1
$\begingroup$

The question is: Consider the family $F$ of circles in the $xy$ plane, $(x-c)^2+y^2=c^2$ tangent to the $y$ axis at the origin. Find a differential equation that is satisfied by the family of curves orthogonal to $F$.

My thinking: Since the implicit equation represents the level sets of the function $$ f(x,y)=c^2=(x-c)^2+y^2 $$ The gradient of the function $f$ will be perpendicular to its level sets, and therefore orthogonal to the family of curves $F$. This yields $$ \nabla f(x,y)=(0,0)=(2x-2c,2y)\Rightarrow \left(x-\frac{x^2+y^2}{2x},y\right)=(0,0)\\ \Rightarrow \left(\frac{x^2-y^2}{2x},y\right)=(0,0) $$ So we have in differential form $$ \frac{x^2-y^2}{2x}dx+ydy=0\Rightarrow \frac{y^2-x^2}{2xy}=\frac{dy}{dx} $$ But the answer is the negative reciprocal, or perpendicular vector to this one. Why? I assume my reasoning was flawed in the first step, when i took the gradient of $f$ to be perpindicular to the family $F$, but I don't see why.

$\endgroup$
  • $\begingroup$ The gradient of $f$ is not perpendicular to the family $F$. If $f(x,y)$ is a function, then the family of level sets $\{f(x,y) = c\}$ would be orthogonal to the gradient. But your function implicitly already depends on $c$! You cannot let $c$ both be a dependent AND independent variable at the same time. If you want to use the level set formulation you need to first solve $(x-c)^2 + y^2 = c^2$ for $c$ to obtain a correct function $f$ which is independent of $c$. $\endgroup$ – Willie Wong Jul 28 '16 at 21:02
  • $\begingroup$ But if you try to do that, you will notice that your function $f$ is not smooth near the origin. That's exactly as expected! All of your circles intersect the origin, so your family cannot be written as the family of level sets for some smooth function. $\endgroup$ – Willie Wong Jul 28 '16 at 21:04
  • $\begingroup$ hm, ok I think I see. So the family $F$ as a an implicit function of the three variables $c,x,y$? And for this function, the origin is not a regular value? $\endgroup$ – qbert Jul 28 '16 at 21:07
  • $\begingroup$ Yes. Further more, notice that your family of circles are all defined on the right half plane. So you have no information on the left half plane. In particular, your function $f$ will be singular all along the axis $x = 0$. (BTW, the formula for $f$ is $f(x,y) = \frac12 ( x + y^2/x)$ for $x > 0$. On the right half plane this function is smooth and the usual machinery works.) $\endgroup$ – Willie Wong Jul 28 '16 at 21:10
  • $\begingroup$ I follow that it's singular along y=0 but doesn't it break down for x=c by implicit function theorem? $\endgroup$ – qbert Jul 28 '16 at 21:25
2
$\begingroup$

$$(x-c)^2+y^2=c^2$$ or : $$y^2+x^2-2cx=0 \quad\to\quad 2c=\frac{y^2+x^2}{x}=\frac{y^2}{x}+x$$ The differential equation of this family of circles is obtained by differentiation : $$dc=0=2\frac{y}{x}dy-\frac{y^2}{x^2}dx+dx$$ $$2\frac{y}{x}dy=\left(\frac{y^2-x^2}{x^2}\right)dx$$ $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$ This is the equation that you found. But, this differential equation is for the family of circles, not for the family of orthogonal curves.

The differential equation of the family of orthogonal curves is : $$-\frac{dx}{dy}=\frac{y^2-x^2}{2xy}$$ $$\frac{dy}{dx}=-\frac{2xy}{y^2-x^2}$$

$\endgroup$
  • $\begingroup$ What allowed you to set dc=0? Also is there a 2 missing on that side? Sorry I am a bit confused about your implicit differentiation, especially since c is not a constant $\endgroup$ – qbert Jul 28 '16 at 23:32
  • $\begingroup$ I am wondering if you used tangency at the origin $\endgroup$ – qbert Jul 28 '16 at 23:32
  • $\begingroup$ For each circle $c$ is constant. The solution of a first order differential equation is a set of equations with a constant $c$ in it. Each value of the constant generates a curve, of course without change of $c$ along the curve. Thus, when going backwards, differentiating the equation of a curve in order to come back to the differential equation, $c$ is constant. $\endgroup$ – JJacquelin Jul 29 '16 at 5:26
  • $\begingroup$ It doesn't matter if the curve is tangent or isn't at origin or somewhere else. In the case considered, the tangency is a consequence of the form of the given equation. If the equation was different and curves without tangency, we would obtain a family of orthogonal curves without tangency. In the wording of the question, the specification " tangent to the $y$ axis at the origin" is superfluous (redondant) since the equation $(x-c)^2+y^2=c^2$ implies this property. So, implicitly, the property of tangency is used as soon as we use the equation $(x-c)^2+y^2=c^2$. $\endgroup$ – JJacquelin Jul 29 '16 at 5:40
  • $\begingroup$ Ok this makes sense to me. Thank you for clarifying! $\endgroup$ – qbert Jul 29 '16 at 16:27
0
$\begingroup$

@H. H. Rugh an equivalent way to yours that one could find in an old textbook of mine is:

"if a family (F) of curves is characterized as the set of solutions of differential equation $y'=f(x,y)$, the family of orthogonal curves to all curves of (F) is solution of the differential equation

$$-\dfrac{1}{y'}=f(x,y)$$

(of course, in this kind of books, it was implicit that the special case $y'=0$ had to be considered apart...)

$\endgroup$
  • $\begingroup$ What book did you find this in? I could use some more practice materials/exposition $\endgroup$ – qbert Jul 28 '16 at 21:08
  • $\begingroup$ It was in a French textbook edited in the 50s. But you can find this recipe in other places, for example here or here $\endgroup$ – Jean Marie Jul 28 '16 at 21:16
  • $\begingroup$ Thank you! I won't hold my breath for finding the textbook :p $\endgroup$ – qbert Jul 28 '16 at 21:17
0
$\begingroup$

The differential of the equation is $2(x-c)\;dx+2y \;dy=0$ and describes the tangents of a curve at a point $(x,y)$ and with $c=c(x,y)$ verifying $2cx =x^2+y^2$. By exchanging coefficients and changing sign on one (so as to get the orthogonal vector) we get $-2y \; dx + 2(x-c) \; dy=0$ or $0= -2yx \; dx + (2x^2-2xc) \;dy = -2yx \; dx + (x^2-y^2)\; dy$. Thus,

$$2yx \; dx + (y^2-x^2)\; dy =0$$ which describes the normal to a curve at $(x,y)$ and whence integrates to the wanted orthogonal curves.

The ODE is $\frac{dy}{dx}=\frac{2yx}{x^2-y^2}$, but is somewhat singular.

The above differential form integrates to a familiy of circles centered on the $y$-axis and going through the origin. It is dual to the given family and are also called "Apollonian circles".

More precisely, $x^2+(y-b)^2=b^2$ has as differential $2x\; dx+2(y-b)\; dy=0$ or using the implied expression for $b$: $0=2xy \; dx + (2y^2-2yb) \;dy=2xy\; dx + (y^2-x^2)\; dy$ i.e. the above differential.

$\endgroup$
  • $\begingroup$ I solved for c is the only difference other than taking the negative reciprocal $\endgroup$ – qbert Jul 28 '16 at 20:41
  • $\begingroup$ But you are taking the total derivative of a function whose level sets are represented by F no? $\endgroup$ – qbert Jul 28 '16 at 20:43
  • $\begingroup$ Yes, a tangent vector, say $(v_x,v_y)$, to $F$ at $(x, y)$ must be in the kernel of the 1st differential, i.e. $2(x-c)v_x + 2 y v_y=0$. A normal vector is in the kernel of the 2nd differential. So when you integrate this 2nd differential you get curves that will always be normal to $F$ at each point $(x,y)$ (when defined). $\endgroup$ – H. H. Rugh Jul 28 '16 at 20:48
  • $\begingroup$ Sorry I'm struggling to reconcile the comment with the answer. You are saying dotting the ranger vector with the gradient should be zero, I agree. But this should mean exactly that the differential is normal to the surface no? $\endgroup$ – qbert Jul 28 '16 at 20:52
  • $\begingroup$ You are right. Should insert the value of $c$ afterwards. Have edited the reply. $\endgroup$ – H. H. Rugh Jul 29 '16 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.