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

Question

Starting from the point

(3, - 2, - 1)

, reparametrize the curve

x (t) = (3 - 3 t, - 2 - t, - 1 - t)

in terms of arclength.

Answer 1

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

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

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