# All in the family

1. Oct 27, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
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.

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

3. 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 ?

2. Oct 27, 2013

### brmath

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.

3. Oct 27, 2013