# Orthogonal curves

1. Nov 7, 2006

### minase

We were given a graded assigment and one of the question asks.
Prove that all curves in the family
y1=-.5x^2 + k (k any constant) are perpendicular to all curves in the family
y2=lnx+ c (c any constant) at their points of intersection.

I found the derivatives of y1 and y2 and they are reciprocal to each other. But it is asking for points of intersection of the curves. Can curves be perpendicular to each other. The graded assigment is due tomorrow if possible if you give an explanation right about now it would be really great.

2. Nov 8, 2006

### driscoll79

Yes, curves can be orthogonal to one another, and you've already determined that these two curves are in fact orthogonal to one another because their respective derivatives are lines with negative reciprocal slopes. You can determine the points of intersection of the curves by setting them equal to one another. Don't set both constants = 0 though, since lnx will never intersect with -.5x^2 unless at least one of them has a vertical shift.

3. Nov 8, 2006

### HallsofIvy

Staff Emeritus
Are you asking if it is possible for curves to be perpendicular to one another? Of course it is! For example the circle with center at (1, 0) and radius 1 and the circle with center at (0,1) and radius 1 are perpendicular to one another. Their tangent lines at (0,0) are perpendicular.

You are dealing with 2 families of curves which means that there is a curve of each family passing through every point in the plane. Every point is a "point of intersection" so you don't need to worry about that. Since each curve is given as a function: y(x)= ..., the derivative only depends on x so as long as the derivatives are correct for each x, the families are perpendicular.

By the way, you say "I found the derivatives of y1 and y2 and they are reciprocal to each other." For two curves to be perpendicular, the slopes of their tangent lines must be negative reciprocals. Is that what you meant?

4. Jan 9, 2007

### michellelover

I have the same exact problem. Can someone please show me how to solve
-.5x^2=lnx. Thanks.

5. Jan 9, 2007

### HallsofIvy

Staff Emeritus
If you have "the exact same problem", then you do NOT need to solve that equation. The problem asks you to show that the curves are perpendicular at every point of intersection, NOT to find the points of intersection of two curves with the same k. The first equation gives a family of parabolas so that there is one curve of that family through every point in the plane. The second equation, x must be positive, gives a family of curves through every point in the half plane, x> 0. Every point in the right half-plane, x>0, is a point of intersection!

6. Jan 9, 2007

### michellelover

Can you still show me how to do it .Please. Thanks.