# Homework Help: Integration of composite derivative of function

1. Nov 19, 2012

### Bonaparte

1. The problem statement, all variables and given/known data
While I was in school I thought to see what a function whose derivative is always perpendicular to another functions derivative would look like, so for example for X^2 we have -ln(x)/2.

2. Relevant equations
Well all those integration tables I guess

3. The attempt at a solution
Obviously their product is -1, so if were looking for g(x),
g(x) = ∫(-1/f'(x))dx
However, I don't seem to be able to integrate this.
I need to have a ln(f'(x)), who's integral is f''(x)/f'(x), but dividing by f''(x) or try to subtract it just makes things much tougher. This isn't homework, but it would be very nice if I was to be helped. Please don't give solution, just clue.

Thanks, Bonaparte

2. Nov 19, 2012

### HallsofIvy

It sounds like you are looking for "orthogonal trajectories", a family of curves that are always perpendicular to another, given, family of curves.

Yes, to find the orthogonal trajectories of $y= x^2+ c$, where c is a constant, we differentiate to get $y'= 2x$ and the constant has disappeared. Now, any curve perpendicular to that has $y'= -1/(2x)$ and the integral of that is -(1/2)ln(x)+ C.

On the other hand, if the first family were given by $y= cx^2$, so that the constant is multiplied rather than added, $y'= 2cx$ and to get rid of the constant we have to write $y'/y= 2cx/x^2= 2/x$ so that $y'= 2/(xy)$. Now to find the orthogonal trajectories, we look at $y'= -2/(xy)$ so that $yy'= -2/x$ or $ydy= (-2/x)dx$ and, integrating $(1/2)y^2= -2 ln(x)+ C$.

But there is no general way to solve that problem for all x. It depends to much upon the specific f. "Orthogonal trajectories" is typically a subject for a "Differential Equations" course.

3. Nov 19, 2012

### Bonaparte

Could you explain how you got from 2cx/x^2 = 2/x? and then when you had y'/y=2/x, and multiplied both sides by y, how did you get 2/xy? should it not be 2y/x?
I actually thought you would have 2cx*g'x = -1, g'x=-1/2cx
gx = (-1/(2c))*ln(x)+C?
Thanks, Bonaparte

4. Nov 19, 2012

### haruspex

HofI meant to write $y'/y= 2cx/(cx^2)= 2/x$, so as you say, $y' = 2y/x$. For the orthogonals I get $y' = -x/(2y)$, leading to ellipses.

5. Apr 12, 2013

### Bonaparte

Why is it not possible to deriviate 2cx and relate to c as if it is a constant? So you have y=2cx^2
y'=(2cx), g'=-1/(2xc), so g= ln((2xc)/-2c'+C, wouldn't this be the answer?

Thanks, you are all great :)
Bonaparte

6. Apr 13, 2013

### haruspex

It is only constant in regard to one of the original family of curves. A curve in the orthogonal family runs across these, so the set of intersections treats this as a parameter. Eliminating the constant, as HofI did, gets around this problem by extracting a general truth about the family as a whole.