Find the Orthogonal Trajectories For The Family of Curves


by EmmanuelD
Tags: curves, family, orthogonal, trajectories
EmmanuelD
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#1
Sep15-11, 10:13 PM
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!!
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Dick
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#2
Sep15-11, 10:32 PM
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Your first line is wrong. If y=C*x^5+3 then y-3=C*x^5. What is the correct solution for C?
EmmanuelD
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#3
Sep15-11, 10:40 PM
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.

Dick
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#4
Sep15-11, 10:43 PM
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Find the Orthogonal Trajectories For The Family of Curves


Quote Quote by EmmanuelD View Post
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.
EmmanuelD
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#5
Sep15-11, 10:50 PM
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.
EmmanuelD
EmmanuelD is offline
#6
Sep15-11, 11:00 PM
P: 10
Quote Quote by EmmanuelD View Post
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!!! :)
Dick
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#7
Sep15-11, 11:03 PM
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Quote Quote by EmmanuelD View Post
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'??
EmmanuelD
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#8
Sep15-11, 11:04 PM
P: 10
So, I got:

x^2+5y^2-30y=C

If you solved the problem, is this correct?

Thanks!
EmmanuelD
EmmanuelD is offline
#9
Sep15-11, 11:13 PM
P: 10
Quote Quote by Dick View Post
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 :)

I reposted with my answer.

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 (?)
Dick
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#10
Sep15-11, 11:19 PM
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Quote Quote by EmmanuelD View Post
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.
EmmanuelD
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#11
Sep15-11, 11:26 PM
P: 10
Quote Quote by Dick View Post
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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