[Math] Orthogonal trajectories for a given family of curves

calculusordinary differential equations

Find the orthogonal trajectories for the given family of curves:

$$y^2=Cx^3-2$$

Derivative with respect to x, and finding the value of C:

$$2yy' = C3x^2 $$
$$C=\frac{y^2+2}{x^3}$$

Replacing C, and solving for y':

$$2yy' = \frac{3y^2+6}{x}$$
$$y'=\frac{3y^2+6}{2xy}$$

Finding the orthogonal and separating terms:

$$y' = -\frac{2xy}{3y^2+6}$$
$$\frac{(3y^2+6)}{y}y'=-2x$$
$$\frac{(3y^2+6)}{y}dy=-2xdx$$

Integrating:

$$\int{\frac{(3y^2+6)}{y}dy}=\int{-2xdx}$$
$$\frac{3y^2}{2}+6log(y)=-x^2+C$$
$$C=\frac{3y^2}{2}+6log(y)+x^2$$

Was I close? We weren't given solutions, so I'd appreciate a check on my answer. 🙂

Best Answer

I've plotted the results on Mathematica and it looks OK.

there's the small detail that the log is missing an absolute value, i.e. it's $6\log |y|$.

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[Math] Orthogonal Trajectories

Suppose your two curves are defined by $x^2+y^2=ax$ and $x^2+y^2=by$ , with $a,b$ real constants. To compare the tangent line slopes at given points for each curve, we differentiate the first equation to find that $$ 2x + 2yy' = a~~\text{and so}~~y' = \frac{a-2x}{2y}~~, $$ and the second to find that $$ 2x + 2yy' = by' ~~\text{and so}~~ y' = \frac{2x}{b-2y}~~. $$ These answers are different from the ones you got -- note that what I did was group terms containing $y'$ on one side of the equation, and divide through by whatever factor accompanied it.

We are in business so long as the product of these two quantities, for a given pair $(x,y)$ where the curves intersect, is $-1$. So, multiply one by the other and we get $$ \frac{2x(a-2x)}{2y(b-2y)} = \frac{x(a-2x)}{y(b-2y)} ~~. $$ I believe the problem you ran into is that it is not clear (in an algebraic sense anyways) that these factors should cancel in any way. But remember -- since we are looking at a point where the two curves we were given intersect, we may apply both of those equations. Where does $a-2x$ appear?

Well we have $x^2 + y^2 = ax$ , so that $ax - x^2 = y^2$ , $ax - 2x^2 = y^2 - x^2$ , and $x(a-2x) = y^2-x^2$. Now bear with me, while this may not look simpler, observe that by the same token, $$ x^2 + y^2 = by ~~\text{means that}~~ by-2y^2 = x^2-y^2 ~~\text{and so}~~ y(b-2y) = x^2-y^2 ~~. $$ The factors on top and bottom are indeed the same aside for a sign switch, so the two slopes are negative reciprocal.

BTW: A helpful, and even pretty exercise is to actually plot some of these orthogonal curves. In this case, you'll notice that the first family is circles with an $x$-offset of the center from the origin, while the second family is circles with a $y$-offset. Is it clear why these are orthogonal?

[Math] How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$

Let $u = y+x$, $v = y-x$, we have

$$\begin{align} u' = y'+1 = & \frac{2x^2-2y^2}{x^2 - 2xy - y^2+2} = \frac{-2uv}{x^2-2xy-y^2+2}\\ v' = y'-1 = & \frac{4xy-4}{x^2 -2xy- y^2+22} = \frac{u^2 - v^2-4}{x^2-2xy-y^2+2}\\ \end{align}$$

From this, we get $$u^2 \left( \frac{v^2}{u} + u +\frac{4}{u} \right)' = u^2\left(\frac{v^2}{u}\right)' + (u^2 - 4)u' = 2uvv' + (u^2 - v^2 - 4)u' = 0$$ and hence for some integration constant $K$, one has $$\frac{v^2}{u} + u +\frac{4}{u} = 2K \;\;\iff\;\; v^2 + u^2 + 4 = 2K u \;\;\iff\;\; x^2 + y^2 + 2 = K(x+y) $$ i.e the orthogonal trajectories is another family of circles.

Update

There is a pure geometric argument why the orthogonal trajectories are circles.

Let $p = (-1,-1), q = (1,1)$. Let $\mathscr{C}$ be a circle centered at $p$ with radius $|pq| = 2\sqrt{2}$. Consider the inversion $\mu$ with respect to circle $\mathscr{C}$, we have:

$$\mu(p) = \infty\quad\text{ and }\quad \mu(q) = q.$$

This means under $\mu$, the family of circles passing through $p$ and $q$ get mapped to the family of straight lines passing through $q$. The orthogonal trajectories of this family of straight lines is the family of circles centered at $q$.

It is known that inversion is conformal. i.e. preserve angles and hence map orthogonal trajectories to orthogonal trajectories. This means the orthogonal trajectories of the family of circles passing through $p$ and $q$ are the inverse image of $\mu$ of the family of circles centered at $q$.

It is also known that inversion $\mu$ map circles not passing through $q$ to circles. As a result, the family of orthogonal trajectories we seek are circles themselves.

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[Math] Orthogonal Trajectories

[Math] How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$

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