What is the arclength of $(t^2lnt,lnt^2)$ on $t in [1,2]$ ?

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Answer 1 · 2016-10-17T17:54:20

Performed by WolframAlpha
$int_1^2 sqrt(4/t^2+(t+2 t log(t))^2) dt = 3.19515$

Let L be the parametric arclength, then the formula is:

$L = int_a^b sqrt((dx/dt)^2 + (dy/dt)^2)dt$

$x = t^2ln(t)$
$dx/dt = 2tln(t) + t$
$y = ln(t^2)$
$dy/dt = 2/t$
$a = 1$
$b = 2$

Substitute the above into the formula:

$L = int_1^2 sqrt((2tln(t) + t)^2 + (2/t)^2)dt$

$int_1^2 sqrt(4/t^2+(t+2 t log(t))^2) dt = 3.19515$

What is the arclength of (t^2lnt,lnt^2)(t^2lnt,lnt^2) on t in [1,2]t in [1,2]?