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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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    yeah. Typo again sorry2012-09-30

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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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    @JavierBadia: Oh yes, I should mention that. Thanks for pointing it out.2012-09-30