Show that all normals to $\gamma(t)=(\cos(t)+t\sin(t),\sin(t)-t\cos(t))$ are the same distance from the origin.

Question

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!

$$$$ $$$$ 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?

Answer 1

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.)

Answer 2

Did you try parameterizing the normal? You will be able to minimize the norm over $t$ and show the minimum value doesn't depend on $t_0$. Click on the link below to see what I mean. I parameterized the normal for you in slot #5

https://www.desmos.com/calculator/rrfe1s94dm

If you do this, you'll find that any normal line is one unit away from the origin.

Hope this helps.

Answer 3

We have

$$ \gamma(t) = \left[\begin{array}{cc}\cos t & -\sin t\\ \sin t& \cos t\end{array}\right]\cdot \left[\begin{array}{c} 1 \\ -t\end{array}\right] = R(t)\cdot \left[\begin{array}{c} 1 \\ -t\end{array}\right] $$

the tangent vector

$$ \dot\gamma = \dot R\cdot\left[\begin{array}{c} 1 \\ -t\end{array}\right]+ R\cdot \left[\begin{array}{c} 0 \\ -1\end{array}\right] $$

so

$$ \eta = \dot R\cdot\left[\begin{array}{c} t \\ 1\end{array}\right]+ R\cdot \left[\begin{array}{c} 1 \\ 0\end{array}\right] $$

is the normal vector because $\dot\gamma\cdot \eta = 0$ The normal lines are thus

$$ n(t,\lambda) = \gamma(t) + \lambda \eta(t) $$

their squared distance to the origin is given by

$$ ||n||^2 = 1+(\lambda-1)^2 t^2 $$

which has a minimum at $\lambda = 1$ with minimum value $||n|| = 1$

Answer 4

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{t\sin t}{t\cos t}$$

The equation of the normal will be $$\dfrac{y-(\sin t-t\cos t)}{x-(\cos t+t\sin t)}=-\dfrac{\cos t}{\sin t}$$

$$\iff x\cos t+y\sin t-1=0$$

The distance from the origin $$\dfrac{\left|0\cdot\cos t+0\cdot\sin t-1\right|}{\sqrt{\cos^2t+\sin^2t}}=?$$

Answer 5

$$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{t\sin t}{t\cos t}$$ The normal has a slope that is its negative reciprocal $ =\dfrac{-\cos t }{\sin t}$

The equation of the normal is:

$$ \dfrac{y-y_1}{x-x_1}=\dfrac{-\cos t}{\sin t}$$

$$\dfrac{y-(\sin t-t\cos t)}{x-(\cos t+t\sin t)}= \dfrac{\cos t}{\sin t}$$

Cross-multiply, transpose and simplify to get

$$ x\cos t+y\sin t =1 = p $$

Recognize this is a standard tangent polar-normal form of a straight line where the tangent has slope $\tan t$ to positive $x$ axis and constant minimum perpendicular/normal pedal distance $p=1$

Alternately convert this to polar coordinates $(r,\theta =t )$ ( working is left to you as simple exercise)

$$ r= p \sec ( \theta - \alpha) $$

which has minimum pedal distance $r=p,$ at $ ( \theta = \alpha) $

The given parametrization belongs to an involute formed by a taut string end-point starting at $(1,0)$ and rotating in the anti-clockwise direction. The normal to involute always has minimum distance to the base circle ( terminology used in gear design) $ r=p=1,$ with Pythagoras triangle property

$$ \sqrt{r^2-T^2} = p. $$

Answer 6

As a variant: The point $\gamma(t)$ is obtained by taking the point on the circle $r(t)=(\cos(t), \sin(t))$ and following circle's tangent line at $r(t)$, which we denote by $L$, and which is in the direction $Rot_{90} r(t)=(\sin(t), -\cos(t))$, for time $t$. Thus after we compute the tangent to $\gamma(t)$ to have direction parallel to $(\cos(t), \sin(t))$ i.e. to $r(t)$, we know that the normal to $\gamma$ at $\gamma(t)$ is parallel to $L$ and passes through $\gamma(t)$, which $L$ also does. So the normal line is nothing but $L$ itself. Of course $L$ is tangent to the unit circle (at $r(t)$), and so is unit distance from the origin.