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Orthogonal Trajectory. Calc IV.

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Find orthogonal trajectories for y=clnx, y=cex, y=sin(x)+cx

    2. Relevant equations

    Simple integration for the most part.

    3. The attempt at a solution

    I'm fairly confident on the first two, its the third that's giving me trouble.

    First, y=cln(x)
    c=[itex]\frac{y}{ln(x)}[/itex]
    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{c}{x}[/itex]
    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{y}{xln(x)}[/itex]=m1
    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-xln(x)}{y}[/itex]=m2
    [itex]\int[/itex]-xln(x)dx=[itex]\int[/itex]ydy using IBP
    [itex]\frac{-lnxx^2}{2}[/itex]-[itex]\int[/itex][itex]\frac{-x}{2}[/itex]dx
    c+[itex]\frac{-lnxx^2}{2}[/itex]+[itex]\frac{x^2}{4}[/itex]=[itex]\frac{y^2}{2}[/itex]

    Second, y=cex
    c=[itex]\frac{y}{e^x}[/itex]
    [itex]\frac{dy}{dx}[/itex]=cex
    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{e^xy}{e^x}[/itex]=y=m1
    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-1}{y}[/itex]=m2
    [itex]\int[/itex]ydy=[itex]\int[/itex]-1dx
    [itex]\frac{y^2}{2}[/itex]+x=c

    Third, y=sin(x)+cx2
    c=[itex]\frac{y-sin(x)}{x^2}[/itex]
    [itex]\frac{dy}{dx}[/itex]=cos(x)+2xc
    [itex]\frac{dy}{dx}[/itex]=cos(x)+[itex]\frac{2(y-sin(x))}{x}[/itex]

    I've been unable to make any more progress.
     
  2. jcsd
  3. Sep 6, 2012 #2
    Can nobody give me an idea of where I should go from here? I don't know if I'm having a simple algebra problem or what.
     
  4. Sep 7, 2012 #3
    Why won't any one help me?! I need to know how to solve this problem before class!
     
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