# Find the Orthogonal Trajectories For The Family of Curves

by EmmanuelD
Tags: curves, family, orthogonal, trajectories
 P: 10 Hello, forum! I'm a newbie here. I've been visiting this site for a while but just recently joined. Anyways, I was wondering if anyone could help with this problem. I can find the orthogonal trajectories, however, this one is killing me because there is a constant. Allow me to type it below: 1. The problem statement, all variables and given/known data Find the orthogonal trajectories for the family of curves: y=C*(x^5) - 3 2. Relevant equations F(x,y,C)=0 and G(x,y,K)=0 1. Determine the differential equation for the given family F(x,y,C)=0 2. Replace y' in that equation by -(1/y'); the resulting equation is the differential equation for the family of orthogonal trajectories. 3. Find the general solution of the new differential equation. This is the family of orthogonal trajectories. 3. The attempt at a solution Given: y=C*(x^5) - 3 --> C=(y/(x^5))-3 y'=5*(x^4)*C - 3 Now, substituting C: y'=5*(x^4)*[(y/(x^5))-3] y'=(5y/x)-15x^4 Conclusion: 1. Am I even on the right track? 2. Is "separable" differentiation/integration the only method that applies? Thank you so much for taking the time to review this!!
 Sci Advisor HW Helper Thanks P: 25,228 Your first line is wrong. If y=C*x^5+3 then y-3=C*x^5. What is the correct solution for C?
 P: 10 I apologize; I just edited the post. Given: y=C*x^5 - 3 Solution for C: C=y/x^5 + 3 Then, as shown above, I sub'd in the value for C in the y': y'=5x^4(y/x^5 + 3) y'=5y/x + 15x^4 Now, I can't factor out a "y" to make this D.E. separable.
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Find the Orthogonal Trajectories For The Family of Curves

 Quote by EmmanuelD I apologize; I just edited the post. Given: y=C*x^5 - 3 Solution for C: C=y/x^5 + 3 Then, as shown above, I sub'd in the value for C in the y': y'=5x^4(y/x^5 + 3) y'=5y/x + 15x^4 Now, I can't factor out a "y" to make this D.E. separable.
That's STILL wrong. y=C*x^5-3 -> y+3=C*x^5 -> C=(y+3)/x^5. That's very different from your solution.
 P: 10 Or, taking a different route (inspired by your initial reply); C=(y-3)/x^5 Sub'd into y' y'=5x^4((y-3)/x^5)) y'=5/x(y-3) 1/(y-3) dy = 5/x dx ln(y-3) = 5*ln(x) + C y-3=x^5 + C ...but this still doesn't look right to me; unless I'm missing something.
P: 10
 Quote by EmmanuelD Or, taking a different route (inspired by your initial reply); C=(y-3)/x^5 Sub'd into y' y'=5x^4((y-3)/x^5)) y'=5/x(y-3) 1/(y-3) dy = 5/x dx ln(y-3) = 5*ln(x) + C y-3=x^5 + C ...but this still doesn't look right to me; unless I'm missing something.
Oops!! I forgot what I was doing here. Back to business:

y'=-1/y'

Don't lose hope on me, I'm just burned out.

I'm going to try this on paper.

Thanks for the help thus far!!! :)
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P: 25,228
 Quote by EmmanuelD Or, taking a different route (inspired by your initial reply); C=(y-3)/x^5 Sub'd into y' y'=5x^4((y-3)/x^5)) y'=5/x(y-3) 1/(y-3) dy = 5/x dx ln(y-3) = 5*ln(x) + C y-3=x^5 + C ...but this still doesn't look right to me; unless I'm missing something.
Exponentiating ln(y-3)=5*ln(x)+C gives you y-3=C*x^5. Don't forget your rules of exponentiation. You might now notice you are right back at your starting point. Hmmm. Didn't you forget to change the y' to -1/y'??
 P: 10 So, I got: x^2+5y^2-30y=C If you solved the problem, is this correct? Thanks!
P: 10
 Quote by Dick Exponentiating ln(y-3)=5*ln(x)+C gives you y-3=C*x^5. Don't forget your rules of exponentiation. You might now notice you are right back at your starting point. Hmmm. Didn't you forget to change the y' to -1/y'??
Hehe, yes, I did in fact forget :)

y'=-x/5(y-3)

(y-3) dy = -x/5 dx

y2/2 - 3y = -x^2/10 +C

10*(x^2/10 + y^2/2 - 3y) = C

Ans: x^2 + 5y^2 - 30y = C (?)
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P: 25,228
 Quote by EmmanuelD So, I got: x^2+5y^2-30y=C If you solved the problem, is this correct? Thanks!
I think it's close. Is your given y=C*x^5-3 or y=C*x^5+3? I think the sign on the 3 may have gotten flipped around. Check it.
P: 10
 Quote by Dick I think it's close. Is your given y=C*x^5-3 or y=C*x^5+3? I think the sign on the 3 may have gotten flipped around. Check it.
Grr! Yes, thank you!!

I notice I make many silly mistakes when I rush through.

So, it's ...+30y = C :]

Thanks a BUNCH! I know it might be simplistic but our professor stressed how easy it was and just did one example (without the constant).

So, not so bad after all.

Again, thank you SO much!!!!!!!

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