1. ## Orthogonal curves?

This is just crazy. My prof never went over this stuff and I have been searching on how to do these. I don't even know how to plot them. Please help!

Are the curves orthogonal:
x^2+y^2=k where k>0
xy=c where c does not = 0.

Thanks so much for the help!

2. Originally Posted by starbuck
This is just crazy. My prof never went over this stuff and I have been searching on how to do these. I don't even know how to plot them. Please help!

Are the curves orthogonal:
x^2+y^2=k where k>0
xy=c where c does not = 0.

Thanks so much for the help!
Two families of curves are orthogonal if their derivatives are negative reciprocals of each other. Using implict differentation we get

$\displaystyle 2x+2y\frac{dy}{dx}=0 \iff \frac{dy}{dx}=-\frac{x}{y}$

and

$\displaystyle (1)y+x\frac{dy}{dx}=0 \iff \frac{dy}{dx}=-\frac{y}{x}$

Just for fun we can solve for the orthogonal family to the circle

$\displaystyle 2x+2y\frac{dy}{dx}=0 \iff \frac{dy}{dx}=-\frac{x}{y}$

So taking the negative reciprocal of the slope we get

$\displaystyle \frac{dy}{dx}=\frac{y}{x} \iff \frac{dy}{y}=\frac{dx}{x}$

Integrating both sides we get

$\displaystyle \ln|y|=\ln|x|+c \iff y=e^cx \iff y=mx$

So the orthogonal family is the set on lines that pass through the point (0,0)

3. Thanks, I just have a quick question. Why did you set the second one (xy=C) to zero when C could not equal 0? Can you still draw trajectories for this problem, or is that only if they are orthogonal?

4. Originally Posted by starbuck
Thanks, I just have a quick question. Why did you set the second one (xy=C) to zero when C could not equal 0? Can you still draw trajectories for this problem, or is that only if they are orthogonal?

The derivative of a constant is 0 for any constant c zero or not. You could plot them but the would not be orthogonal.

5. Ahhh, okay. Thanks soooooooooooo much.