### Arc Length of a Curve - Reparametrization a Curve (Example 2)

Reparametrize the curve r(t) with respect to arc length, starting from t = 0 to a direction of increasing t, where $$r(t) = 2ti + (1 - 3t)j + (5 + 4t)k$$.

Arc length is given by:

$s(t) = L = \int_0^t {\left| {r'(u)} \right|} du$

Note that the equation above is for r(t), but you're not technically allowed to have the same variable for r as what is in the integral boundaries (t). Even though it says r(u), it really is just r(t).

We start by finding the derivative of r(t).

$r'(t) = 2i - 3j + 4k$

Then substitute it into the arc length formula.

$s(t) = \int_0^t {\sqrt {{{(2)}^2} + {{( - 3)}^2} + {{(4)}^2}} du}$

$s(t) = \sqrt {29} (t - 0)$

$s(t) = \sqrt {29} t$

Rearrange in terms of t, and substitute it into r(t). That's just what arc length parametrization is.

$t = \frac{s}{{\sqrt {29} }}$

And our original r(t) is:

$r(t) = 2ti + (1 - 3t)j + (5 + 4t)k$

Substituting in we obtain:

$r\left( {t(s)} \right) = \frac{{2s}}{{\sqrt {29} }}i + \left( {1 - \frac{{3s}}{{\sqrt {29} }}} \right)j + \left( {5 + \frac{{4s}}{{\sqrt {29} }}} \right)k$

Done! Sometimes the final substitution gets messy arithmetically.