# Curve tangent is orthogonal to curve at a point

1. Aug 20, 2014

### mahler1

The problem statement, all variables and given/known data.

Let $C$ be a curve that doesn't pass through the origin and let $P$ be the closest point on the curve to the origin. Prove that the tangent to $C$ at $P$ is orthogonal to the vector $P$.

The attempt at a solution.

Suppose $P=\gamma(t_0)$, I want to show that $<\gamma'(t_0),\gamma(t_0)>=0$. I am pretty lost with the exercise and I don't know why they mention the curve doesn't pass through the origin or the fact that $P$ is the closest point to the origin, are those hypothesis necessary?. I would appreciate some suggestions and maybe an intuitive idea of why these two vectors are orthogonal. By the way, in my class we are always working with curves parametrized by the arc lenght, so maybe I have to use the fact that $|\gamma'|=1$.

2. Aug 20, 2014

### haruspex

How would you express the distance from the origin, O, to a point on C in terms of γ(t)? How would you then express the condition for an extremum of that function?
If C passes through the origin, where will P be? What meaning would you attach to the notion of a vector orthogonal to OP in that case?

3. Aug 20, 2014

The claim seems false to me. Don't you have to add that $C$ is closed?

4. Aug 20, 2014

### ehild

If the curve passes through the origin which point is closest to the origin?

The points of the curve C are at some distance from the origin. P is the closest, that is, the distance between P and the origin is the shortest. You have to name a special point somehow...
The vector pointing to some point of the curve is $\vec r(t)$ where t is the arc length. What is that distance of that point from the origin in terms of $\vec r(t)$ ?

ehild

5. Aug 20, 2014

### haruspex

You would certainly need that it is differentiable everywhere and does not suddenly stop. Only needs to be closed if you count closure at infinity.

6. Aug 20, 2014

### mahler1

Following your suggestions (haruspex,ehild), the distance between any point of the curve $\gamma(t)$ and the origin is $||\gamma(t)-(0,0)||$. At the point $\gamma(t_0)=P$, this function has a minimum and as the norm is a monotone increasing function, I can look at the function $||\gamma(t)-(0,0)||^2=<\gamma(t),\gamma(t)>$. Since $\gamma(t_0)$ is a minimum, then $(<\gamma(t_0),\gamma(t_0)>)'=0$. But then$0=(<\gamma(t_0),\gamma(t_0)>)'=<\gamma'(t_0),\gamma(t_0)>+<\gamma(t_0),\gamma'(t_0)>=2<\gamma'(t_0),\gamma(t_0)>$. From here it follows $\gamma'(t_0)$ is orthogonal to $\gamma(t_0)$.

Is this correct?

7. Aug 20, 2014

### ehild

Add, that γ'(t) is tangent to the curve γ(t), so γ(t0)γ'(t0)=0 involves that γ(t0) is perpendicular to he tangent of the curve at point P.

ehild

8. Aug 20, 2014

At the risk of being too obvious, say the planar curve $\gamma(t) = (1+t, 0), \, t \in [0,1].$