# Parameterized curve vs. Integral(Please look at my result)

1. Feb 10, 2009

### Superboy1234

1. The problem statement, all variables and given/known data

Hello just found this forum and have question if I have done my homework correctly

Anyway let $$\theta: I \to \mathbb{R}^3$$ be a parameterized curve.
Next assume that $$[a,b] \subset I$$ and put $$\theta(a) = l$$(1) and $$\theta(b) = m$$

Then show for any constant vector $$\vec{u}$$,$$|\vec{u}| = 1$$.

that the inequality

$$(m-l) \cdot \vec{u} = \int_{a}^{b} \vec{\theta(t)} \cdot \vec{u(t)} \ dt \leq \int_{a}^{b} |\vec{\theta'(t)}| dt$$(1)

3. The attempt at a solution

Solution:
If I look at the dot-product of $$\frac{d}{dt}(\vec{\theta} \cdot \vec{u}}) = \vec{\theta(t)^{\prime}} \cdot \vec{u(t)}$$

Next by Fundamental theorem of Calculus leftside of (1) is true.

The inequality however is a bit tricky for me. I seem to remember an inequality
which states that the dot product of two vectors a and b can be written as
$$\vec{a} \cdot \vec{b} \leq |\vec{a}| \cdot |\vec{b}|$$ which transfered to Our situation gives
$$\vec{\theta'(t)} \cdot \vec{u(t)} \leq |\vec{\theta'(t)}| \cdot |\vec{u(t)}|$$ and since $$|\vec{u}| = 1$$, then this is equivalent to $$\theta'(t) \cdot \vec{u(t)} \leq |\theta'(t)|$$ and finally by the rules regarding our Riemann Integrals $$\int_{a}^{b} f \leq \int_{a}^{b} g$$, then the integrals $$\int_{a}^{b} \ldots dt \leq \int_{a}^{b} \ldots \ dt$$ is also true.

There is a second part as well

If I set $$\vec{u} = \frac{l-m}{|l-m|}$$

Then show that $$|\theta(b) - \theta(a)| \leq \int_{a}^{b} |\vec{\theta(t)^{\prime}| dt$$

If I insert $$\vec{u}$$ into lefthand side of the inequality I get $$(l-m) \cdot \frac{l-m}{|l-m|} = |l-m|$$ since $$(l-m) \cdot (l-m) = |l-m|^2$$

From this it should follow that the length of the curve from l to m is a straight line joining these points. But how exactly does it do that?
Is it something to do with if I on my curve draw a straight line from l to m and use this line as a side in a triangle and by using pythagoras then claim that line previously mentioned always will be shorter than the arc length from l to m?

Cheers Superboy

Last edited: Feb 10, 2009