2
$\begingroup$

My textbook asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to that moment.$^*$ To compensate, the textbook switched to polar coordinates and started solving it that way, which was what I have shown in the calculations below.

$*$ For further elaboration, the authors got $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$ from $x^2 + y^2 = 2Cx$. They then said, "Unfortunately, the variables cannot be separated, so without additional techniques for solving differential equations we can go no further in this direction. However, if we use polar coordinates, the equation of the family can be written as $r = 2Ccos(\theta)$". And then they continue with their calculations, as stated below.

The solution that I am getting is different from that of the textbook. All of the (similar) previous problems that I have completed have been correct, so if there are errors in my understanding of this concept, I cannot detect them.


My Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

We need to eliminate the arbitrary constant $C$ because we don't just want the orthogonal trajectory for a single curve -- we want the orthogonal trajectories for the entire family of curves; therefore, we want $\dfrac{dr}{d\theta}$ in terms of $r$ and $\theta$.

$\dfrac{r}{2\cos(\theta)} = C$

$\therefore \dfrac{dr}{d\theta} = -2\left( \dfrac{r}{2cos(\theta)} \right) sin(\theta)$

$= \dfrac{-rsin(\theta)}{cos(\theta)}$

The orthogonal trajectories will have a slope which is the negative reciprocal of the slopes of the family of curves:

$\therefore \dfrac{-d\theta}{dr} = \dfrac{-rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{d\theta}{dr} = \dfrac{rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{dr}{d\theta} = \dfrac{cos(\theta)}{rsin(\theta)}$

$\implies dr(r) = \dfrac{cos(\theta)}{sin(\theta)} (d\theta)$

And we can now proceed with separation of variables...


Textbook's Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

After eliminating C we arrive at

$\dfrac{rd\theta}{dr} = \dfrac{-cos(\theta)}{sin(\theta)}$

as the differential equation of the given family. Accordingly,

$\dfrac{rd\theta}{dr} = \dfrac{sin(\theta)}{cos(\theta)}$

is the differential equation of the orthogonal trajectories. In this case, the variables can be separated, yielding

$\dfrac{dr}{r} = \dfrac{cos(\theta) d\theta}{sin(\theta)}$

And it then proceeds with integration ...


I'm wondering if both solutions (mine and the textbook) are correct? Or have I made an error? If I've made an error, I would appreciate it if people could please take the time to carefully explain the reasoning behind it. I have only just begun studying differential equations (chapter 1), so any explanation would have to be very elementary.

$\endgroup$
  • $\begingroup$ I don't follow either of these. $dr/d\theta$ is not the slope of the curve in the $xy$-plane. Do they mean orthogonal trajectory in the $r\theta$-plane? $\endgroup$ – Ted Shifrin Mar 10 '17 at 18:58
  • $\begingroup$ @TedShifrin Yes, they've done this by converting the original equation of the family of curves, which was $x^2 + y^2 = 2Cx$, into polar coordinates. $\endgroup$ – The Pointer Mar 10 '17 at 19:00
  • $\begingroup$ @TedShifrin I've updated the OP for clarity. $\endgroup$ – The Pointer Mar 10 '17 at 19:06
2
$\begingroup$

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r \frac1{\text{original }dr/d\theta}.$$ (Note that this is replacing $\tan\phi$ with $\tan(\phi+\pi/2) = -1/(\tan\phi)$.)

$\endgroup$
  • $\begingroup$ thanks for the elaborate response, Ted. This question was posed at the very beginning (chapter 1) of my differential equations book, so I assume that the author wanted to simplify it instead of assuming prior knowledge. Are you saying that both solutions (mine and the author's) are incorrect? The textbook is, "Differential Equations with Applications and Historical Notes, 3rd edition", by Simmons and Finlay. Can you please be more precise with regards to what's wrong with my/the textbooks calculations? I've only just begun studying differential equations so I really don't understand much. $\endgroup$ – The Pointer Mar 10 '17 at 19:24
  • $\begingroup$ Simmons is very astute, so I would trust him. I don't have the book to check. However, he is certainly not explaining what's going on here. Perhaps he is not claiming to have justified his solution. I'll add on a bit more. $\endgroup$ – Ted Shifrin Mar 10 '17 at 19:31
  • $\begingroup$ Thanks, Ted; I look forward to it. For further elaboration, the authors got $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$ from $x^2 + y^2 = 2Cx$. They then said, "Unfortunately, the variables cannot be separated, so without additional techniques for solving differential equations we can go no further in this direction. However, if we use polar coordinates, the equation of the family can be written as $r = 2Ccos(\theta)$". And then they continue with their calculations, as stated above. $\endgroup$ – The Pointer Mar 10 '17 at 19:36
  • $\begingroup$ In respect to your last edit: that was my point exactly. $\endgroup$ – Rafa Budría Mar 10 '17 at 19:36
  • $\begingroup$ @RafaBudría: Well, more than a cursory assertion is needed here. This is far from well-known. (I've taught calculus and multivariable calculus for over 40 years and certainly didn't know this off the top of my head.) $\endgroup$ – Ted Shifrin Mar 10 '17 at 19:37
0
$\begingroup$

Edited (Not the slope, but the orthogonal direction)

The correct is the one from your textbook. In polars, the perpendicular is got by means $\dfrac{1}{r}\dfrac{\mathrm dr}{\mathrm d\theta}\to-r\dfrac{\mathrm d\theta}{\mathrm dr}$

Added

Suppose the position vector for each curve points with parameter $t$: $(r,\theta)=(r(t),\theta(t))$

It's known the vector tangent to this curve is:

$$\frac{\mathrm d\mathbf v}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat r}+\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}$$

An orthogonal vector to this one is:

$$\mathbf v_p=r\frac{\mathrm d\mathbf \theta}{\mathrm dt}\mathbf{\hat r}-\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat\theta}$$

You can now consider eliminating the parameter for the curve:

$$\frac{\mathrm dr/\mathrm dt}{r\mathrm d\mathbf \theta/\mathrm dt}=\frac{\mathrm dr}{r\mathrm d\theta}=f(r,\theta)$$

Now,

$$-\frac{r\mathrm d\theta}{\mathrm dr}=f(r,\theta)$$

is true of the orthogonal curves to the original one.


$r=2C\cos\theta$, $t=\theta$, $(r(t),\theta(t))=(2C\cos(t),t)$

$$\frac{\mathrm d\mathbf v}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat r}+\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}=-2C\sin(t)\mathbf{\hat r}+1·2C\cos(t)\mathbf{\hat\theta}$$

$$\frac{\mathrm dr/\mathrm dt}{r\mathrm d\mathbf \theta/\mathrm dt}=\frac{-2C\sin(t)}{2C\cos(t)}=\frac{-\sin\theta}{\cos\theta}=\frac{\mathrm dr}{r\mathrm d\theta}$$

and

$$\frac{r\mathrm d\theta}{\mathrm dr}=\frac{\sin\theta}{\cos\theta}$$

$\endgroup$
  • $\begingroup$ Again I complain that this is not the slope of the curve in the $xy$-plane. $\endgroup$ – Ted Shifrin Mar 10 '17 at 19:02
  • $\begingroup$ You are right. But, anyway, the book get the right equation. $\endgroup$ – Rafa Budría Mar 10 '17 at 19:16
  • $\begingroup$ Can you please explain why the slope is $\dfrac1r\dfrac{dr}{d\theta}$? I gave a counterexample in my answer. Am I being stooopid? $\endgroup$ – Ted Shifrin Mar 10 '17 at 19:19
  • $\begingroup$ Not the slope. I am possibly mistaken, but I think you need only the orthogonal to the tangent and you get so. $\endgroup$ – Rafa Budría Mar 10 '17 at 19:19
  • $\begingroup$ But it's not orthogonal in the $xy$-plane!! Is it? Why? My counterexample doesn't fit your formula, does it? $\endgroup$ – Ted Shifrin Mar 10 '17 at 19:19

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.