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I know that by Frenet-Serret, we have (I know this is only for curves parametrized by arclength, but since every plane curve can be reparametrized by arclength, there's no loss of generality):

$t'(s) = k(s)n(s) \Rightarrow t''(s) = k(s)n'(s)$ (since $k(s)$ is constant)

$n'(s) = -k(s)t(s)$

Since the curvature is constant, I can call $k(s) = c$ and get:

$t''(s) = -c^2 t(s)$

$t''(s) + c^2t(s) = 0$

which has solutions:

$t(s) = A_1 \cos(cs) + A_2 \sin(cs)$ (where I also assume $||t|| = ||n|| = 1$)

so

$a(s) = \frac{1}{c} (A_1 \sin(cs) - A_2 \cos(cs)) + A_3$ is the form all regular curve planes are (I forgot to add regular in the beginning, but it's in the exercise)

"Squaring" both sides, we get:

$||a(s) - A_3||^2 = \frac{1}{c^2} ||A_1||^2 \Rightarrow ||a(s) - A_3||^2 = \frac{1}{c^2} \Rightarrow ||a(s) - A_3|| = \frac{1}{c}$, which is clearly the equation of a circle (I used Gribouillis' answer in the middle of these steps)

Matheus Andrade
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    I don't understand how you replace $t(s)$ with $a(s)$ in the second line. Why not write directly $t'' = -c^2 t$ ? After solving this, you need to add the condition that $|t| = 1$, because $t$ and $n$ need to be orthogonal. – Gribouillis Dec 15 '17 at 06:38
  • You're quite right, I've edited my post to reflect that. Thanks! Am I correct now? – Matheus Andrade Dec 15 '17 at 06:49
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    No, you can write $t''=k n'$ only after saying that $k$ is constant, because this is obtained by derivating $t' = k n$. Also note that $t$ is a vector, which means that $K_1$ and $K_2$ are vectors. You need to compute $|t|$ at the end, in order to obtain a condition of $K_1$ and $K_2$. – Gribouillis Dec 15 '17 at 06:58
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    @Gribouillis: The title idicates that $k$ is constant. – Robert Lewis Dec 15 '17 at 07:01
  • @Gribouillis Forgive my naivety, but how do I compute $||t||$? I can imagine setting $A_1 = (a_1, b_1)$ and $A_2 = (a_2, b_2)$ and proceeding from that, but is there a better way? – Matheus Andrade Dec 15 '17 at 07:05

2 Answers2

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You have $$ 1 = \|t\|^2 = (A_1 \cos(cs) + A_2\sin(cs))^2 = A_1^2 \cos^2(cs) + A_2^2 \sin^2(cs) + 2 \cos(cs)\sin(cs)(A_1\cdot A_2) $$ Differentiating gives (we assume $c\not=0$) $$ (A_2^2 - A_1^2)2\cos(cs)\sin(cs)+ A_1\cdot A_2 2(\cos^2(cs) - \sin^2(cs)) = 0 $$ which you can rewrite as $$ (A_2^2 - A_1^2)\sin(2cs)+ 2 A_1\cdot A_2 \cos(2cs) = 0 $$ derivating again gives $$ (A_2^2 - A_1^2)\cos(2cs)- 2 A_1\cdot A_2 \sin(2cs) = 0 $$ It follows that $A_1^2 = A_2^2$ and $A_1\cdot A_2 = 0$. Inserting this in the first equation above shows that $A_1$ and $A_2$ are two orthogonal vectors with norm $1$.

With very little work, you can now conclude that the curve is a circular arc.

Gribouillis
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  • After some time thinking I got to my edited answer, which shows it is indeed a circle. Does it check out? – Matheus Andrade Dec 15 '17 at 08:03
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    No, $|a(s) - A_3|^2 = \frac{1}{c^2}|A_1|^2 = \frac{1}{c^2}$. One gets a circular arc centered at $A_3$ with radius $r = \frac{1}{c}$. – Gribouillis Dec 15 '17 at 08:07
  • I see my mistake. But shouldn't the radius be $\frac{1}{\sqrt{c}}$? My understanding is that if we have

    $||a(t) - v|| = r^2$, that is a circle centered at v with radius r

     – Matheus Andrade Dec 15 '17 at 08:15
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    No, it is either $|a(t) - v|=r$ or $|a(t)-v|^2 = r^2$. Remember that $|a(t) - v| = \text{dist}(a(t), v)$ – Gribouillis Dec 15 '17 at 08:17
  • I understand now. I appreciate your patience and thoroughness in the explanations. Thank you! – Matheus Andrade Dec 15 '17 at 08:19
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The OP's solution appears essentially correct to Yours Truly, though as suggested by Gribouillis in his comment it might be wise to say a bit more about $A_1, A_2$ and $K_1$, $K_2$, $K_3$.

Having typed these words, I present a more geometrically-flavored solution which does not rely so much on differential equations or their exact solutions. To wit:

Suppose $\gamma(s)$ is our curve, parametrized by arc-length, and that

$\kappa(s) = 0; \tag 1$

then

$T'(s) = \kappa(s)N(s) = 0, \tag 2$

so

$T(s) = T_0, \tag 3$

a constant. Then

$\dot \gamma(s) = T(s) = T_0, \tag 4$

whence

$\gamma(s) - \gamma(s_0) = \displaystyle \int_{s_0}^s \dot \gamma(u) \; du = \int_{s_0}^s T_0 du = T_0(s - s_0), \tag 5$

so

$\gamma(s) = T_0 (s - s_0) + \gamma(s_0), \tag 6$

which is the equation of a straight line.

If, on the other hand, we have constant

$\kappa(s) = \kappa > 0, \tag 7$

we consider the curve $c(s)$ defined by

$c(s) = \gamma(s) + \kappa^{-1} N(s); \tag 8$

we have

$\dot c(s) = \dot \gamma(s) + \kappa^{-1} \dot N(s) = T(s) + \kappa^{-1} (-\kappa T(s)) = T(s) - T(s) = 0; \tag 9$

it thus follows that the "curve" $c(s)$ is constant:

$c(s) = c_0; \tag{10}$

then (8) becomes

$c_0 = \gamma(s) + \kappa^{-1}N(s), \tag{11}$

or

$\gamma(s) - c_0 = -\kappa^{-1}N(s), \tag{12}$

and thus

$\Vert \gamma(s) - c_0 \Vert = \Vert -\kappa^{-1} N(s) \Vert = \kappa^{-1} \Vert N(s) \Vert = \kappa^{-1}, \tag{13}$

since $\Vert N(s) \Vert = 1$. (13) indicates that every point or $\gamma(s)$ is the same distance $\kappa^{-1}$ from $c_0$; thus it is the equation of a circle, which in fact $\gamma(s)$ is.

We see that every constant curvature planar curve is either a straight line or a circle.

Robert Lewis
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