Lesson 3a - Orthogonal Trajectories

# Lesson 3a - Orthogonal Trajectories -...

This preview shows pages 1–4. Sign up to view the full content.

Orthogonal Trajectories O t F il f C A t f d fi d b ti One parameter Family of Curves: A set of curves defined by an equation g(x,y,c)=0 where c is a parameter. That is to say an equation that only has x, y, and c. Orthogonal Trajectory: Given a one parameter family of curves g(x,y,c)=0, a relation f(x,y)=0 that intersects g(x,y,c)=0 at a right angle (regardless of the l f ) ll d h l value of c) is called an orthogonal trajectory.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Finding Orthogonal Trajectories Ri ht l th t th l f f( ) t i i t ( ) t b Right angle means that the slope of f(x,y) at any given point (x,y) must be negative reciprocal of the slope of g(x,y,c) at the same point (x,y). This means to find an orthogonal trajectory of g(x,y,c) we must: Step 1: Derive g(x,y,c) and solve for y’. Step 2: Take the negative reciprocal of y’, this will be our new slope of our orthogonal trajectory. Step 3: Solve the differential equation y’=(negative reciprocal slope) to get your orthogonal trajectory.
Example 1 D t i th diff ti l ti th t t th th l t j t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern