# Arc Length

3.b Find the arc length of the given curve

α(t) = (t, 1, (1/6)t^3 + (1/2)t^-1) from t = 1 to t = 3.

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?

5. Let α(t) = (e^t, e^-t, root2*t). Calculate first derivative, norm, and re-parametrize alpha by its arc length, starting at t = 0.

α'(t) = (e^t, -e^-t, root2)

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

## Answers and Replies

LCKurtz
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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.

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?

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.

α'(t) = (e^t, -e^-t, root2)

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

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

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##.

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

Last edited:
For the arc length I got

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

LCKurtz
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For the arc length I got

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

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##.

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##.

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.

LCKurtz
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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.

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

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

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?

LCKurtz
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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?

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.