The arclength of a parametrized segment (integration).

[Yami]

Homework Statement

For two points p and q in ℝ^n, use the formula (20.3) to check that the arclength of the parametrized segment from p to q is ||p - q||.

Homework Equations

Formula (20.3):
A smooth parametrized path $$\gamma: [a, b]→ℝ^n$$is rectifiable, and its arclength l is given by
$$l = \int_{a}^{b}||\gamma '||$$


Norm of a point x = (x_1, ... , x_n) in ℝ^n is defined in this book as
$$||x|| = \sqrt{x_1^2 + ... + x_n^2}$$

The Attempt at a Solution

$$\gamma: [q, p]→ℝ^n$$
is defined as
$$\gamma (t) = (\gamma _1 (t), \gamma _2 (t), ... , \gamma _n (t))$$
$$t \in [q, p]$$
then
$$\gamma '(t) = (\gamma _1 '(t), ... , \gamma _n '(t))$$
$$||\gamma '(t)|| = \sqrt{(\gamma _1 '(t))^2+ ... + (\gamma _n '(t))^2}$$

$$l = \int_{q}^{p} \sqrt{(\gamma _1 '(t))^2+ ... + (\gamma _n '(t))^2} dt$$

I can't figure out how to integrate this though to get to ||p - q||.

 

[LCKurtz]

You need to write out the actual parameterization. Start with $p=(p_1,p_2,...,p_n)$ and $q = (q_1,q_2,...q_n)$. Write out the parameterization of the line pq and work out the integral.

 

[Yami]

I just realized p and q are in $ℝ^n$. So [p, q] isn't an interval I can integrate over.
I just found the equation I should probably use: $\gamma:[0,1]→ℝ^n$ defined as $\gamma (t) = tq + 1(1 - t)p$ for $t \in [0, 1]$.
Thanks for the hint.

 

[LCKurtz]

I just realized p and q are in $ℝ^n$. So [p, q] isn't an interval I can integrate over.
I just found the equation I should probably use: $\gamma:[0,1]→ℝ^n$ defined as $\gamma (t) = tq + 1(1 - t)p$ for $t \in [0, 1]$.
Thanks for the hint.

Good start. Now calculate $|\gamma'(t)|$ and work out the $t$ integral. You can do it by components or work at this level.