Fdtxjr

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

My attempt:

Let $vecp=(cos(t_0)+t_0sin(t_0),sin(t_0)-t_0cos(t_0))$ be any arbitrary point for $t_0inmathbbR$. Then the tangent vector at $vecp$ is given by $dotgamma(t_0)=(t_0cos(t_0),t_0sin(t_0))implies$ 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 $vecp:$ $$y-(sin(t_0)-t_0cos(t_0))=-cot(t_0)(x-cos(t_0)-t_osin(t_0))implies$$ $$cot(t_0)x+y+(2t_0cos(t_0)+cot(t_0)cos(t_0)-sin(t_0))=0$$

Recall that the distance from $Ax+By+C=0$ and $Q(x_0,y_0)$ is: $$|l,Q|=fracsqrtA^2+B^2$$

Hence $$|l,Q|=fracsqrt4t_0^2cos^2(t_0)+cot^2(t_0)cos^2(t_0)+sin^2(t_0)sqrtcot^2(t_0)+1$$

How can I proceed from here? Thanks in advance!

$$$$
$$$$
Further progress:

Following the advice of user429040, the parametric form of any normal line is: $$mathscr l=(x(t_0)-tt_0sin(t_0), y(t_0)+tt_ocos(t_o))$$

The goal is to now minimize the norm of this parametric line over $t$, and show that this minimum does not depend on $t_0$:

$$|mathscr l|=|(x(t_0)-tt_0sin(t_0), y(t_0)+tt_ocos(t_o))|=((t-1)^2(t_0^2+1))^frac12impliesmin|mathscr l|=0.$$

However, we can graphically confirm and confirm by Prof. Blatter's answer above that this conclusion is incorrect. Where do I go from here?

At the curve point $gamma(t)$ we have the tangent vector $dotgamma(t)=(tcos t,tsin t)$. Turn this vector counterclockwise $90^circ$, and obtain $(-tsin t, tcos t)$. 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
$$nu:quad umapstonu(u)=gamma(t)+u,n(t)=bigl(cos t+(t-u)sin t, sin t-(t-u)cos tbigr) .$$
In order to determine the distance of $nu$ from the origin $O$ we have to determine the point $P$ on $nu$ for which $vecOPperp n(t)$. This means that we have to find the $u$-value for which $nu(u)perp n(t)$, or
$$nu(u)cdot n(t)=-sin tbigl(cos t+(t-u)sin tbigr)+cos tbigl(sin t-(t-u)cos tbigr)=0 .$$
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.)

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.

We have

$$
gamma(t) = left[beginarraycccos t & -sin t\ sin t& cos tendarrayright]cdot left[beginarrayc 1 \ -tendarrayright] = R(t)cdot left[beginarrayc 1 \ -tendarrayright]
$$

the tangent vector

$$
dotgamma = dot Rcdotleft[beginarrayc 1 \ -tendarrayright]+ Rcdot left[beginarrayc 0 \ -1endarrayright]
$$

so

$$
eta = dot Rcdotleft[beginarrayc t \ 1endarrayright]+ Rcdot left[beginarrayc 1 \ 0endarrayright]
$$

is the normal vector because $dotgammacdot eta = 0$ 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 $||n|| = 1$

$$dfracdydx=dfracdfracdydtdfracdxdt=dfractsin ttcos t$$

The equation of the normal will be
$$dfracy-(sin t-tcos t)x-(cos t+tsin t)=-dfraccos tsin t$$

$$iff xcos t+ysin t-1=0$$

The distance from the origin $$dfracsqrtcos^2t+sin^2t=?$$

$$dfracdydx=dfracdy/dtdx/dt=dfractsin ttcos t$$
The normal has a slope that is its negative reciprocal $ =dfrac-cos t sin t$

The equation of the normal is:

$$ dfracy-y_1x-x_1=dfrac-cos tsin t$$

$$dfracy-(sin t-tcos t)x-(cos t+tsin t)= dfraccos tsin t$$

Cross-multiply, transpose and simplify to get

$$ xcos t+ysin t =1 = p $$

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

$$ sqrtr^2-T^2 = p. $$

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.