# How to find the orthogonal trajectories

## Understanding a family of curves and their trajectories

Given a family of curves, like ???y=kx???, we can choose different values for ???k??? to write the equations of some of the curves in the family.

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Each of the values of ???k??? above give a different curve that is part of the family of curves given by ???y=kx???. If we graph each of the curves above together on the same graph, we get

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Sometimes you’ll be asked to find the orthogonal trajectories to a family of curves. The orthogonal trajectories are the curves that are perpendicular to the family everywhere. In other words, the orthogonal trajectories are another family of curves in which each curve is perpendicular to the curves in original family. In the example below we’ll show how to use calculus to find the orthogonal trajectories, but for now we’ll give away the answer so that we can sketch the family of orthogonal trajectories and see that they are perpendicular to the original family.

The curves in blue are from the original family ???y=kx???. The curves in green are

???x^2+y^2=1???

???x^2+y^2=2???

???x^2+y^2=3???

???x^2+y^2=4???

which are four of their orthogonal trajectories, part of their whole family of orthogonal trajectories given by ???x^2+y^2=C???. Notice how each green circle is perpendicular to every blue line, wherever a green and blue curve intersect one another.

Now that we have an idea of what we’re trying to find, let’s try an example where we use calculus to show that the orthogonal trajectories are given by ???x^2+y^2=C???.

The orthogonal trajectories are the set of curves that are always perpendicular to a particular family of curves

## Step-by-step example of how to find the orthogonal trajectories

Example

Find the orthogonal trajectories to the family of curves.

???y=kx???

We always start by using implicit differentiation to take the derivative of both sides, and then we’ll solve for ???dy/dx???.

???(1)\frac{dy}{dx}=k(1)???

???\frac{dy}{dx}=k???

Once we’ve got an equation for ???dy/dx???, we’ll go back to the original equation and solve it for ???k???, so that we can plug a value in for ???k??? that’s in terms of ???x??? and ???y???.

???y=kx???

???k=\frac{y}{x}???

Plug the value for ???k??? into the equation for ???dy/dx???.

???\frac{dy}{dx}=k???

???\frac{dy}{dx}=\frac{y}{x}???

Remember that ???dy/dx??? is the slope of the family of curves ???y=kx???, so the equation we just found represents the slope of the family everywhere.

If we want to find the orthogonal trajectories, and we know that they’re perpendicular to our family everywhere, then we want a slope for the orthogonal trajectories that is perpendicular to the slope of the original family. To find a perpendicular slope, we take the negative reciprocal (flip it upside down and add a negative sign).

Slope of the original family: ???\frac{dy}{dx}=\frac{y}{x}???

Slope of the orthogonal trajectories: ???\frac{dy}{dx}=-\frac{x}{y}???

If we treat the slope of the orthogonal trajectories as a separable differential equation, we can separate variables and integrate both sides in order to find the equation of the family of orthogonal trajectories. We’ll start by separating variables.

???\frac{dy}{dx}=-\frac{x}{y}???

???dy=-\frac{x}{y}\ dx???

???y\ dy=-x\ dx???

Then we’ll integrate both sides.

???\int y\ dy=\int -x\ dx???

???\frac12 y^2=-\frac12 x^2+C???

???\frac12 x^2+\frac12 y^2=C???

???x^2+y^2=2C???

The ???2??? can be absorbed into the constant, so the equation becomes

???x^2+y^2=C???

This is the equation of the family of orthogonal trajectories, which is the family of curves that are perpendicular to the original family.

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