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Orthogonal Trajectory. Calc IV.

  • Thread starter Gummy Bear
  • Start date
  • #1
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Homework Statement



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

Homework Equations



Simple integration for the most part.

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.
 

Answers and Replies

  • #2
18
0
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.
 
  • #3
18
0
Why won't any one help me?! I need to know how to solve this problem before class!
 

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