# Starting from the point \left(3,−2,−1\right), reparametrize the curve x

Suman Cole 2021-09-30 Answered
Starting from the point $$\displaystyle{\left({3},−{2},−{1}\right)}$$, reparametrize the curve $$\displaystyle{x}{\left({t}\right)}={\left({3}−{3}{t},−{2}−{t},−{1}−{t}\right)}$$ in terms of arclength.

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Asma Vang

Step 1
First of all, the derivative of x is
$$\displaystyle{x}'{\left({t}\right)}={\left(-{3},-{1},-{1}\right)}$$
Its norm is
$$\mid x'\left(t\right)\mid =\sqrt{\left(-3\right)^{2} +\left(-1\right)^{2} +\left(-1\right)^{2} }= \sqrt{9+1+1}= \sqrt{11}$$
Now we find
$$s\left(t\right)=\int_{t_{0}}^{t} \mid x'\left(u\right) \mid du$$,
where $$\displaystyle{x}{\left({t}_{{{0}}}\right)}$$ is the starting point So, we must have that
$$\displaystyle{x}{\left({t}_{{{0}}}\right)}={\left({3},−{2},−{1}\right)}$$
which yields
$$\displaystyle{t}_{{{0}}}={0}$$
Therefore,
$$s\left(t\right)=\int_{0}^{t} \mid x'\left(u\right) \mid du=\int_{0}^{t}\sqrt{11}du=\sqrt{11}t$$
Therelore, we set
$$\displaystyle{t}={\frac{{{s}}}{{\sqrt{{{11}}}}}}$$
This means that the reparametrization of x in terms of arclength is
$$\widetilde{x}\left(s\right)=x\left(\frac{s}{\sqrt{11}}\right)=\left(3-\frac{3s}{\sqrt{11}}, -2-\frac{s}{\sqrt{11}}, -1-\frac{s}{\sqrt{11}}\right)$$
To write it clearly,
$$\widetilde{x}\left(s\right)=\left(3-\frac{3s}{\sqrt{11}}, -2-\frac{s}{\sqrt{11}}, -1-\frac{s}{\sqrt{11}}\right)$$