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Homework Help: Question about graphing

  1. Sep 19, 2006 #1
    hi guys this is my first time posting on here hope it's in the right section
    well heres my problem i have a ploted an inverse relationship ( hyperbole ) on my graph now my teacher wants me to make the graph into a linear relationship, he also gave us an equation which is y=k/x, now how would i go about doing this? i will provide more information if this is not enough...

    thanks
     
  2. jcsd
  3. Sep 20, 2006 #2

    andrevdh

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    Homework Helper

    Welcome mushroom,

    I will try to answer your question.

    The relationship between the variables of a linear graph, say [itex]u,v[/itex], is of the form

    [tex]v = grad\ u + v_o[/tex]

    where [itex]grad[/itex] is the gradient of the graph and [itex]v_o[/itex] is the value of the dependent variable when the independent variable is zero. That is [itex]v[/itex] when [itex]u=0[/itex].

    A special case arises when [itex]v_o = 0[/itex] then whe have

    [tex]v = grad\ u[/tex]

    a directly proportional relationship with a straight line graph through the origin.

    By plotting your [itex]y[/itex] values against [itex]u = \frac{1}{x}[/itex] values the relationship take on the same form:

    [tex]y = grad\ u[/tex]

    so that we get a directly proportional relationship between [itex]y[/itex] and [itex]\frac{1}{x}[/itex].
     
    Last edited: Sep 20, 2006
  4. Sep 20, 2006 #3

    HallsofIvy

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    Andrevdh's point was, that if y= k/x so that plotting y against x gives a hyperbola, then plotting y against 1/x rather than x (so the X-coordinate of the point is 1/x rather than x) gives the straight line graph y= kX.

    I would also recommend noting that log(y)= log(k/x)= log(k)- log(x). If you graph log(y) against log(x) you get a straight line. In fact it used to be (probably still is, but computers have changed so many things) possible to buy "log-log" graph paper which was ruled so that marking "3" on an axis you were actually marking "log(3)".
     
  5. Sep 20, 2006 #4

    nazzard

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    Gold Member

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