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Working backwards

  1. Aug 31, 2006 #1
    I've taken Calculus 1, but it was a few years ago, so bear with me. I understand how to use derivitaves to find critical numbers, relative max's and min's, points of inflections, incresing, decreasing. all that good stuff.

    my question is, if I have two points I want a x^3 line pass through, how can i accomplish this?
     
  2. jcsd
  3. Aug 31, 2006 #2

    VietDao29

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    y = x3 is a curve, not a line.
    I don't really get what you mean. y = x3 is a specific curve, i.e you cannot change the set of points that it passes through.
    If you want to get 2 points on the curve, just choose 2 arbitrary x1, and x2 values, then from there find the corresponding y1, and y2. And you'll have 2 points that the curve passes through.
     
  4. Aug 31, 2006 #3
    no, i want to MODIFY a x^3 equation to MAKE it pass through the points i already have set.
     
  5. Aug 31, 2006 #4
    the first point is 0,0, and the 2nd point could be anything, how to i alter the curve to make it pass through my 2nd point?
     
  6. Aug 31, 2006 #5

    chroot

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    You're trying to curve-fit a cubic to some known data set.

    Look up cubic spline interpolation. Numerical Recipes in C has a section on it, with the algorithm coded in C.

    http://www.library.cornell.edu/nr/cbookcpdf.html

    If you only have two points, one of which is (0, 0) and the other is (x0, y0), just solve this equation for a:

    [itex]a x_0^3 = y_0[/itex]

    - Warren
     
  7. Sep 3, 2006 #6

    HallsofIvy

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    There are many different ways to do this. For example, as chroot said, if you want a formula of the form y= ax3, which necessarily passes through (0,0) for all a, just select a so that y0= ax3: That is [tex]y= \left(\frac{y_0}{x_0^3}\right)x^3[/tex] passes through (0,0) and (x0,y0).

    Or, you could alter y= x3 to look like y= x3+ ax. In order to have y= y0 when x= x0 we must have [itex]y_0= x_0^3+ ax_0[/itex] or, solving for a, [itex]a= \frac{y_0- x_0^3}{x_0}[/itex]. That is, the graph of [itex]y= x^3+ \frac{y_0- x_0^3}{x_0}x[/itex] passes through (0,0) and (x0,y0). There are many other possiblities. The choice is essentially arbitrary unless you have other conditions to fulfill.
     
    Last edited: Sep 8, 2006
  8. Sep 7, 2006 #7
    well....i guess I've forgotten a lot of calculus...what does the I in those equations stand for? Integral?
     
  9. Sep 8, 2006 #8

    HallsofIvy

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    ?? What I are you talking about? Which response does this relate to?
     
  10. Sep 8, 2006 #9
    nevermind...the X's in those equations look like I's...

    that's retarded
     
  11. Sep 8, 2006 #10

    chroot

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    Mm... that's how all x's are typeset in every math book I've ever seen.

    - Warren
     
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