Calc III: Arc Length

**saxyliz** · September 27th 2008, 05:04 PM

Find the arclength of the curve:
$r(t) = <6t^2, 2sqrt(6)t, ln(t)>$ when
t is less than or equal to 2, but greater than or equal to 1.

I know that the arc length formula is the integral of the magnitude of the derivative of the function r, but I don't know how to integrate it to find the arc length.

Thank you!

**Chris L T521** · September 27th 2008, 05:35 PM

Originally Posted by saxyliz

Find the arclength of the curve:
$r(t) = <6t^2, 2sqrt(6)t, ln(t)>$ when
t is less than or equal to 2, but greater than or equal to 1.

I know that the arc length formula is the integral of the magnitude of the derivative of the function r, but I don't know how to integrate it to find the arc length.

Thank you!

$r(t) = \left<6t^2, 2\sqrt 6 t, \ln(t)\right>$

therefore, $x(t)=6t^2$ , $y(t)=2\sqrt 6 t$ and $z(t)=\ln t$

Thus, $\frac{\,dx}{\,dt}=12t$ , $\frac{\,dy}{\,dt}=2\sqrt 6$ and $\frac{\,dz}{\,dt}=\frac{1}{t}$

Thus, $L=\int_1^2\sqrt{144t^2+24+\frac{1}{t^2}}\,dt$

Note that $144t^2+24+\frac{1}{t^2}=144t^2\cdot{\color{red}\fr ac{t^2}{t^2}}+24\cdot{\color{red}\frac{t^2}{t^2}}+ \frac{1}{t^2}=\frac{144t^4+24t^2+1}{t^2}=\frac{(12 t^2+1)^2}{t^2}$

Thus, $L=\int_1^2\sqrt{\frac{(12t^2+1)^2}{t^2}}\,dt=\int_ 1^2\left(12t+\frac{1}{t}\right)\,dt$

Can you take it from here?

--Chris

**saxyliz** · September 27th 2008, 06:13 PM

Wow! Thank you so much! I got to that integral, but I had no idea how to solve it, I found the answer from there.

Thank you again!

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