Arc Length

1. Sep 3, 2014

Shackleford

Of course, I need to find the first derivative and integrate its norm.

α'(t) = (1, 0, (1/2)t^2 - (1/2)t^-2)

∫ [1 + (1/4)t^4 + (1/4)t^-4]^(1/2) dt, t = 1 to t = 3.

Have I simply forgotten useful integrals?

α'(t) = (e^t, -e^-t, root2)

∫ [e^2u + e^-2u + 2]^(1/2) du, u = 0 to u = t.

2. Sep 3, 2014

LCKurtz

With over 1500 posts to this group, I would say it is time you learned to type simple equations like yours in Latex to make them more readable for everyone.

Apparently you have forgotten algebra too. $(a-b)^2\ne a^2+b^2$. Once you calculate the norm correctly, use algebra to make a perfect square under the square root.

Again, you might work some algebra on the square root. Arc length is usually denoted with $s$.

3. Sep 3, 2014

Shackleford

Eh. You're right. 3.b is not a mystery anymore. I'll see what I can do with 5.

Last edited: Sep 3, 2014
4. Sep 4, 2014

Shackleford

For the arc length I got

s(t) = $e^t - e^{-t}$

5. Sep 4, 2014

LCKurtz

You can include the whole equation in itex brackets so it looks like this:

$s(t) = e^t - e^{-t}$

Your problem asked for the arc length from between $t=1$ to $t=3$. I think what you have calculated is the arc length measuring from $0$ to $t$.

6. Sep 4, 2014

Shackleford

Thanks for the tip.

That's right. The second problem wants me to parametrize the curve in terms of arc length. In this case, I need to find t(s) to get $α(t(s))$.

I suppose I simply need to take the logarithm of both sides and go from there.

7. Sep 4, 2014

LCKurtz

Or, if you are comfortable with hyperbolic functions, you might notice$$s(t)=2\sinh(t)$$and use the inverse hyperbolic sine function.

8. Sep 4, 2014

Shackleford

I haven't dealt much with hyperbolic functions, but I did get

$t(s) = sinh^{-1}\frac{s}{2}$.

Why is the "s" so tiny?

9. Sep 4, 2014

LCKurtz

Maybe because the itex tag is for inline tex and it's trying to conserve space. If you put it on its own line with tex tags instead it looks like this$$t(s) = sinh^{-1}\frac{s}{2}$$Then again, maybe not. It still looks small compared to the $2$. Something to do with the default fonts I suppose.