What is the arclength of $f(t) = (t-2,1/(t-2))$ on $t in [0,1]$ ?

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Answer 1 · 2016-10-16T21:42:22

Use the equation:

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

$L ≈ 1.13209$

The arc length, L, for parametric equations is:

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

$dx/dt = 1$
$dy/dt = -(t - 2)^-2$
$a = 0$
$b = 1$

$L = int_0^1sqrt((1)^2 + (-(t - 2)^-2)^2)dt$

$L = int_0^1sqrt(1 + (t - 2)^-4)dt$

$L ≈ 1.13209$

What is the arclength of f(t) = (t-2,1/(t-2))f(t) = (t-2,1/(t-2)) on t in [0,1]t in [0,1]?