn that, but waiting is pobably the best way to anyway. My school's pretty fast on things. It shouldn't be long. I hope.
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Math makes me happy.
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I love math too. 
(Interesting how the UK uses "math" in plural form though, heheh)
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Yes, it is quite interesting.
After I made it I was like... "ohhhh most people are from the US aren't they... oh well, screw american spelling."
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LOL!
It's true, you never hear it plural in the US.
The funny thing was before I took the thumbcode, to doublecheck I actually went to dictionary.com to look up the spelling for it 
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Yeah...
Ah well... Close enough.
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if i was a DA subscriber i'd put it in my journal but i'm not so ya...
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when i (hopefully) get one i will put it in! but who knows how long it'll be until i get one...i'll probably have forgotten by then
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okay!
yeah... true. hope you get one anyway. subbies just rock XD
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the thing that bothers me the most is the fact that i have quite a bit of money but i can't buy one cause i don't have a credit card
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ahh..
that totally sucks. i hope you can get one soon anyhow. get a credit card!!!!
^^
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do you have to be a certain age to get a credit card?
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umm...
i think... if you get like a bank account... you can get a credit card... i dunno. i think my friend has one and she's like 13... but i'm not sure.
(i just payed my parents to do it for me =.
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oh...
it's parent pestering time
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