Arc length and straight lines(Need clarification please)

[Cauchy1789]

Homework Statement

Here is difficult one guys,

Lets imagine that an object movement along a curve is described by the parameterized function called

$$\omega: I \rightarrow \mathbb{R}^3$$ which moves on the interval $$[a,b]\subset I$$. and this depended on motor which supplies the constant effect of |v| = 1.

With this in mind show that

$$(\omega(b) - \omega(a)) \cdot v = \int_{a}^{b} \omega(t)' \cdot v dt \leq \int_{a}^{b} |\omega(t)'| dt$$

Homework Equations

The Attempt at a Solution

From what I learned in Calculus is relatively easy to show that according to the fundametal theorem of Calculus which states that $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$ where F is the anti-derivate of f.

such that $$(\omega(b) - \omega(a)) = \int_{a}^{b} \omega'(t) dt$$ where $$\omega(t)$$ is the anti-derivative, and since the movement depends of the constant, then both on side are the same aren't they?

If I expand the inequaliy then $$v \cdot(\omega(b) - \omega(a)) \leq |\omega(b) - \omega(a)|$$ which is only true if $$|v| \leq 1$$

Haven't I covered what needs to be covered in this?

Sincerely
Cauchy

 

[gabbagabbahey]

and since the movement depends of the constant, then both on side are the same aren't they?

Careful, just because $||\vec{v}||$ is a constant, doesn't mean that $\vec{v}$ is constant --- It can still point in different directions at different points along the curve. (I'm using \vec to make it clear which quantities are vectors!)

Hint (1): Use the product rule for $$\left(\vec{\omega}(t)\cdot\vec{v}(t))'$$
Hint (2): $$\vec{A} \cdot \vec{B}=||\vec{A}||*||\vec{B}||\cos\theta$$ , where $\theta$ is thew angle between the vectors.

 

[Cauchy1789]

Careful, just because $||\vec{v}||$ is a constant, doesn't mean that $\vec{v}$ is constant --- It can still point in different directions at different points along the curve. (I'm using \vec to make it clear which quantities are vectors!)

Hint (1): Use the product rule for $$\left(\vec{\omega}(t)\cdot\vec{v}(t))'$$
Hint (2): $$\vec{A} \cdot \vec{B}=||\vec{A}||*||\vec{B}||\cos\theta$$ , where $\theta$ is thew angle between the vectors.

Hi gab and thank you for your answer,

Then using hint(1) and hint (2)

$$\omega(t) \cdot v(t)' + \omega(t)' \cdot v(t) = \omega'(t) \cdot v(t) \leq |\omega'(t)|$$

Is this what you mean?

Sincerely
Cauchy

 

[gabbagabbahey]

Hi gab and thank you for your answer,

Then using hint(1) and hint (2)

$$\omega(t) \cdot v(t)' + \omega(t)' \cdot v(t) = \omega'(t) \cdot v(t) \leq |\omega'(t)|$$

Is this what you mean?

Sincerely
Cauchy

Careful, you need to justify why $$\vec{\omega}(t)\cdot\vec{v}(t)'=0$$...Hint: is $$\vec{v}(t)'$$ (a)tangent to the curve? (b)normal to the curve (c) some combination of the above?

 

[Cauchy1789]

Careful, you need to justify why $$\vec{\omega}(t)\cdot\vec{v}(t)'=0$$...Hint: is $$\vec{v}(t)'$$ (a)tangent to the curve? (b)normal to the curve (c) some combination of the above?

if $$\omega(t) \cdot v(t) = 0$$ then v(t) is perpendicular to the omega function? and thus the normal vector. Isn't it?

where $$v(t) \cdot(\omega(b)-\omega(a)) = 0$$

this is what you mean?

Sincerely

Cauchy

 

[gabbagabbahey]

if $$\omega(t) \cdot v(t) = 0$$ then v(t) is perpendicular to the omega function? and thus the normal vector. Isn't it?

You are missing the primes in this expression and you have the if statement backwards...you should have that:

If $$\vec{v}(t)'$$ is perpendicular to $$\vec{\omega}(t)$$ then $$\vec{\omega}(t) \cdot \vec{v}(t)' = 0$$

But why is $$\vec{v}(t)'$$ is perpendicular to $$\vec{\omega}(t)$$ in this case?

P.S. You really need to make it clear which quantities are vectors and which are scalars; your original post was very confusing.

 

[Cauchy1789]

You are missing the primes in this expression and you have the if statement backwards...you should have that:

If $$\vec{v}(t)'$$ is perpendicular to $$\vec{\omega}(t)$$ then $$\vec{\omega}(t) \cdot \vec{v}(t)' = 0$$

But why is $$\vec{v}(t)'$$ is perpendicular to $$\vec{\omega}(t)$$ in this case?

Hi again,

That can only be because v'(t) its the normal for the omega function?

So if I have understood this correctly then v'(t) is the normal vector for the omega function. Then reversed $$\omega(t)'$$ is the normal vector for $$v(t)$$ ? Because v is constant?

Sincerely
Cauchy

 

[gabbagabbahey]

Hi again,

That can only be because v'(t) its the normal for the omega function?

No. $$\vec{v}(t)$$ represents the velocity of the object correct?...So $$\vec{v}(t)'$$ represents___?

 

[Cauchy1789]

No. $$\vec{v}(t)$$ represents the velocity of the object correct?...So $$\vec{v}(t)'$$ represents___?

the speed.

oh then the right hand side of the equality sign is the function which defines the speed and the left hand the verlocity??

Sincerely Cauchy

 

[gabbagabbahey]

The time derivative of velocity is not the speed. I think you need to reread the chapter on vector calculus in your textbook.

 

[Cauchy1789]

The time derivative of velocity is not the speed. I think you need to reread the chapter on vector calculus in your textbook.

The time derivative is $$\frac{d\omega}{dt} = \omega(t)'$$ isn't it?


Sincerely

Cauchy

 

[gabbagabbahey]

The time derivative is $$\frac{d\omega}{dt} = \omega(t)'$$ isn't it?


Sincerely

Cauchy

That's the time derivative of omega...what's the time derivative of $$\vec{v}(t)$$?

 

[Cauchy1789]

That's the time derivative of omega...what's the time derivative of $$\vec{v}(t)$$?

That must be $$\frac{dv}{dt} = v'(t)$$

The first part of the original eqn:
$$(\omega(b) - \omega(a)) \cdot v(t) = \omega(b) \cdot v(t) - \omega(a) \cdot v(t)$$

But what I don't get here is that this then according should equal $$= \int_{a}^{b}( \omega(t) \cdot v(t))' dt = \omega(b) \cdot v(t) - \omega(a) \cdot v(t) = v(t) \cdot (\omega(b) - \omega(a))$$

because omega(t) is the anti-derivative of omega(t)'.

Isn't this pretty much it for equality part?

Next I end up with $$(\omega(b) - \omega(a)) \cdot v(t) = \omega(b) \cdot v(t) - \omega(a) \cdot v(t) = \int_{a}^{b}( \omega(t) \cdot v(t))' dt = \omega(b) \cdot v(t) - \omega(a) \cdot v(t) = v(t) \cdot (\omega(b) - \omega(a)) \leq |\omega(b) - \omega(a)|$$

But then this must mean that the left hand side of the inequlity is depending on the verlocity vector while the right hand is not? Or in which direction should I go here?

Sincerely
Cauchy

 

[gabbagabbahey]

That must be $$\frac{dv}{dt} = v'(t)$$

Sure, but what quantity does this represent? Position? Velocity? Acceleration? Speed? The number of Newfies it takes to screw in a light bulb?

But what I don't get here is that this then according should equal $$= \int_{a}^{b}( \omega(t) \cdot v(t))' dt = \omega(b) \cdot v(t) - \omega(a) \cdot v(t) = v(t) \cdot (\omega(b) - \omega(a))$$

Hold on, doesn't the fundamental theorem of calculus tell you that

$$\int_{a}^{b}( \vec{\omega}(t) \cdot \vec{v}(t))' dt = \vec{\omega}(b) \cdot \vec{v}(b) - \vec{\omega}(a) \cdot \vec{v}(a)$$

Something is clearly missing/unclear in your original post...For $\vec{\omega}(b) \cdot \vec{v}(b) - \vec{\omega}(a) \cdot \vec{v}(a)$ to be equal to $\left(\vec{\omega}(b)- \vec{\omega}(a)\right) \cdot \vec{v}(t)$, $\vec{v}(t)$ would have to be constant...but that wasn't what I inferred from your OP.

Are you told that $\vec{v}(t)$ is a constant or that $\vec{v}(a)=\vec{v}(b)$, or are you actually asked to show that

$$\vec{\omega}(b) \cdot \vec{v}(b) - \vec{\omega}(a) \cdot \vec{v}(a)=\int_{a}^{b}\vec{\omega}(t)' \cdot \vec{v}(t) dt \leq \int_a^b|\vec{\omega}(t)|dt$$

?

Your original post is very unclear...

 

[Cauchy1789]

Hi again,

I am very sorry if my op has been unclear.

I am told to show the later.

I will repost it again now.

Let $$\omega: I \rightarrow \mathbb{R}^3$$ be a parameterized curve. Let $$[a,b]\subset I$$ and set $$\omega(b) = s$$ and $$\omega(a) = t$$.

1) Show that, for any constant vector v, |v| = 1,

$$(t-s) \cdot v(t)= \int_{a}^{b} \omega(t)' \cdot v(t) dt \leq \int_{a}^{b}|\omega(t)'| dt$$

2) Set $$v = \frac{t-s}{|t-s|}$$

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

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

Solution part(1)

$$(s-t) \cdot v(t) = \int_{a}^{b} \omega'(t) \cdot v(t) \ dt$$

Because by the fundamental theorem of Calculus.

$$\int_{a}^{b} \omega'(t) \cdot v(t) \ dt = \int_{a}^{b} (\omega(t) \cdot v(t))' dt = \omega(b) \cdot v(b) - \omega(a) \cdot v(a)$$ and since as |v| = 1 is constant vector v, then the lefthand side and the righthand side of the equality equal each other.

But does that apply for the inequality as well? Thats what I am not sure of.

Sincerely
Cauchy

 

[gabbagabbahey]

Okay...a constant $$\vec{v}(t)$$ makes much more sense! That means that $$\vec{v}(t)'=0$$ and hence

$$\left(\vec{\omega}(t)\cdot\vec{v}(t)\right)'= \vec{\omega}(t)'\cdot\vec{v}(t)+\vec{\omega}(t)\cdot\vec{v}(t)'=\vec{\omega}(t)'\cdot\vec{v}(t)+\vec{\omega}(t)\cdot(0)=\vec{\omega}(t)'\cdot\vec{v}(t)$$

Now for the inequality part just use the fact that $$\vec{A} \cdot \vec{B}=||\vec{A}||*||\vec{B}||\cos\theta$$

$$\implies \vec{\omega}(t)'\cdot\vec{v}(t)=\left(||\vec{\omega}(t)'||*||\vec{v}(t)\right)||\cos\theta=||\vec{\omega}(t)'||*(1)\cos\theta=||\vec{\omega}(t)'||\cos\theta$$

theta is the angle between omega'(t) and v, which is always between 0 and pi right? what does that make the min and max values of cos(theta)?

 

[Cauchy1789]

Okay...a constant $$\vec{v}(t)$$ makes much more sense!


That means that $$\vec{v}(t)'=0$$ and hence

$$\left(\vec{\omega}(t)\cdot\vec{v}(t)\right)'= \vec{\omega}(t)'\cdot\vec{v}(t)+\vec{\omega}(t)\cdot\vec{v}(t)'=\vec{\omega}(t)'\cdot\vec{v}(t)+\vec{\omega}(t)\cdot(0)=\vec{\omega}(t)'\cdot\vec{v}(t)$$

Now for the inequality part just use the fact that $$\vec{A} \cdot \vec{B}=||\vec{A}||*||\vec{B}||\cos\theta$$

$$\implies \vec{\omega}(t)'\cdot\vec{v}(t)=\left(||\vec{\omega}(t)'||*||\vec{v}(t)\right)||\cos\theta=||\vec{\omega}(t)'||*(1)\cos\theta=||\vec{\omega}(t)'||\cos\theta$$

theta is the angle between omega'(t) and v, which is always between 0 and pi right? what does that make the min and max values of cos(theta)?

hi again and thanks for taking time :)

cos has max in cos(0) = 1 and min in cos(pi) = -1.

and this basicly proves that the inequality is true?

Sincerely Cauchy

 

[gabbagabbahey]

Opps...the max angle between two vector is pi/2 not pi, so cos\theta has min of 0 and max of 1 ...therefor

$$\cos\theta\leq1$$

So, $||\vec{\omega}(t)'||\cos \theta\leq$ ____?

 

[Cauchy1789]

Opps...the max angle between two vector is pi/2 not pi, so cos\theta has min of 0 and max of 1 ...therefor

$$\cos\theta\leq1$$

So, $||\vec{\omega}(t)'||\cos \theta\leq$ ____?

That must be then

$||\vec{\omega}(t)'||\cos \theta\leq ||\vec{\omega}(t)'||$

please excuse me if I am misunderstanding you but cos(theta) <= 1 shows then that inequality is true if the vector v is constant?

Cauchy

 

[gabbagabbahey]

That must be then

$||\vec{\omega}(t)'||\cos \theta\leq ||\vec{\omega}(t)'||$

Right, so $\vec{\omega}(t)'\cdot\vec{v}(t)\leq$___?

 

[Cauchy1789]

Right, so $\vec{\omega}(t)'\cdot\vec{v}(t)\leq$___?

Hi

That must be $\vec{\omega}(t)'\cdot\vec{v}(t)\leq \|\vec{w(t)'}\|$ ?

Cauchy

 

[gabbagabbahey]

Hi

That must be $\vec{\omega}(t)'\cdot\vec{v}(t)\leq \|\vec{w(t)'}\|$ ?

Cauchy

Well, in post #16 I showed $\vec{\omega}(t)'\cdot\vec{v}(t)=||\vec{\omega}(t)'||\cos\theta$

And in post #19 You showed that $||\vec{\omega}(t)'||\cos \theta\leq ||\vec{\omega}(t)'||$

So yes, when you put those two relations together you get $\vec{\omega}(t)'\cdot\vec{v}(t)\leq||\vec{\omega}(t)'||$

So, now what can you say about $$\int_a^b \vec{\omega}(t)'\cdot\vec{v}(t)dt$$ ?

 

[Cauchy1789]

Well, in post #16 I showed $\vec{\omega}(t)'\cdot\vec{v}(t)=||\vec{\omega}(t)'||\cos\theta$

And in post #19 You showed that $||\vec{\omega}(t)'||\cos \theta\leq ||\vec{\omega}(t)'||$

So yes, when you put those two relations together you get $\vec{\omega}(t)'\cdot\vec{v}(t)\leq||\vec{\omega}(t)'||$

So, now what can you say about $$\int_a^b \vec{\omega}(t)'\cdot\vec{v}(t)dt$$ ?

Thats its

that $$\vec{\omega}(b)\cdot\vec{v}(b) - \vec{\omega}(a)\cdot\vec{v}(a)$$ ?

Cauchy

 

[gabbagabbahey]

No.

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

 

[Cauchy1789]

No.

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

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

 

[gabbagabbahey]

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?

 

[Cauchy1789]

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]

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)

 

[Cauchy1789]

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]

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

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

 

[Cauchy1789]

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