 # Starting from the point \left(3,−2,−1\right), reparametrize the curve x Suman Cole 2021-09-30 Answered
Starting from the point $\left(3,-2,-1\right)$, reparametrize the curve $x\left(t\right)=\left(3-3t,-2-t,-1-t\right)$ in terms of arclength.
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Step 1
First of all, the derivative of x is
${x}^{\prime }\left(t\right)=\left(-3,-1,-1\right)$
Its norm is
$\mid {x}^{\prime }\left(t\right)\mid =\sqrt{{\left(-3\right)}^{2}+{\left(-1\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{9+1+1}=\sqrt{11}$
Now we find
$s\left(t\right)={\int }_{{t}_{0}}^{t}\mid {x}^{\prime }\left(u\right)\mid du$,
where $x\left({t}_{0}\right)$ is the starting point So, we must have that
$x\left({t}_{0}\right)=\left(3,-2,-1\right)$
which yields
${t}_{0}=0$
Therefore,
$s\left(t\right)={\int }_{0}^{t}\mid {x}^{\prime }\left(u\right)\mid du={\int }_{0}^{t}\sqrt{11}du=\sqrt{11}t$
Therelore, we set
$t=\frac{s}{\sqrt{11}}$
This means that the reparametrization of x in terms of arclength is
$\stackrel{~}{x}\left(s\right)=x\left(\frac{s}{\sqrt{11}}\right)=\left(3-\frac{3s}{\sqrt{11}},-2-\frac{s}{\sqrt{11}},-1-\frac{s}{\sqrt{11}}\right)$
To write it clearly,
$\stackrel{~}{x}\left(s\right)=\left(3-\frac{3s}{\sqrt{11}},-2-\frac{s}{\sqrt{11}},-1-\frac{s}{\sqrt{11}}\right)$