Question

Q. Reparameterize the curve w.r.t arc length parameter s.
γ(t)=(2t,1-3t,5+4t)

Answer 1

Answer on Question #72687, Math / Geometry

Reparameterize the curve w.r.t arc length parameter $s$: $y(t) = (2t, 1 - 3t, 5 + 4t)$.

Answer:

Let's figure out for what value of $t$ will we get the point $(0, 1, 5)$. First, we consider, $t = 0$.

$$2t = 0 \rightarrow t = 0$$

If $t = 0$, then we need $1 - 3(0) = 1$ and $5 + 4(0) = 5$, which are both true. So, this happens at $t = 0$. This is our initial value. Now, let's figure out $s(t)$.

First,

$$y'(t) = \langle 2 + (-3) + 4 \rangle$$ $$s(t) = \int_{0}^{t} \sqrt{(2)^2 + (-3)^2 + (4)^2} \, d\tau$$ $$s(t) = \int_{0}^{t} \sqrt{4 + 9 + 16} \, d\tau$$ $$s(t) = \int_{0}^{t} \sqrt{29} \, d\tau$$ $$s(t) = t \sqrt{29}$$

Once we have our expression $s(t) = t\sqrt{29}$ we can solve for $t$.

We get

$$t = \frac{s}{\sqrt{29}}$$

We get final answer.

$$y(s) = \frac{2}{\sqrt{29}} s, \quad 1 - \frac{3}{\sqrt{29}} s, \quad 5 + \frac{4}{\sqrt{29}} s$$

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