# Closest line from a point to a curve in R^2

1. Jan 19, 2016

### RBG

Given a parametrized curve $X(t):I\to\mathbb{R}^2$ I am trying to show given a fixed point $p$, and the closest point on $X$ to $p$, $X(t_0)$, the line between the point and the curve is perpendicular to the curve. My only idea so far is to show that $(p-X(t))\cdot(\frac{X'(t)}{||X'(t)||})=0$. But in general, I don't see why this would be true? It seems clear geometrically, but obviously that's not an argument. Any hints?

2. Jan 19, 2016

### Simon Bridge

Use the definition of "perpendicular".

3. Jan 19, 2016

### RBG

Isn't that the dot product is zero? That or that the slopes of the tangent lines are inverse reciprocals of one another. But I don't see how the latter definition can be applied...

4. Jan 19, 2016

### Simon Bridge

Dot products of what?
How can a line be perpendicular to a curve?

5. Jan 19, 2016

### Krylov

I take it that you assume the curve is smooth. Even then, note that the closest point may not be unique. Also, when $I$ is not compact, a closest point is not guaranteed to exist, as $t \mapsto \|p - X(t)\|$ need not assume its infimum over $I$ then.

6. Jan 19, 2016

### RBG

Dot product of the tangent vector, right? So above $(p-X(t_0))$ is the vector between point and curve and $X'(t_0)$ is the tangent vector. But I don't see why should $p\dot X'(t_0)-X(t_0)X'(t_0)=0$

7. Jan 19, 2016

### Simon Bridge

... given a point on the curve, how would you tell that it is the closest point?
Given the curve C and a point P, how would you usually go about finding the closest point on C to P?

8. Jan 19, 2016

### RBG

We are assuming $X(t)$ is a regular parametrized curve and $t_0$ is not an endpoint of $I$.

9. Jan 19, 2016

### Simon Bridge

... why not just work through the expression that arises from the definition you just used?

10. Jan 19, 2016

### RBG

You would take the derivative of $||p-X(t)||$ and minimize it. Then check which points are minimal, right?

11. Jan 19, 2016

### Ray Vickson

That is how I would do it, but I would make the problem easier by minimizing $|| p - X(t)||^2$ instead of $||p - X(t)||$. These problems are equivalent, in the sense that their $t$-solutions are the same.

12. Jan 19, 2016

### RBG

I really don't understand what you mean by this. I should use the fact that I am minimizing $|p-X(t)||$ somehow to reduce the dot product?

13. Jan 19, 2016

### RBG

OOOOOOHHHHH... duh. Nevermind. Right... Thanks! Just do the calculation of taking the derivative