Shortest distance between two points

[brotherbobby]

TL;DR Summary

I am trying to convince myself that the shortest distance between two points on a plane is indeed along a straight line and not the distance along a pair of lines that join them via a third intermediary point as shown below.

I copy and paste the diagram I drew for the problem to the right. The two points in question are A and B. The vertical from B is $d$ units long and is $s$ units away from A.

What is the shortest way to go from A to B?

We expect it to be a distance of (say) $y=\sqrt{s^2+d^2}$.

Thus, if we travel from A to B via an intermediary point C at a distance $x$ from A and extremise the path $\text{AC}\rightarrow \text{CB}$, we should obtain $\boxed{\boldsymbol{x=0}}$ for shortest path.

Let's see.



The distance $\small{\text{AC}+\text{CB}=y=x+\sqrt{d^2+(s-x)^2}\Rightarrow \dfrac{dy}{dx}= 1-\dfrac{s-x}{\sqrt{d^2+(s-x)^2}}}$.

For minimum (extremum, more generally), $\small{\dfrac{dy}{dx}=0\Rightarrow \dfrac{s-x}{\sqrt{d^2+(s-x)^2}}=1\Rightarrow (s-x)^2=d^2+(s-x)^2}$.

The last expression is not true for all $x$. I was hoping I'd get an expression that would say $x=0$.

Where am I going wrong?

 

[Orodruin]

Your expression for the distance is incorrect. It should be:
$$
y = |x| + \sqrt{d^2 + (s-x)^2}
$$
This is not differentiable at $x = 0$. Here is a plot of $y(x)$ for $d = 2$ and $s = 1$:

 

[Orodruin]

Apart from that, there is the triangle inequality, which is a basic result in geometry:
$$
A + B \geq C
$$
which is true for any sides of a triangle.

Edit: This is easy to show in any inner product space. Given an inner product and vectors $\vec v = \vec u + \vec w$, it holds that
$$
v^2 = \vec v^2 = \vec u^2 + \vec w^2 + 2\vec u \cdot \vec w \leq u^2 + v^2 + 2uv = (u+v)^2
$$

 

[brotherbobby]

our expression for the distance is incorrect. It should be:
y=|x|+d2+(s−x)2

I didn't understand the $|x|$ bit.

(I am aware that it is not differentiable at $x=0$)

 

[brotherbobby]

You meant to say $A+B\ge C$, right?

 

[Baluncore]

The first law of the butterfly, states simply that, the shortest distance between two points is a zigzag line.

 

[Orodruin]

I didn't understand the $|x|$ bit.

(I am aware that it is not differentiable at $x=0$)

Your expression says that if $x$ is negative, then the line segment $AC$ has negative length. The actual length of the line segment $AC$ is the absolute value of $x$ (or $\sqrt{x^2}$ if you will, but that's the same for real numbers).

View attachment 345212
You meant to say $A+B\ge C$, right?

Yes, I thought I edited that out before anyone saw it ... brain jumped to Pythagoras' theorem for some reason. (Also, using the quote feature is more efficient than screen shots)

 

[sophiecentaur]

The first law of the butterfly, states simply that, the shortest distance between two points is a zigzag line.

The crow has more sense.

 

[brotherbobby]

Your expression says that if x is negative, then the line segment AC has negative length. The actual length of the line segment AC is the absolute value of x (or x2 if you will, but that's the same for real numbers).

Yes, pardon me. So it's $|x|$, and yes that invalidates my procedure even if I had put the $|x|$ correctly owing to the lack of differentiability.

Also, using the quote feature is more efficient than screen shots

I could never use the "Quote" feature here. I copy the statement and select "Reply" instead. However, copying is poor with mathematics. Let me show you.

Edit: This is easy to show in any inner product space. Given an inner product and vectors v→=u→+w→, it holds that

An example of how it looks. You, being the writer of the statement, can barely understand what's written. Not to mention that the second line which was entirely mathematicsl did not even appear.

 

[brotherbobby]

Edit: This is easy to show in any inner product space. Given an inner product and vectors v→=u→+w→, it holds that

Ok, still no luck with "Insert Quotes". The inline mathematics is garbled. The displaymath line does not appear.
A screenshot seems the only option for now.

 

[Orodruin]

I could never use the "Quote" feature here. I copy the statement and select "Reply" instead. However, copying is poor with mathematics. Let me show you.

Use the ”Reply” feature and delete any part you don’t want to quote.

Like so:

Yes, pardon me. So it's $|x|$, and yes that invalidates my procedure even if I had put the $|x|$ correctly owing to the lack of differentiability.

Not invalidate. It is still important to conclude that the function has no local extrema where differentiable. Once you know that, check the point where it is not differentiable.

 

[brotherbobby]

Apart from that, there is the triangle inequality, which is a basic result in geometry:
$$
A + B \geq C
$$
which is true for any sides of a triangle.

Edit: This is easy to show in any inner product space. Given an inner product and vectors $\vec v = \vec u + \vec w$, it holds that
$$
v^2 = \vec v^2 = \vec u^2 + \vec w^2 + 2\vec u \cdot \vec w \leq u^2 + v^2 + 2uv = (u+v)^2
$$

Let's see how this works without me deleting anything. Only by clicking on Post Reply am I going to find out.

Yes looks alright (edit).