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

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To find the arc length of the curve ( f(t) = (\sqrt{t^2 - 2t + 1}, t^2 - 2t + 1) ) on the interval ([0, 1]), we'll use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} , dt ]

First, we need to find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ), then we'll plug them into the formula and integrate over the interval ([0, 1]).

[ \frac{dx}{dt} = \frac{d}{dt}(\sqrt{t^2 - 2t + 1}) = \frac{1}{2\sqrt{t^2 - 2t + 1}} \cdot (2t - 2) = \frac{t - 1}{\sqrt{t^2 - 2t + 1}} ]

[ \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t + 1) = 2t - 2 ]

Now, we'll plug these into the formula for arc length:

[ L = \int_{0}^{1} \sqrt{\left(\frac{t - 1}{\sqrt{t^2 - 2t + 1}}\right)^2 + (2t - 2)^2} , dt ]

This integral represents the arc length of the curve ( f(t) ) on the interval ([0, 1]).

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