# Arc Length Parameterization

• Oct 4th 2009, 08:26 AM
TheUnfocusedOne
Arc Length Parameterization
Well here I am again, I'm completely lost on how to do this problem.

it is as followed:
Find an arc length parametrization of r(t) = < t2, t3 >.

The answer is in the form R(s)=< , >

Ive been following my instructions in the book, but this problem is must tricker than the book version.

First i found

$\displaystyle ||r'(t)|| = t sqrt(4+9t)$

I'm being told to then integrate this to find the arc length function, which is bound over the limits of 0->t

$\displaystyle s(t) = 1/27 (4+9t^2)^3/2 + 8$

changing this function to be g(s) we get this horrific equation of

$\displaystyle sqrt(((27s-8)^2/3)-4)/9)$

then you plug that into the equation up top, and it fails

what do i do?
• Oct 4th 2009, 09:17 AM
Jester
Quote:

Originally Posted by TheUnfocusedOne
Well here I am again, I'm completely lost on how to do this problem.

it is as followed:
Find an arc length parametrization of r(t) = < t2, t3 >.

The answer is in the form R(s)=< , >

Ive been following my instructions in the book, but this problem is must tricker than the book version.

First i found

$\displaystyle ||r'(t)|| = t sqrt(4+9t)$

I'm being told to then integrate this to find the arc length function, which is bound over the limits of 0->t

$\displaystyle s(t) = 1/27 (4+9t^2)^3/2 + 8$

changing this function to be g(s) we get this horrific equation of

$\displaystyle sqrt(((27s-8)^2/3)-4)/9)$

then you plug that into the equation up top, and it fails

what do i do?

I don't see a problem except for your latex and a sign (-8)
$\displaystyle s(t) = \int_0^t u\sqrt{9u^2+4}\,du = \frac{\left(9t^2+4\right)^{3/2} - 8}{27}$

and solving for t gives

$\displaystyle t = \sqrt{\frac{\left(27s+8\right)^{2/3} - 4}{9}}$
• Oct 4th 2009, 10:51 AM
TheUnfocusedOne
hmm...
i think there must be something wrong with the computer then

can anyone second this?

nm, it worked

thanks!