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$$r(t) = \langle 2t, t^2, \ln t \rangle.$$

I know that to find arclength you do $$L = \int_a^b \|r'(t)\| ~dt.$$

I found $r'(t)$ to be $r'(t) = \left\langle 2, 2t, \dfrac{1}{t} \right\rangle$.

To find $\|r'(t)\|$ I did

$$\sqrt{(2)^2 + (2t)^2 + \dfrac{1}{t^2}}$$

but how do I integrate that?

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Isn't the term that you want to integrate just $|2t+t^{-1}|$? And the fact that $\ln t$ is defined means that $t$ must be positive; so you can lose the $| |$ signs. – user22805 Sep 30 '12 at 0:57
I'm not sure what you mean. – 40Plot Sep 30 '12 at 1:02
oh! my bad. I fixed it now. There was a typo in the question: there were 3 components. – 40Plot Sep 30 '12 at 1:05
I mean, isn't $\sqrt{4+4t^2+t^{-2}}$ exactly the same thing as $|2t+t^{-1}|$. Try squaring the latter; you'll see. – user22805 Sep 30 '12 at 1:05
$r'(t)$ should be $\left\langle 2, 2t, \dfrac{1}{t} \right\rangle$ – Paul Sep 30 '12 at 1:07
up vote 2 down vote accepted

As you have shown, $$\|r'(t)\|=\sqrt{4t^2+4+\frac{1}{t^2}} =\sqrt{\left(2t+\frac{1}{t}\right)^2}=2t+\frac{1}{t}.$$ Therefore, the arclength is given by $$L=\int_a^b \|r'(t)\| ~dt=\int_a^b\left(2t+\frac{1}{t}\right)dt =(t^2+\log t)\Big|_a^b=b^2-a^2+\log b-\log a.$$

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You should mention that you can drop the absolute value because $t$ has to be positive. – Javier Sep 30 '12 at 1:37
@JavierBadia: Oh yes, I should mention that. Thanks for pointing it out. – Paul Sep 30 '12 at 23:56

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