Find Orthogonal Family of Curves to (x^2)/(a^2) + (y^2)/(b^2) = 1

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SUMMARY

The discussion focuses on finding the orthogonal family of curves to the ellipse defined by the equation (x^2)/(a^2) + (y^2)/(b^2) = 1. The derivative dy/dx is calculated as -(xb^2)/(ya^2), and the orthogonal curves are derived by taking the opposite reciprocal, resulting in dy/dx = (ya^2)/(xb^2). After integration, the relationship ln(y) = (a^2/b^2)ln(x) is established, leading to the final form y = x^(a^2/b^2) * e^C, where C is a constant arising from integration.

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Homework Statement


For the family of curves given by (x^2)/(a^2) + (y^2)/(b^2) = 1, find the family of curves that are orthogonal to these.


Homework Equations


Here I will number my steps
1. Differentiate
2. opposite reciprocal
3. integrate
4. simplify



The Attempt at a Solution



1. dy/dx = -(xb^2)/(ya^2) this what I have for my derivative
2. dy/dx = (ya^2)/(xb^2 ) flip and sign.
3. After integration I have (1/a^2) lny = (1/b^2)lnx
4. I just tried to simplify it

So from
(1/a^2) lny = (1/b^2)lnx
ln(y) = (a^2/b^2)ln(x)
I just got ride of ln with e.
y = x^(a^2/b^2) but I feel like I should have c in my equation immediately after integration. So rather
just left as it was ?
 
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You're nearly there. The C arises during integration -- you should get ## log y = \frac{a^2}{b^2}logx + C.## When you exponentiate you get ##y = x^{ \frac{a^2}{b^2}} * e^C. ## Of course ##e^C## is just some constant.

One thing I would suggest is -- don't call 2 different things dy/dx. It's passable here because the problem is very simple. But if you had something more complex, and you discover on page 6 that you made an error on page 3, and then you can't figure out what dy/dx is supposed to be -- that is a mess. If in the equation for the orthogonal family you change the y to w, and write dw/dx ... then all is clear.
 
Good advice thx.
 

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