Find the orthogonal trajectory for x - y = cx^2

Davidloke

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Can someone help to to confirm my answer for this!

Find the orthogonal trajectory for x-y=cx^2

I got something like ln|2|Sqrt(16x)....arctan....
 
Can someone help to to confirm my answer for this!

Find the orthogonal trajectory for x-y=cx^2

I got something like ln|2|Sqrt(16x)....arctan....
What does orthogonal trajectory mean?

How did you get your answer?

Please share your work.
 
What does orthogonal trajectory mean?

How did you get your answer?

Please share your work.


Suppose we have a curve. F(x,y,c)=0 The curve consists of a parameter like x^2+y^2=k
k can take the value of 0,1,2,3,4,5,....
This is an example a contour or level curves in multi-variable calculus!

As for differential equation it is the same too. F(x,y,c)=0. We already have the one parameter family curves.

Through differentiating the curves we will have a gradient that is f(x,y) just for clarity purpose!

The gradient is a tangent line that is parallel to the cuves at point(x,y)

We know that a perpendicular point at the point can be found using the negative reciprocal of f(x,y) this will not work for vertical tangent because we know that vertical tangent has infinite value.

Then, we have a perpendicular line that pass through the curve at that point is given by differential equation of dy/dx=-1/f(x,y)

Solving this will yield us the orthogonal trajectory equation of family solutions.
we will have infinite number of straight lines penetrating perpendicularly though all points that are on the curves.

I think what I claim in the above is almost complete... for the definition of orthogonal trajectory!


So this question is to find the orthogonal trajectory of the curves!

x-y=c(x^2)

My working is

Step 1: finding tangent line

differentiating w.r.t x

to find the differential equation

We have dy/dx=(2y-x)/x

Step 2: Orthogonal tangent line!

So here it is. with negative reciprocal of

dy/dx=(-x)/(2y-x)

Step 3: solve the differential equation

Through inspection we have the homogeneous equation to solve

My final answer is ln2(sqrt(16x^2-8xy+5x^2))-1/2arctan(y/x-1/4)

There is no answer for this question behind the book! The way I do step 3 is very messy!
 
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