r'(t) = <t^2, 2t,0> |r'(t)| = sqrt( t^4 + 4t^2) Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it?
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Originally Posted by linalg123 r'(t) = <t^2, 2t,0> |r'(t)| = sqrt( t^4 + 4t^2) Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it? Integral , click "show steps"
$\displaystyle \int_0^2 \sqrt{t^4+4t^2}dt = \int_0^2 t \sqrt{t^2+4} dt$ make the substitution $\displaystyle u = t^2 + 4 $ and the integral should be easy.
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