Maths questions

Discussion in 'Physics & Math' started by Adam, May 13, 2002.

  1. Adam §Þ@ç€ MØnk€¥ Registered Senior Member

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    I have some homework/study questions I've got for maths, and I'd like a few pointers on graphing things please.

    Find the slope of y = x^2 -1 at (2,3). I got 15 for the slope. But how does that number translate into a graphed line? What does that mean when sketched?

    Remember, I never really studied any maths, just getting into it.
     
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  3. Merlijn curious cat Registered Senior Member

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    to get the slope of a function at any particular point one has to differentiate the function.
    here y = x^2 -1;
    then the diffentiated function is dx/dy = 2*x (sometimes notation is y'=2*x). Now, how you get from the original function y=x^2-1 to the function y'=2x, is a bit too complicated to write down here, without knowledge of what you know already. if you like, I can tell you later.

    Anyway, at point (2,3) the x-value is x=2, so the slope has a value of y'=dx/dyx = 2*2 = 4. (not 15 ... can you tell me how you got that?)

    this value (y'=4) gives the amount of increase in value y per increase of value x. Thus it is a measure of how steep the line is. (a value of 0 means it is a horizontal line)
     
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  5. Adam §Þ@ç€ MØnk€¥ Registered Senior Member

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    Oh, never mind, I see the error I made. Very silly.

    Thanks.
     
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  7. c'est moi all is energy and entropy Registered Senior Member

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    """how you get from the original function y=x^2-1 to the function y'=2x, is a bit too complicated to write down here"""

    it's actually not complicated at all

    what you do is

    n.X^m ------> (m.n) X^(m-1)

    at least that's hat I remember of those integrales
     
  8. thed IT Gopher Registered Senior Member

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    1,105
    FWIW, it's a differenatial, not an integral.

    The integral of 0->∞ y = x^2 -1 dx = x^3/3-x.

    Th"being pedantic" ed
     
  9. James R Just this guy, you know? Staff Member

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    Here's how you get from y=x^2 - 1 to the slope:

    y'(x) = dy/dx = limit(h->0) {[y(x+h) - y(x)]/h}

    so...

    y' = limit(h->0) {[(x+h)^2 - 1 - (x^2-1)]/h]}

    y' = limit(h->0) {[x^2 +2hx +h^2 - 1 - x^2 + 1]/h}

    y' = limit(h->0) [2x + h]

    y' = 2x
     
  10. Adam §Þ@ç€ MØnk€¥ Registered Senior Member

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    New question. There are two ball bearings. One has radius "r", the other has radius "r + 1.2mm". I'm to use the chain method to find total volume when "r = 3.3mm".

    Apparently I should start with this:

    Total Volume = 4/3 pi r^3 + 4/3 pi (r + 1.2)^3

    Where, if everyone uses the same notation as I am being taught, 4/3 pi is "u'.
     
  11. Mallory Knox Banned Banned

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    I think he meant REALLY get from one to the other.
     
  12. ChristCrusher Registered Senior Member

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    HAHAHA

    that integral, as you defined it on the left hand side of the equality, is infinity ...


    lmfao....
     
  13. Merlijn curious cat Registered Senior Member

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    I know. I am a mathematics teacher, so I am not satisfied with just saying how things are done. I also want to tell why they are.
    I thought you would all understand that.

    What is the use of knowledge if there is no insight?
    (PS is that what you meant Mallory?)
     
  14. Mallory Knox Banned Banned

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    Exactly.

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