Straight lines as shortest connections

[t_n_p]

Homework Statement

http://img193.imageshack.us/img193/1066/65157866.png

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!

 

[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 [tex] \gamma . v [/tex] this gives

[tex] \frac{d}{dt} (\gamma . v) = \frac{d \gamma}{dt}.v + \gamma.\frac{dv}{dt} = \frac{d \gamma}{dt}.v +0 [/tex]

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

 

[t_n_p]

thanks for the reply, the product rule makes a lot 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

 

[D H]

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

How? I'm quite confused by this part

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.

 

[t_n_p]

Ah, you mean like [tex]\vec{PQ}[/tex]?

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?

 

[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.

 

[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

 

[lanedance]

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

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|

 

[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

 

[lanedance]

ok, so to double check..

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

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?

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

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

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

ok

-----------------------------
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

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

 

[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!

 

[t_n_p]

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

which essentially leads me to:

(PQ)²/|PQ| <= ∫ |dγ(t)/dt| dt for a-> b

confused!