What is the arclength of $f (t) = (sin t \cdot cos 2 t, sin t)$ $f(t) = (sint*cos2t,sint)$ on $t \in [- π, 0]$ $t in [-pi,0]$ ?

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Answer 1 · 2018-06-10T09:13:47

$\approx 3.943805190587053$ $approx 3.943805190587053$

We have
$x (t) = sin (t) cos (2 \cdot t)$ $x(t)=sin(t)cos(2*t)$
$x'(t)=cos(t)cos(2t)-2sin(t)sin(2t)$

$y(t)=sin(t)$
$y'(t)=cos(t)$
so we have to integrate

$int_(-pi)^0sqrt((cos(t)cos(2t)-2sin(t)sin(2t))^2+cos^2(t))dt$
i got only a numerical result

$approx 3.943805190587053$

What is the arclength of f(t) = (sint*cos2t,sint)f(t)=(sint⋅cos2t,sint)f(t) = (sint*cos2t,sint) on t in [-pi,0]t∈[−π,0]t in [-pi,0]?