# Finding Orthogonal Trajectories (differential equations)

• Westlife
In summary, to find the orthogonal trajectories of the given equation, we can substitute the new independent variable w = x^2 and new dependent variable z = y^2, and then take the derivative with respect to w to find the equations for w' and z'. This will lead to the final solution for the orthogonal trajectories.
Westlife

## Homework Statement

Find Orthogonal Trajectories of ##\frac{x^2}{a}-\frac{y^2}{a-1}=1##
Hint
Substitute a new independent variable w
##x^2=w##
and an new dependent variable z
##y^2=z##

## The Attempt at a Solution

substituting ##x## and ##y## I get
$$\frac{w}{a}-\frac{z}{a-1}=1\quad /a(a-1)$$
$$w(a-1)-za=a(a-1)\quad /'$$
$$w'(a-1)-z'a=0$$

Here I wanted to figure out what ##w'##and ##z'## are and here is where I get confused. How do I get them?
since w is now the new independent variable is ##w'= \frac{dw}{dw}=1## and ##z'=\frac{dz}{dw}## or does mean mean something else ?

Westlife said:

## Homework Statement

Find Orthogonal Trajectories of ##\frac{x^2}{a}-\frac{y^2}{a-1}=1##
Hint
Substitute a new independent variable w
##x^2=w##
and an new dependent variable z
##y^2=z##

## The Attempt at a Solution

substituting ##x## and ##y## I get $$\frac{w}{a}-\frac{z}{a-1}=1\quad /a(a-1)$$ $$w(a-1)-za=a(a-1)\quad /'$$ $$w'(a-1)-z'a=0$$

Here I wanted to figure out what ##w'##and ##z'## are and here is where I get confused. How do I get them?
since w is now the new independent variable is ##w'= \frac{dw}{dw}=1## and ##z'=\frac{dz}{dw}## or does mean mean something else ?
Hello @Westlife .

Since the new independent variable is ##\ w\,,\ ## it follows that you take the derivative with respect to ##\ w\,.\ ## That is to say, take ##\ \frac{d}{dw}\,.##

So, yes, you get
##\displaystyle \ (a-1)-a\,z'=0 \text{ , where } z'=\frac{dz}{dw} \,.##​

## 1. What is an orthogonal trajectory in the context of differential equations?

An orthogonal trajectory is a curve that intersects another curve at a right angle. In the context of differential equations, it refers to a solution curve that is perpendicular to the family of curves described by the given differential equation.

## 2. Why is finding orthogonal trajectories important in mathematics?

Finding orthogonal trajectories allows us to better understand the relationship between different curves and can provide more information about a given system. It also has practical applications in fields such as physics and engineering.

## 3. What is the process for finding orthogonal trajectories?

To find orthogonal trajectories, we first need to find the general solution to the given differential equation. Then, we use the fact that the slopes of orthogonal curves are negative reciprocals to find the equation of the orthogonal trajectory. This process may involve using techniques such as implicit differentiation or substitution.

## 4. Can all differential equations have orthogonal trajectories?

No, not all differential equations have orthogonal trajectories. For a differential equation to have orthogonal trajectories, it must be of a certain form, such as a first-order or second-order linear differential equation.

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

Orthogonal trajectories have practical applications in fields such as fluid mechanics, where they can help determine the flow of fluids and the paths of particles. They can also be used in optics to calculate the path of light rays and in electrical engineering to determine the path of electric fields.

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