# Arc length and straight lines(Need clarification please!)

gabbagabbahey
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Then thinking about it it must be:

$\int_a^b \vec{\omega}(t)'\cdot\vec{v}(t)dt \leq \int_a^b ||\vec{\omega}(t)'|| dt$??

Sincerely Cauchy

Yes, exactly!

And that's exactly what you were trying to show isn't it?

Yes, exactly!

And that's exactly what you were trying to show isn't it?

I surely is.

So just to recap then proving part(1) I partion the equation into vector equtions, and then use the fact about th constant vector.

Thank You.

I hope I don't sound completely stupid, been away from the studies for 6 month do a irregular heartrytm, but that a longer story. I can see that I need to re-read some stuff.

Anyway part(2) Do You have an idear howto get started here?

From I can figure that v is presented a united vector composed by the components of omega. What need to show here is that the dot-product between omega prime and the distance from omega(b) to omega(a) is less or equal to each other??

Sincerely
Cauchy

gabbagabbahey
Homework Helper
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2) Set $$\vec{v} = \frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|}$$

and show $$|\vec{\omega(b)}-\vec{\omega(a)}| \leq \int_{a}^{b} ||\vec{\omega}(t)'||dt$$

that is that the shortest length between omega(a) and omega(b) is a straight line.

I've added vector arrows to this to make it more clear.....do exactly what they tell you to.

Set $$\vec{v} = \frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|}$$ and plug it into your result from (1)

I've added vector arrows to this to make it more clear.....do exactly what they tell you to.

Set $$\vec{v} = \frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|}$$ and plug it into your result from (1)

Hi again You mean

$\frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} \cdot (t-s)= \int_a^b \vec{\omega}(t)'\cdot \frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} dt \leq \int_a^b ||\vec{\omega}(t)'|| dt$??

and then just expand??

Cauchy

gabbagabbahey
Homework Helper
Gold Member
Hi again You mean

$\frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} \cdot (t-s)= \int_a^b \vec{\omega}(t)'\cdot \frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} dt \leq \int_a^b ||\vec{\omega}(t)'|| dt$??

and then just expand??

Cauchy

Sure, but all you really need is

$$\frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} \cdot (\vec{t}-\vec{s}) \leq \int_a^b ||\vec{\omega}(t)'|| dt$$

Sure, but all you really need is

$$\frac{\vec{t}-\vec{s}}{|\vec{t}-\vec{s}|} \cdot (\vec{t}-\vec{s}) \leq \int_a^b ||\vec{\omega}(t)'|| dt$$

Hi again and one time more thank you for your reply,

What I am going to end up here just to clarify is that some of the terms on the lefthand side of inequality will eat each other and I will end up with a formula which resembles the arc-length formula?

Sincrely

Cauchy