What is the arclength of $(sint/(t+cos2t),cost/(2t))$ on $t in [pi/12,pi/3]$ ?

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Answer 1 · 2016-10-14T04:15:56

Use $L = int_(pi/12)^(pi/3)sqrt((dx/dt)^2+ (dy/dt)^2)dt~~ 2.34$

Given:
$x = sin(t)/(t + cos(2t))$
$y = cos(t)/2t$

$dx/dt = (sin(t) (2 sin(2 t)-1)+cos(t) (t+cos(2 t)))/(t+cos(2 t))^2$

$dy/dt = -(t sin(t)+cos(t))/(2 t^2)$

$L = int_(alpha)^(beta)sqrt((dx/dt)^2+ (dy/dt)^2)dt$

$L = int_(pi/12)^(pi/3)sqrt(((sin(t) (2 sin(2 t)-1)+cos(t) (t+cos(2 t)))/(t+cos(2 t))^2)^2 + (-(t sin(t)+cos(t))/(2 t^2))^2)dt$

$L ~~ 2.34$

What is the arclength of (sint/(t+cos2t),cost/(2t))(sint/(t+cos2t),cost/(2t)) on t in [pi/12,pi/3]t in [pi/12,pi/3]?