# Orthogonal trajectories

1. Jun 4, 2005

### RadiationX

find the orthogonal trajectories of the following

(a) $$x^2y=c_1$$

(b) $$x^2+c_{1}y^3=1$$

for part (a) i've found $$y=\frac{1}{2}\log{|x|} + C_2$$

for part (b) if i solve this integral this should be the O.T.

$$\frac{3}{2}\int{(\frac{1}{x^2}-1)}dx= \frac{y^2}{2}$$

is this correct?

2. Jun 5, 2005

### RadiationX

this is very unusual that nobody has posted a response to this thread.

3. Jun 5, 2005

### HallsofIvy

Staff Emeritus
There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

Your answer to part (a) is incorrect. We might be able to help you more if you showed how you got that.

Differentiating x2y= c1 with respect to x gives
2xy+ x2y'= 0: we have removed the constant c1. The point is that any member of that family of trajectories must satisfy that differential equation which is the same as y'= -y/x. We know that the slopes, m1 and m2 of two lines y= m1x+ b1 and y= m1x+ b2 must satisfy m1m2= -1 or m2= -1/m1.
Any curve orthogonal to every member of the family given must have, at each point, y'= dy/dx= -1/(-y/x)= x/y. That separates as ydy= xdx. What is the general solution to that?

For (b) $$x^2+c_{1}y^3=1$$
it's a little harder to eliminate c1 but if you rewrite this as $$x^2y^{-3}+ c_1= y^{-3}$$, differentiating with respect to x will eliminate c1.

4. Jun 5, 2005

### RadiationX

There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

I appreciate all the help that i've received on this site. Phs. forums is a great resource. But i posted my initial post yesterday and my second post today. I just thought that it was unusual for a post not to be responded to in the amount of time that my post was up. I'm glad you had a good time hiking unlike myself who has spent all of today as well as yesterday getting prepared for the upcoming week.

5. Jun 5, 2005

### HallsofIvy

Staff Emeritus
Oops! My bad: there were 12+ about three hours between your first and second posts! Hope my response helped.

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