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

Aug 17, 2017

$\approx 1.47894$

#### Explanation:

As this curve is $1 - t , {\left(1 - t\right)}^{2}$ (the $x$ coordinate is $1 - t$, rather than $t - 1$ because the former is positive in $\left[0 , 1\right]$, the curve is the same as the segment of the parabola $y = {x}^{2}$ between $\left(0 , 0\right)$ and $\left(1 , 1\right)$. The length of this segment is

$L = {\int}_{0}^{1} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = {\int}_{0}^{1} \sqrt{1 + 4 {x}^{2}} \mathrm{dx} = 2 {\int}_{0}^{1} \sqrt{{x}^{2} + \frac{1}{2} ^ 2} \mathrm{dx}$

Using the standard integral
$\int \sqrt{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{2} x \sqrt{{x}^{2} + {a}^{2}} + {a}^{2} / 2 \ln | x + \sqrt{{x}^{2} + {a}^{2}} |$
this becomes

$L = x \sqrt{{x}^{2} + \frac{1}{4}} + \frac{1}{4} \ln | x + \sqrt{{x}^{2} + \frac{1}{4}} | {|}_{0}^{1}$
$= \sqrt{\frac{5}{4}} + \frac{1}{4} \ln \left(1 + \sqrt{\frac{5}{4}}\right) + \frac{1}{4} \ln 2$
$= \sqrt{\frac{5}{4}} + \frac{1}{4} \ln \left(2 + \sqrt{5}\right) \approx 1.47894$