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How do you add vectors that are in polar form? watch

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    (Original post by FireGarden)
    1) You've assumed constant acceleration - what more needs to be said!? Integrate for one equation, and integrate again for a second, and then rearrange some stuff for the others.

    2) Well, cause aside, I would've thought your teacher would've explained it's a constant because, doing the experiment, you get a linear relationship between force and extension. I guess you might question the generalisation to other materials..
    Yeah, it's easy to integrate... but 95% of my class still aren't aware that this is possible because they never though of putting the rates of change of maths together with the rates of change in physics. All I'm saying is that my physics teacher could have said "Oh, and if you're doing calculus, you can derive these equations from integrating a constant acceleration" and BOOM, instant inspiration and wannabe physicists.

    and yeah, it looks linear on a graph, but mechanically, in the molecules/atoms of the spring itself, why is it constant; what is it about the atoms and/or orientation of the spring that make it linear -- i still don't understand (being nit-picky I know, but I could bring up many more examples)
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    (Original post by Callum Scott)
    I have students in my physics class that thought the maths was too hard last year, but because they find it interesting, plan on doing it in uni..and I'm there like... "*****?! Do you even know what a second order differential equations is?" *snaps fingers*

    But all joking aside, people seriously consider physics when they're crap at maths because they aren't being told that physics is basically 100% maths at uni, or being taught in that way. My physics teacher winces when he has to say the world maths because people moan about it all of the time
    This is so true! And unfortunately not everyone has someone that knows about this or advices them about the fact that physics is full of mathematics.
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    (Original post by gagafacea1)
    This is so true! And unfortunately not everyone has someone that knows about this or advices them about the fact that physics is full of mathematics.
    Something you'd expect the teachers to be shoving down their throats before they decide to waste 2 years of their life going down a dead end
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    (Original post by Callum Scott)
    Yeah, it's easy to integrate... but 95% of my class still aren't aware that this is possible because they never though of putting the rates of change of maths together with the rates of change in physics. All I'm saying is that my physics teacher could have said "Oh, and if you're doing calculus, you can derive these equations from integrating a constant acceleration" and BOOM, instant inspiration and wannabe physicists.
    This is so true! That is how I got to love physics, when I saw maths come to life. But then I watched it die in agony last year :'(
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    (Original post by Callum Scott)
    Suvat equations can be derived from calculus
    I'm just going to show it because:
    1) I literally tell everyone I meet and I just can't get enough :ahee:
    2) LaTex is just so goddamn beautiful
    3) Calculus in LaTex, is just heaven
    ...
    Spoiler:
    Show
    Let x represent the position of a particle along an axis; the instantaneous velocity of this particle is thusly noted to be v = \frac{d}{dt}x (rate of change of x with respect to time).The acceleration is denoted:

    \frac{d}{dt}v = \frac{d^2}{dt^2}x

    If we assume that the acceleration is a constant a, then:

    a = \frac{d^2}{dt^2}x \quad\therefore\displaystyle\int a\:dt = \frac{d}{dt}x = at + C = v

    Since C is the v value when t = 0 (the v intercept), it can be considered the 'initial' velocity; denoted u

    \frac{d}{dt}x =v = at + u\

\



\displaystyle\int \frac{d}{dt}x\:dt = x = \displaystyle\int (at + u)\:dt = \frac{at^2}{2} + ut + C

    Again, since C is where the x value lies when t = 0, it can be considered the 'initial position', denoted x_0

    \therefore\: x = \frac{at^2}{2}+ ut + x_0 \leftrightarrow a\neq f(t)
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    (Original post by gagafacea1)
    This is so true! That is how I got to love physics, when I saw maths come to life. But then I watched it die in agony last year :'(
    It's just something I have to endure until Uni I guess :T
    Chemistry still annoyed me though, haha. The Boltzmann distribution looked like a cewl-curve...but we were never given the equation for it?!

    \displaystyle\int^{\infty}_{0} f(x)\:dx = N Where N is the total number of molecules.

    \displaystyle\lim_{x\to 0} f'(x) = 0\quad\&\: f'(M) = 0 where M was the most common molecular kinetic energy

    ...I mean, come on chemistry. I was frothing at the mouth for that equation
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    (Original post by Callum Scott)
    It's just something I have to endure until Uni I guess :T
    Chemistry still annoyed me though, haha. The Boltzmann distribution looked like a cewl-curve...but we were never given the equation for it?!

    \displaystyle\int^{\infty}_{0} f(x)\:dx = N Where N is the total number of molecules.

    \displaystyle\lim_{x\to 0} f'(x) = 0\quad\&\: f'(M) = 0 where M was the most common molecular kinetic energy

    ...I mean, come on chemistry. I was frothing at the mouth for that equation
    OMG Same!! It looks similar to \frac{\ln(x+1)}{x+1} with a little bit of stretching.
    Though after looking it up:
    f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}

    (shamelessly copied from wikipedia)
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    (Original post by gagafacea1)
    OMG Same!! It looks similar to \frac{\ln(x+1)}{x+1} with a little bit of stretching.
    Though after looking it up:
    f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}

    (shamelessly copied from wikipedia)
    If you're feeling short-changed by the non-mathematical approach to A level physics, you would probably benefit from reading at least vol I of the Feynman Lectures on Physics. They are online these days (http://www.feynmanlectures.caltech.edu/) and it's well within the capacity of a medium to strong A level student to follow, almost in its entirety, particularly as Feynman tends to develop the maths as he goes along as well, rather than expecting you to know it.

    In fact, I suspect there are quite a few A level students who could get a lot out of vol. II (Electromagnetism) as well - Feynman explains stuff very clearly.
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    (Original post by atsruser)
    If you're feeling short-changed by the non-mathematical approach to A level physics, you would probably benefit from reading at least vol I of the Feynman Lectures on Physics. They are online these days (http://www.feynmanlectures.caltech.edu/) and it's well within the capacity of a medium to strong A level student to follow, almost in its entirety, particularly as Feynman tends to develop the maths as he goes along as well, rather than expecting you to know it.

    In fact, I suspect there are quite a few A level students who could get a lot out of vol. II (Electromagnetism) as well - Feynman explains stuff very clearly.
    WOW!! Oh my god thank you so much ! I never knew about this. Callum Scott , you might like this!
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    (Original post by ghostwalker)
    Your question is akin to asking what is the sum of two complex numbers given in polar form. Polar form is useful if you want to multiply or divide them, but rather a pig for adding and subtracting.
    In "Surely You're Joking, Mr Feynman", Feynman talks about his time working on a mechanical calculator for the army:

    One day the lieutenant came by and asked us a simple question: “Suppose that the observer is not at the same location as the gunner—how do you handle that?”
    We got a terrible shock. We had designed the whole business using polar coordinates, using angles and the radius distance. With X and Y coordinates, it’s easy to correct for a displaced observer. It’s simply a matter of addition or subtraction. But with polar coordinates, it’s a terrible mess!
    (They had been keeping the Lieutenant in the dark about what they were doing because of fears the officers would poke their noses into things they didn't understand).
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    (Original post by gagafacea1)
    This brings up a good question, when DO you use polar form? For complex numbers it's easy to see the benefits, but I'm struggling to find a use for them when it comes to vectors.
    There was a whole chapter in the OLD M6 Edexcel module.

    look at some questions in
    http://madasmaths.com/archive/maths_...ral_forces.pdf
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    (Original post by DFranklin)
    In "Surely You're Joking, Mr Feynman", Feynman talks about his time working on a mechanical calculator for the army:



    (They had been keeping the Lieutenant in the dark about what they were doing because of fears the officers would poke their noses into things they didn't understand).
    something about this reply; it makes me happy
    But I'm glad to know that even the great mr Feynman doesn't know an easier way to do it
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    (Original post by gagafacea1)
    WOW!! Oh my god thank you so much ! I never knew about this. Callum Scott , you might like this!
    I've watched some of his lectures, and little clips where he explains rubber hysteresis, train wheels and stuff, but never was I aware of an entire transcript for every lecture he's done :badger:
 
 
 
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