Show that all normals to $\gamma(t)=(\cos(t)+t\sin(t),\sin(t)-t\cos(t))$ are the same distance from the origin.
My attempt:
Let $\vec{p}=(\cos(t_0)+t_0\sin(t_0),\sin(t_0)-t_0\cos(t_0))$ be any arbitrary point for $t_0\in\mathbb{R}$. Then the tangent vector at $\vec{p}$ is given by $\dot\gamma(t_0)=(t_0\cos(t_0),t_0\sin(t_0))\implies$ the slope of the tangent vector at any point is given by $m=\tan(t_0)\implies$ the slope of any normal line is given by $m_{\perp}=-\cot(t_0)$. Now we calculate the normal line at any point $\vec{p}:$ $$y-(\sin(t_0)-t_0\cos(t_0))=-\cot(t_0)(x-\cos(t_0)-t_o\sin(t_0))\implies$$ $$\cot(t_0)x+y+(2t_0\cos(t_0)+\cot(t_0)\cos(t_0)-\sin(t_0))=0$$
Recall that the distance from $Ax+By+C=0$ and $Q(x_0,y_0)$ is: $$|l,Q|=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$
Hence $$|l,Q|=\frac{\sqrt{4t_0^2\cos^2(t_0)+\cot^2(t_0)\cos^2(t_0)+\sin^2(t_0)}}{\sqrt{\cot^2(t_0)+1}}$$
How can I proceed from here? Thanks in advance!
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Further progress:
Following the advice of user429040, the parametric form of any normal line is: $$\mathscr l=(x(t_0)-tt_0\sin(t_0), y(t_0)+tt_o\cos(t_o))$$
The goal is to now minimize the norm of this parametric line over $t$, and show that this minimum does not depend on $t_0$:
$$|\mathscr l|=|(x(t_0)-tt_0\sin(t_0), y(t_0)+tt_o\cos(t_o))|=((t-1)^2(t_0^2+1))^\frac{1}{2}\implies\min|\mathscr l|=0.$$
However, we can graphically confirm and confirm by Prof. Blatter's answer above that this conclusion is incorrect. Where do I go from here?
Best Answer
At the curve point $\gamma(t)$ we have the tangent vector $\dot\gamma(t)=(t\cos t,t\sin t)$. Turn this vector counterclockwise $90^\circ$, and obtain $(-t\sin t, t\cos t)$. When $t>0$ therefore the unit normal at $\gamma(t)$ is $n(t)=(-\sin t, \cos t)$. This allows to obtain the normal $\nu$ at $\gamma(t)$ in the parametric form $$\nu:\quad u\mapsto\nu(u)=\gamma(t)+u\,n(t)=\bigl(\cos t+(t-u)\sin t,\ \sin t-(t-u)\cos t\bigr)\ .$$ In order to determine the distance of $\nu$ from the origin $O$ we have to determine the point $P$ on $\nu$ for which $\vec{OP}\perp n(t)$. This means that we have to find the $u$-value for which $\nu(u)\perp n(t)$, or $$\nu(u)\cdot n(t)=-\sin t\bigl(\cos t+(t-u)\sin t\bigr)+\cos t\bigl(\sin t-(t-u)\cos t\bigr)=0\ .$$ This simplifies to $u=t$, so that we obtain $P=\nu(t)=(\cos t,\sin t)$. This shows that $|OP|=1$, independently of $t>0$. The case $t<0$ is of course analogue. (At $\gamma(0)$ the curve has a singularity.)