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Show that the curve defined by $$\gamma(s):=\left(a\cos\left(\frac{s}{a}\right),a\sin\left(\frac{s}{a}\right)\right)$$

Is on a circle with the radius $a$ and the center $(0,0)$,also show that $\gamma(s)$ has been parameterized by its arc length.

I know that the parametric equation of a circle with radius $a$ and center $(x_0,y_0)$ is : $$x=x_{0}+a\cos\left(t\right)$$ $$y=y_{0}+a\sin\left(t\right)$$ If we denote the parametric equation of a circle with $\gamma(t)=\left(x_{0}+a\cos\left(t\right),y_{0}+a\sin\left(t\right)\right)$,then we have:

$$\frac{d\gamma(t)}{dt}=\left(-a\sin\left(t\right),a\cos\left(t\right)\right)$$

$$\left\Vert \frac{d\gamma(t)}{dt}\right\Vert=\sqrt{a^{2}}=a$$

So the arc length is :

$$s=\int_{0}^{t}\left\Vert \frac{d\gamma(τ)}{dt}\right\Vert dτ=a\int_{0}^{t} \;dτ$$

Which implies $s=at$.

And if we parameterize the circle by its arc length then:

$$\gamma(s)=\left(a\cos\left(\frac{s}{a}\right),a\sin\left(\frac{s}{a}\right)\right)$$

Which is the given parametrization.

But I have not shown that the center is at the origin.

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    $\begingroup$ Who don't you start from your particular problem instead of the general equation? First, $x(t)^2+y(t)^2=constant$ and you are done. Then, check that the velocity is one, and... voilà $\endgroup$ Commented Jun 30, 2020 at 19:00

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You are given the parametrization in the form $x(s) = a\cos\left(\frac s a\right) = 0 + a\cos\left(\frac s a\right)$, $y(s) = a\sin\left(\frac s a\right)=0 + a\sin\left(\frac s a\right)$. According to what you say you know about parametrizing circles, that $x(t) = x_0 + a\cos(t)$ and $y(t) = y_0 + a\sin(t)$ parametrizes a circle centered at $(x_0,y_0)$, then you should be able to conclude that $s\mapsto \big(x(s),y(s)\big)$ parametrizes a circle centered at $(x_0,y_0) = (0,0)$ of radius $a$.

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Note that $x(s)=a\cos\frac s a $ and $y(s)=a\sin\frac s a $, which leads to

$$x^2+ y^2 = a^2\left(\cos^2\frac s a + \sin^2\frac s a \right)=a^2$$ i.e. a circle of the radius $a$ and center at origin.

Also note $x=a\sin\theta $ and $y=a\sin\theta$, which implies that the polar angle $\theta$ is given by $\theta = \frac s a$, or $s= a\theta$. Thus, $s$ is the arc length corresponding to the circle sector angle $\theta $ and $\gamma (s) $ is parameterized in arc length.

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  • $\begingroup$ @45465 - you did, but could have been shown more simply per $t=\frac s a$ directly. $\endgroup$ Commented Jul 9, 2020 at 11:58

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