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

Question

Charles Arocha · Accepted Answer

#approx 1.47894# As this curve is #1-t, (1-t)^2# (the #x# coordinate is #1-t#, rather than #t-1# because the former is positive in #[0,1]#, the curve is the same as the segment of the parabola #y=x^2# between #(0,0)# and #(1,1)#. The length of this segment is #L = int_0^1 sqrt{1+(dy/dx)^2} dx = int_0^1 sqrt{1+4x^2} dx = 2int_0^1 sqrt{x^2 +1/2^2} dx# Using the standard integral #int sqrt{x^2+a^2 }dx = 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+1/4}+1/4 ln| x+sqrt{x^2+1/4}||_0^1 # # = sqrt{5/4}+1/4ln(1+sqrt{5/4})+1/4 ln 2# # = sqrt{5/4}+1/4ln(2+sqrt{5}) approx 1.47894#

Aurora Alvarado · Answer

undefined 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 $[0, 1]$.