# Can Orthogonal Trajectories have Multiple Solutions? And how do I know it's correct as an autodidact without teacher?

TLDR: Are there different solutions to orthogonal trajectories, and if so, how can I as an autodidact check whether I'm doing things correctly or not?

The reason I'm asking this is because the textbook and the solution of someone with a PhD in Mathematics have different answers (or at least, someone claiming to have it). I'll show both and then my approach.

Problem: find the orthogonal trajectory of $$r = c(1 + \cos(θ))$$

1. Textbook solution without work: $$r = c(1 - \cos(θ))$$
2. PhD Mathematics solution: $$(1)\ \frac{dr}{dθ} = -c\sin(θ)$$ $$(2)\ \frac{1}{r}\frac{dr}{dθ} = -\frac{c}{r}\sin(θ)$$ $$(3)\ \frac{1}{r}\frac{dr}{dθ} = \frac{r}{c\sin(θ)}$$ $$(4)\ \frac{dr}{r^2} = \frac{1}{c}\csc(θ)dθ$$ Solve r by seperating variables $$(5)\ \frac{dr}{r^2} = \frac{1}{c}\csc(θ)dθ$$ After integrating both sides $$(6)\ -\frac{1}{r} = -\frac{1}{c}ln|\csc(θ) + \cot(θ)|- C$$ $$(7)\ r = -\frac{1}{\frac{1}{c}ln|\csc(θ) + \cot(θ)| + C}$$

Which is a different answer. Now I'm writing down how I did it.

1. My Solution: $$(1)\ c = \frac{r}{1+\cos(θ)}$$ $$(2)\ c^{-1}r = 1 + \cos(θ)$$ $$(3)\ \frac{d}{dr}(c^{-1}r) = (1 + \cos(θ))\frac{d}{dr}$$ $$(4)\ c^{-1} = -\sin(θ)\frac{dθ}{dr}$$ $$(5)\ -\frac{1}{c\sin(θ)} = \frac{dθ}{dr}$$ After substituting into c, multiplying both sides by r to get into the right form and simplifying we get $$(6)\ -\frac{1+\cos(θ)}{\sin(θ)} = -\frac{dr}{rdθ}$$ $$(7)\ \frac{\sin(θ)}{1+\cos(θ)} = \frac{rdθ}{dr}$$ $$(8)\ \frac{1+\cos(θ)}{\sin(θ)}dθ = \frac{1}{r}dr$$ $$(9)\ \int \frac{1}{\sin(θ)} + \frac{\cos(θ)}{\sin(θ)}dθ = \int \frac{1}{r}dr$$ After integration $$(10)\ θ\arcsin(θ) + \sqrt{1-θ^2} + ln|\sin(θ)| + C_1 = \ln|r| + C_2$$ After isolating $$(11)\ r = e^{θ\arcsin(θ) + \sqrt{1-θ^2} + ln|\sin(θ)| + C}$$

My approach looks really bad, but its more about whether different solutions are possible and if it is how I will know whether my solution is a correct one. I'm self-studying and the books I'm using didn't mention multiple solutions being possible, so I'm a bit lost.

• Why not draw them to see whether they look orthogonal? Oct 1 at 12:15
• The PhD answer seems to have the wrong sign for $r$. Reversing the sign, I get something that looks orthogonal to $r = c(1 + \cos(\theta))$, but only if I use the same value of $c$ in both formulas. Your answer crosses the curve at a slightly different angle and is only defined in the sector of the plane where $-1\leq\theta\leq1$. Here are the graphs: desmos.com/calculator/d3mbuifwrq Oct 1 at 15:49
• Perhaps my answer here will help you understand better. I don't see how you're actually dealing with the orthogonal trajectory at all. Oct 2 at 0:46
• Doesn't autodidact mean without a teacher? Wait ... I said the same thing you said and I'm being redundant. Carry on. Oct 2 at 3:10

I'm not familiar with any problem of "orthogonal trajectories". It seems to me that if you just have one trajectory and you want another orthogonal to it, you start anywhere you want on one side of the first trajectory and construct a curve to anywhere you like on the other side. The only constraint is that the constructed curve must be orthogonal to the given one at the intersection point. This allows for infinitely many solutions.

I suspect that the problem you are really supposed to be solving is to find an entire family of non-intersecting curves that is orthogonal to an entire given family of non-intersecting curves. The given family of curves is parameterized by $$c$$, which could be any positive real number, so you start at the origin (which is the only point that is never outside any of the curves) and cross every one of those curves at a right angle. This determines your trajectory uniquely.

If I understand the question correctly -- that is, if it really is supposed to be the "family of curves parameterized by $$c$$" problem and not the "just the trajectory for one particular $$c$$" problem -- then the "PhD solution" is incorrect because it treats $$c$$ as a constant along the new trajectory. It is therefore only solving the "one particular $$c$$" problem. It agrees with the book solution at the point where it crosses the particular curve $$r = c(1 + \cos(\theta))$$ for the value of $$c$$ chosen for the "orthogonal" trajectory. Everywhere else, the trajectories are different.

To solve the "family of curves" problem, I would first find the correct value $$c$$ for the curve passing through an arbitrary point in the plane. But I will first assume that $$c$$ is always positive, for reasons I will discuss later.

Since the family of curves is given by $$r = c(1 + \cos(\theta)), \tag{I}$$ we can only use positive values of $$r$$. For any arbitrary point other than the origin, let $$(r,\theta)$$ be polar coordinates of that point with $$r > 0$$. Then $$c = \frac{r}{1 + \cos(\theta)}, \tag{II}$$ The same as your Equation $$(1)$$. As in the "PhD solution", for the original curve through $$(r,\theta)$$ we differentiate Equation (I) with respect to $$\theta$$ to get $$\frac{\mathrm dr}{\mathrm d\theta} = -\,c\sin(\theta), \tag{III}$$ which is equivalent to your Equation $$(5)$$. Substituting for $$c$$ and dividing by $$r$$, $$\frac1r \frac{\mathrm dr}{\mathrm d\theta} = -\,\frac{\sin(\theta)}{1 + \cos(\theta)}.$$ For an orthogonal curve through the same point we have $$\frac1r \frac{\mathrm dr}{\mathrm d\theta} = \frac{1 + \cos(\theta)}{\sin(\theta)}.\tag{IV}$$ Separate, and integrate: $$\int \frac1r \,{\mathrm dr} = \int \frac{1 + \cos(\theta)}{\sin(\theta)}\,{\mathrm d\theta} . \tag{V}$$

The integral on the left, of course, is just $$\ln(r) + C_1$$. (There is no need to take an absolute value, since we already know $$r > 0$$.) The integral on the right is solved as follows: \begin{align} \int \frac{1 + \cos(\theta)}{\sin(\theta)}\,\mathrm d\theta &= \int \frac{1 + \cos(\theta)}{\sin^2(\theta)}\sin(\theta)\,\mathrm d\theta \\ &= \int \frac{1 + \cos(\theta)}{1-\cos^2(\theta)}\sin(\theta)\,\mathrm d\theta \\ &= \int \frac{1}{1-\cos(\theta)}\sin(\theta)\,\mathrm d\theta . \\ \end{align}

Substitute $$u = 1 - \cos(\theta)$$, so $$\mathrm du = \sin(\theta)$$, and then $$\int \frac{1 + \cos(\theta)}{\sin(\theta)}\,\mathrm d\theta = \int \frac 1u \,\mathrm du = \ln(u) + C_2 = \ln(1 - \cos(\theta)) + C_2.$$ Note that again there is no need to take an absolute value, because $$1 - \cos(\theta)$$ cannot be negative.

Therefore $$\ln(r) = \ln(1 - \cos(\theta)) + C. \tag{VI} \\$$

Taking the exponential of both sides of Equation (VI) and setting $$c' = e^C$$, $$r = c'(1 - \cos(\theta)).$$

Now you can pick any pair of positive numbers $$c$$ and $$c'$$ and you will get two cardioids that intersect in two places other than the origin, and in both of those places the curves will be orthogonal.

Comparing this answer with your solution, we agree up to your Equation $$(9)$$ and my Equation (V). There is only a disagreement in the solution of the second integral. Did you perhaps write $$\dfrac{1}{\sin(\theta)}$$ as $$\sin^{-1}(\theta)$$ and give that to an integral calculator to solve? The function $$\sin^{-1}(\theta)$$ would be interpreted as $$\arcsin(\theta)$$, which has a very different integral from $$\dfrac{1}{\sin(\theta)}$$.

I disagree somewhat with the book solution, by the way. I deliberately chose the name $$c'$$ rather than $$c$$ for the constant in my solution in order to make it clear that the orthogonal curves through any arbitrary point do not necessarily have the same constant in their equations. If the two equations have the same constant, that is, if $$c' = c$$, then the curves can only intersect on the $$y$$-axis due to symmetry.

Concerning how you could know (without a teacher) that the solution $$r = c'(1 - \cos(\theta)$$ is correct, differentiate it as you did for the original equation $$r = c(1 + \cos(\theta)$$ and put it into the form of an equation with $$\frac1r\frac{\mathrm dr}{\mathrm d\theta}$$ on one side. You should then be able to verify that the two curves are orthogonal at their intersection point, that is, if $$\frac1r\frac{\mathrm dr}{\mathrm d\theta}=m$$ for one curve then $$\frac1r\frac{\mathrm dr}{\mathrm d\theta}=-\,\frac1m$$ for the other curve.

Now to discuss why I assume $$c > 0$$. The reason is that if $$c = -c'$$, the curve $$r = c(1 + \cos(\theta))$$ is identical to the curve $$r = c'(1 - \cos(\theta))$$. That is, if we allow $$c$$ to be any non-zero real number, we get a family of curves where the curves in one half of the family are orthogonal to the curves in the other half of the family. There cannot be a third set of curves that is orthogonal to both of these subsets of the family, so this would not be a solvable problem. Therefore the setter of the problem must somehow have assumed $$c > 0$$, perhaps by assuming that $$r \geq 0$$.

Your solution has a typo in $$(3)$$ --- it should be $$\frac{d}{dr}(c^{-1}r) = \frac{d}{dr}(1 + \cos(\theta))$$ --- and a gap between $$(5)$$ and $$(6)$$: substituting $$c$$ in $$(5)$$ and multiplying both sides by $$r$$ one obtains $$-\frac{1+\cos(\theta)}{\sin(\theta)} = r\frac{d\theta}{dr}. \tag{5.5}$$ Then, to obtain the differential equation for the orthogonal curves one replaces $$r\frac{d\theta}{dr}$$ with $$-\frac{1}{r}\frac{dr}{d\theta}$$ in $$(5.5)$$, thus obtaining $$(6)$$. Equations $$(7)-(9)$$ are correct, but there is a serious mistake in $$(10)$$: its LHS is not equal to the LHS of $$(9)$$. Indeed, \begin{align} \frac{d}{d\theta}\left(\theta\arcsin(\theta) + \sqrt{1-\theta^2} + \ln|\sin(\theta)|\right)&=\arcsin(\theta)+\frac{\cos(\theta)}{\sin(\theta)} \neq\frac{1+\cos(\theta)}{\sin(\theta)}. \end{align} Apparently, you confused $$\frac{1}{\sin(\theta)}$$ with $$\sin^{-1}(\theta)=\arcsin(\theta)$$.

The correct result for the integral on the LHS of $$(9)$$ is $$\ln|\csc(\theta)-\cot(\theta)|+\ln|\sin(\theta)|+C_1=\ln|1-\cos(\theta)|+C_1;$$ plugging this result in $$(10)$$ and solving for $$r$$ one finally obtains $$r=C(1-\cos(\theta)),$$ which is the textbook solution.