What is the arclength of $(t^{2} - t - 1, 9 t^{2} - \frac{2}{t})$ $(t^2-t-1,9t^2-2/t)$ on $t \in [1, 4]$ $t in [1,4]$ ?

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Answer 1 · 2018-07-17T14:56:31

$\approx 137.033$ $approx 137.033$

Let $x (t) = t^{2} - t - 1$ $x(t)=t^2-t-1$ then

$x'(t)=2t-1$

$y(t)=9t^2-2/t$

$y'(t)=18t+2/t^2$
then we have to calculate

$int_1^4sqrt((2t-1)^2+(18t+2/t^2)^2)dt$

by a numerical Methode we get $approx 137.033$

What is the arclength of (t^2-t-1,9t^2-2/t)(t2−t−1,9t2−2t)(t^2-t-1,9t^2-2/t) on t in [1,4]t∈[1,4]t in [1,4]?