orthogonal trajectory for x^2 + 4y^2 = c

[jdgamble]

I am having a problem getting an answer. I need the orthogonal trajectory or this equation:

. . .x^2 + 4y^2 = c

. . .dy/dx = -1 / f(x,y)i so...

I use implicit differentiation to get:

. . .2x + 8y(dy/dx) = 0

So (dy/dx) = - x / 4y

Therefore the slope of the perpendicular tangent (orthogonal) is the negative reciprocal (4y / x), right?

The answer I am looking for is "y = cx^4".

I don't know what to do next. Do I integrate? If so, why and how?

Thanks,

 

[royhaas]

Yes, you integrate to get the family of curves. You have $\displaystyle dy/dx = 4y/x$ for the family of orthogonal trajectories. The general solution of this differential equation is $\displaystyle y=cx^4$. Put another way, integrate both sides of $\displaystyle dy/y = (4/x)dx$, then solve for $\displaystyle y$..

 

[jdgamble]

Re;

Okay, so how do I integrate (dy/dx) = 4y/x ?

I imagine that I can do something like:

x^-1 dx = (1/4) y^-1 dy

then what? Isn't the integral (1/x)dx a special case? I'm trying to remember.

 

[galactus]

Separation of variables. Get the respective variables on each side of the equation.

\(\displaystyle \L\\\frac{dy}{dx}=\frac{4y}{x}\)

\(\displaystyle \L\\\frac{dy}{y}=\frac{4}{x}dx\)

Integrate:

\(\displaystyle \L\\\int\frac{1}{y}dy=\int\frac{4}{x}dx\)

\(\displaystyle \L\\ln(y)=4ln(x)+C\)

\(\displaystyle \L\\y=e^{4ln(x)+C_{1}}\)

log property:

\(\displaystyle \L\\y=e^{ln(x^{4})}e^{C}\)

Let $\displaystyle e^{C}=C_{1}$

\(\displaystyle \L\\y=C_{1}x^{4}\)

 

[jdgamble]

orthogonal trajectories

Thank you! That is what I needed. One last question. How did you crate those images?

 

[galactus]

By using LaTex. It is a code one learns to post in these forums. It can be

done in TexAide and copied and pasted, but I type mine.

Click on quote in the upper corner of my post to see the code I used.

Experiment a little. With some practice you'll get the hang of it. It's easy.

There are many tutorials on line.