# Find the length of the parametric curve defined over the given intervalx=3t-6,y=6t+1,0<=t<=1

Find the length of the parametric curve defined over the given interval $x=3t-6,y=6t+1,0\le t\le 1$
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delilnaT
$x=3t-6,y=6t+1,0\le t\le 1$
length=${\int }_{a}^{b}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$
$\frac{dx}{dt}=\frac{d}{dt}\left[3t-6\right]=3\frac{dt}{dt}-d\frac{6}{dt}=3x1-0=3$
$\frac{dy}{dt}=\frac{d}{dt}\left[6t-1\right]=6\frac{dt}{dt}+d\frac{1}{dt}=6x1+0=6$
length=${\int }_{0}^{1}\sqrt{{3}^{2}+{6}^{2}}dt$
$={\int }_{0}^{1}\sqrt{45}dt=\sqrt{45}{\in }_{0}^{1}dt$
$=\sqrt{45}{\left[t\right]}_{0}^{1}$
$=\sqrt{45}\left[1-0\right]=\sqrt{45}$
length=$\sqrt{45}=3\sqrt{5}$