# Straight lines as shortest connections

1. Aug 19, 2009

### t_n_p

1. The problem statement, all variables and given/known data
http://img193.imageshack.us/img193/1066/65157866.png [Broken]

3. The attempt at a solution
Since v is a constant vector, I took it out the front of the integral and used the ftc so the integral of the derivative collapses to γ(t) with terminals a to b. Evaluating leads to γ(b) - γ(a), which simplies to v[Q-P] (using the info provided), notice how this is different to the left hand side of the in/equality (PQ)v.

Confused!

Last edited by a moderator: May 4, 2017
2. Aug 19, 2009

### lanedance

hi tnp

that looks alright to me, and though it works, you need to be a bit careful about interchanging limit operations and changing the integral from a scalar integral to a vector integral

Though it amounts to the same thing, i think a safer way would be to evaluate the derivative of the scalar $$\gamma . v$$ this gives

$$\frac{d}{dt} (\gamma . v) = \frac{d \gamma}{dt}.v + \gamma.\frac{dv}{dt} = \frac{d \gamma}{dt}.v +0$$

then use that in your integral expression with an FTC

I would also read PQ = Q - P so you're on the right track (i would think of the arrow from P to Q)

aopolgies if the latex is messy, still doesn't display right on my computer

3. Aug 19, 2009

### t_n_p

thanks for the reply, the product rule makes alot more sense that what I was doing.

I got down to this...

∫ (dγ(t)/dt · v) dt with limits a to b.

Can I take v out the front as a constant, then use FTC?
That's what I reckon!
then I end up with.......
= v [Q-P]

You also said:
"I would also read PQ = Q - P"

How? I'm quite confused by this part

4. Aug 19, 2009

### D H

Staff Emeritus
Thinking of PQ as a product doesn't make sense in this context because P and Q are points in three space. Think of PQ as the displacement vector from point P to point Q. If you describe P and Q themselves in terms of displacement vectors P and Q from the origin, then PQ is Q-P.

5. Aug 19, 2009

### t_n_p

Ah, you mean like $$\vec{PQ}$$?

Makes sense now!

so that's the LHS done, now to show the RHS inequality...

If I take absolute value of everything, then I end up with..

∫ |dγ(t)/dt| · |v| dt with limits a -> b

But I'm given that |v|=1 in the question, which breaks down to:

∫ |dγ(t)/dt| dt with limits a -> b
= |γ(b)-γ(a)|
= |Q-P|

and since there is no vector v out the front, it is smaller?

Ideas?

Last edited: Aug 19, 2009
6. Aug 19, 2009

### lanedance

PQ = Q-P is a vector so be careful, negative as a description doesn't really apply to it

(Q-P).v is a scalar so can be negative

Also ∫ |dγ(t)/dt| dt with limits a -> b does not equal |γ(b)-γ(a)|

All you need to do to finish the inequality is to concentrate on the middle & right hand side. Note that in comparing 2 integrals with same integraion limts, if you can show one integrand is always less than the other, then the value of integral itself must also be less than the other.

7. Aug 19, 2009

### t_n_p

ah, so this is what I have..

Since |dγ(t)/dt| <= dγ(t)/dt ALWAYS,

then ∫ |dγ(t)/dt| dt with limits a -> b <= ∫ dγ(t)/dt dt with limits a -> b

seems simple enough, thanks for the help guys

8. Aug 19, 2009

### lanedance

not quite,

First you can only compare scalars with <=, so you need to keep the dot product in your equality, at least on one side (dγ(t)/dt is a vector)

Also check the direction of you equalities above. If x is any scalar, then by definition x<=|x|

9. Aug 19, 2009

### t_n_p

ok, so to double check..

∫ |dγ(t)/dt · v| dt with limits a -> b
=∫ |dγ(t)/dt| · |v| dt with limits a -> b

Now since |dγ(t)/dt| · |v| = |dγ(t)/dt| is always <= dγ(t)/dt · v,

then it holds that ∫ dγ(t)/dt · v dt <= ∫ |dγ(t)/dt| dt for a-> b

-----------------------------
But I have a question..
At this line here: Now since |dγ(t)/dt| · |v| = |dγ(t)/dt| is always <= dγ(t)/dt · v

What if dγ(t)/dt is positive, then by dotting it with some constant vector v (say it is also positive), isn't it possible that dγ(t)/dt · v will be greater than |dγ(t)/dt| by a factor of v? I hope that makes sense

10. Aug 19, 2009

### lanedance

this is not true, considering only the integrands, the equality should be
|dγ(t)/dt · v| <= |dγ(t)/dt||v|

to help see this, consider the case when dγ(t)/dt and v, are perpindicular and both non zero, what is the dot product?
the equals sign is correct, based on the fact that |v| = 1

but as above the <= isn't true here, it should reversed. Consider a case where the dot product returns a negative value
ok
rememeber that |v| = 1 is defined in the question, the maximum value dγ(t)/dt · v can have is |dγ(t)/dt|.

This will occur only in the case dγ(t)/dt and v are parallel pointing the same direction

11. Aug 19, 2009

### t_n_p

the dot product when they are perp will be zero, just a silly lapse by me. and the inequality was another silly mistake, since in the question the absolute value was on the right, then I moved it to the left of the inequality without reversing the sign.

everything makes sense now.

There is a second part that reads:

set v = PQ/|PQ| and hence show |γ(b)-γ(a)| <= ∫ |dγ(t)/dt| dt for a-> b (i.e the curve of shortest length from P to Q is a straight line)

If I do similar to what I did before, up to:

v[γ(b)-γ(a)] <= ∫ |dγ(t)/dt| dt for a-> b, what should I do next?

I'm stumped, because in the question they want to show|γ(b)-γ(a)| on the LHS!

Last edited: Aug 19, 2009
12. Aug 20, 2009

### t_n_p

Say now I change v to PQ/|PQ|,