Proving Orthogonal Curves at Intersection Point

In summary, to prove that two curves are orthogonal when they intersect at a specific point, you need to compare the slopes of their tangent lines, which should be negative reciprocals of each other. For smooth curves, this means that their equations should have orthogonal tangent lines. Examples of orthogonal curves can be found by taking the derivative of one curve and setting the slope of the other curve to be the negative reciprocal of the derivative. In differential equations, finding the family of curves orthogonal to a given family involves using the derivative to eliminate constants and finding a relationship between x and y. The family of curves orthogonal to hyperbolas with the axes as asymptotes is the family of hyperbolas with y= x and y= -x as asympt
  • #1
gokugreene
47
0
How do I prove that two curves are orthogonal when they interest each other at a specific point?

Do I just take the derivative of both and compare the slopes?
The slopes should be negative reciprocals of each other, correct?
 
Physics news on Phys.org
  • #2
well since you have good ideas, what would convince you they are right?
 
  • #3
I believe I am correct; however, I do wish to prove it.
I need two equations that are orthogonal.
I can't find any in my Calculus book.

Thanks
 
  • #4
do you mean you want the definition of orthogonal curves? to me it would be as you assume, i.e. at least for smooth curves, that their tangent lines are orthogonal.
 
  • #5
Yea that works, but do you have any equations I could play with?

Thanks :)
 
  • #6
You've got the basic idea, so what's stopping you from developing that idea into an equation?
 
  • #7
Do you mean you want examples?


y= x is orthogonal to y= -x!

Can you find the equation of the line orthogonal to y= x2 at (1,1)?

(Find the derivative of x2 at x= 1. Yes, the slope of a line orthogonal to that is the negative of the reciprocal of the derivative.)

A common problem in differential equations is to find the family of curves orthogonal to a given family: To find the set of all curves orthogonal to the family of curves
xy= a, differentiate with respect to x: y+ xy'= 0 (thus eliminating the constant a).
Then y'= -y/x at each point (except x= 0). Any curve orthogonal to that must have
y'= x/y or yy'= x so (1/2)y2= (1/2)x2+ C or
x2- y2= c (c= -2C). The family of curves orthogonal to the hyperbolas having the axes as asymptotes is the family of hyperbolas having y= x and y= -x as asymptotes.
 

1. What are orthogonal curves?

Orthogonal curves are two curves that intersect at a right angle. In other words, the tangent lines of the curves at the point of intersection are perpendicular to each other.

2. Why is it important to prove that curves are orthogonal at their intersection point?

Proving that curves are orthogonal at their intersection point can help us understand the relationship between the curves and the geometry of the space they exist in. It can also provide useful information for applications in fields such as engineering and physics.

3. What is the process for proving orthogonal curves at their intersection point?

The process for proving orthogonal curves at their intersection point involves finding the equations of the two curves, finding the point of intersection, and then showing that the tangent lines at the point of intersection are perpendicular to each other. This can be done using the slope formula and the product of slopes rule.

4. Can orthogonal curves exist in three-dimensional space?

Yes, orthogonal curves can exist in three-dimensional space. In fact, in three-dimensional space, there can be an infinite number of orthogonal curves that intersect at a single point.

5. How can proving orthogonal curves at their intersection point be applied in real life?

The concept of orthogonal curves at their intersection point has various applications in real life. For example, it can be used in engineering to design structures that can withstand forces from different directions, or in physics to understand the motion of objects in three-dimensional space.

Similar threads

  • Calculus
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
498
Replies
4
Views
1K
Replies
1
Views
3K
  • Calculus
Replies
20
Views
5K
Replies
4
Views
2K
  • Calculus
Replies
3
Views
5K
  • Special and General Relativity
2
Replies
58
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
163
Replies
13
Views
2K

Back
Top