# What are the orthogonal trajectories for x^2y=c_1 and x^2+c_{1}y^3=1?

In summary: Hiking in the Shenandoah National Park was great!In summary, the orthogonal trajectories for the given equations are y = -x/2 + C and x^2y^{-3} + C = y^{-3}, respectively. The first equation was solved by finding the slope of the tangent lines and setting it equal to the negative reciprocal of the slope of the original curve. The second equation was solved by rewriting it to eliminate the constant and then differentiating with respect to x.
find the orthogonal trajectories of the following

(a) $$x^2y=c_1$$

(b) $$x^2+c_{1}y^3=1$$

for part (a) I've found $$y=\frac{1}{2}\log{|x|} + C_2$$

for part (b) if i solve this integral this should be the O.T.

$$\frac{3}{2}\int{(\frac{1}{x^2}-1)}dx= \frac{y^2}{2}$$

is this correct?

this is very unusual that nobody has posted a response to this thread.

There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

Differentiating x2y= c1 with respect to x gives
2xy+ x2y'= 0: we have removed the constant c1. The point is that any member of that family of trajectories must satisfy that differential equation which is the same as y'= -y/x. We know that the slopes, m1 and m2 of two lines y= m1x+ b1 and y= m1x+ b2 must satisfy m1m2= -1 or m2= -1/m1.
Any curve orthogonal to every member of the family given must have, at each point, y'= dy/dx= -1/(-y/x)= x/y. That separates as ydy= xdx. What is the general solution to that?

For (b) $$x^2+c_{1}y^3=1$$
it's a little harder to eliminate c1 but if you rewrite this as $$x^2y^{-3}+ c_1= y^{-3}$$, differentiating with respect to x will eliminate c1.

There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

I appreciate all the help that I've received on this site. Phs. forums is a great resource. But i posted my initial post yesterday and my second post today. I just thought that it was unusual for a post not to be responded to in the amount of time that my post was up. I'm glad you had a good time hiking unlike myself who has spent all of today as well as yesterday getting prepared for the upcoming week.

Oops! My bad: there were 12+ about three hours between your first and second posts! Hope my response helped.

## 1. What are orthogonal trajectories?

Orthogonal trajectories are a set of curves that intersect another set of curves at right angles. In other words, they are a family of curves that are perpendicular to another family of curves.

## 2. What is the significance of orthogonal trajectories?

Orthogonal trajectories have many applications in science and engineering. They are commonly used in fields such as physics, engineering, and mathematics to study and analyze various phenomena, such as electric and magnetic fields, fluid dynamics, and heat transfer.

## 3. How are orthogonal trajectories related to differential equations?

Orthogonal trajectories are closely related to differential equations. In fact, they are often used to solve differential equations. The curves of the orthogonal trajectories can be found by solving a differential equation that represents the relationship between the two families of curves.

## 4. Can orthogonal trajectories exist for any type of curves?

Yes, orthogonal trajectories can exist for any type of curves, as long as they are defined by a differential equation. However, in some cases, it may be difficult or impossible to find the equations for the orthogonal trajectories analytically.

## 5. How are orthogonal trajectories used in real-life applications?

Orthogonal trajectories have many practical applications in fields such as engineering and physics. They are used to study and analyze various phenomena, such as electric and magnetic fields, fluid dynamics, and heat transfer. They are also used in applications such as antenna design, circuit analysis, and surface modeling.

Replies
3
Views
1K
Replies
10
Views
1K
Replies
1
Views
513
Replies
6
Views
702
Replies
9
Views
1K
Replies
28
Views
842
Replies
18
Views
2K
Replies
6
Views
2K
Replies
5
Views
1K
Replies
7
Views
984