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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Expert Answer

Asma Vang
Answered 2021-10-01 Author has 14730 answers

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,
\(\widetilde{x}\left(s\right)=\left(3-\frac{3s}{\sqrt{11}}, -2-\frac{s}{\sqrt{11}}, -1-\frac{s}{\sqrt{11}}\right)\)

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