Solution: We recall that the parametrization $\sigma (t)$ of the circle $x^2 + y^2 = 1$, in the counter-clockwise direction with $\sigma (0) = (1,0)$ is given by \[ \sigma (t) = \left( \cos t, \sin t \right) , 0\leq t \le 2\pi. \] We will find the parametrization of the asked circle in the following steps:

• Orient $\sigma $ in the clockwise direction. For this we will replace $t$ with $-t$ and this will give \[ \gamma _1(t) = \sigma (-t) = \left( \cos t, - \sin t \right). \]
• Now instead of starting from $(1,0)$ we need to start from $(0,1)$, so we need to start from $t = \frac{\pi}{2} $ in $\gamma_1$. That is, \begin{align*} \gamma (t) & = \gamma _1\left( t - \frac{\pi}{2} \right) \\ & = \left( \cos \left( t - \frac{\pi }{2} \right), -\sin \left( t - \frac{\pi }{2} \right) \right) \\ & = \left( \sin t, \cos t \right) \end{align*}

Therefore, the required parametrization will be \[ \gamma (t) = \left( \sin t, \cos t \right),\ 0 \leq t \leq 2\pi. \]