Question

Q. Which of the following curves are regular?
(i) γ(t) = (cos^2 t, sin^2 t) for -∞<t<∞
(ii) γ(t) = (cos^2 t, sin^2 t) for 0<t<π/2
(iii) γ(t) = (t, cosh t) for -∞<t<∞
Find unit speed reparametrisation of regular curve(s).

Answer 1

ANSWER on Question #70833 – Math – Geometry

QUESTION

Which of the following curves are regular?

1)

$$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for} \quad -\infty < t < \infty$$

2)

$$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for} \quad 0 < t < \frac{\pi}{2}$$

3)

$$\gamma(t) = (t, \cosh t), \quad \text{for} \quad -\infty < t < \infty$$

Find unit speed reparametrisation of regular curve(s).

SOLUTION

By the definition, let $\gamma(t)$ be a curve. The velocity vector of $\gamma(t)$ at $t$ is $\gamma'(t)$.

The speed at $t$ is a length $|\gamma'(t)|$.

$$\gamma(t) = (x(t), y(t), z(t)) \rightarrow \gamma'(t) = (x'(t), y'(t), z'(t))$$ $$|\gamma'(t)| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}$$

By the definition, the curve $\gamma(t)$ is regular if all velocity vectors are different from zero, that is,

$$\gamma'(t) \neq 0, \forall t$$

In our case,

1)

$$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for} \quad -\infty < t < \infty$$ $$x(t) = \cos^2 t \rightarrow x'(t) = 2 \cdot (\cos t) \cdot (-\sin t) = -\sin(2t) \leftrightarrow \boxed{x'(t) = -\sin(2t)}$$ $$y(t) = \sin^2 t \rightarrow y'(t) = 2 \cdot (\sin t) \cdot (\cos t) = \sin(2t) \leftrightarrow \boxed{y'(t) = \sin(2t)}$$

Then,

$$\gamma'(t) = (-\sin(2t), \sin(2t))$$ $$\gamma'(\pi k) = \left(-\sin\left(2 \cdot (\pi k)\right), \sin\left(2 \cdot (\pi k)\right)\right) = (0, 0)$$

Conclusion,

$$\gamma'(t) = 0, \forall t = \pi k, k \in \mathbb{Z} \rightarrow$$ $$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for } -\infty < t < \infty \text{ isn't regular}$$

2)

$$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for } 0 < t < \frac{\pi}{2}$$ $$x(t) = \cos^2 t \rightarrow x'(t) = 2 \cdot (\cos t) \cdot (-\sin t) = -\sin(2t) \leftrightarrow x'(t) = -\sin(2t)$$ $$y(t) = \sin^2 t \rightarrow y'(t) = 2 \cdot (\sin t) \cdot (\cos t) = \sin(2t) \leftrightarrow y'(t) = \sin(2t)$$

Since the

$$0 < t < \frac{\pi}{2} \rightarrow 0 < (2t) < \pi$$

As we know

$$\sin(2t) \neq 0, \forall 0 < t < \frac{\pi}{2}$$

Then,

$$\gamma'(t) = (-\sin(2t), \sin(2t)) \rightarrow \gamma'(t) \neq 0, \forall 0 < t < \frac{\pi}{2}$$

Conclusion,

$$\gamma'(t) \neq 0, \forall 0 < t < \frac{\pi}{2} \rightarrow$$ $$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for } 0 < t < \frac{\pi}{2} \text{ is regular}$$

Now, we find the unit speed reparametrisation of regular curve $\gamma(t)$ :

$$\begin{array}{l} c = \int_ {0} ^ {\frac {\pi}{2}} | \gamma^ {\prime} (t) | d t = \int_ {0} ^ {\frac {\pi}{2}} \sqrt {(- \sin (2 t)) ^ {2} + (\sin (2 t)) ^ {2}} d t = \int_ {0} ^ {\frac {\pi}{2}} \sqrt {2} \cdot \sin (2 t) d t \\ = \frac {\sqrt {2}}{2} \cdot (- \cos (2 t)) \Bigg | _ {0} ^ {\frac {\pi}{2}} = - \frac {\sqrt {2}}{2} \cdot \left(\cos \left(2 \cdot \frac {\pi}{2}\right) - \cos (2 \cdot 0)\right) = - \frac {\sqrt {2}}{2} \cdot (\cos (\pi) - \cos (0)) = \\ = - \frac {\sqrt {2}}{2} \cdot (- 1 - 1) = - \frac {\sqrt {2}}{2} \cdot (- 2) = \sqrt {2} \rightarrow \boxed {c = \sqrt {2}} \\ \end{array}$$

Then

$$s (t) = \int_ {0} ^ {t} c d u = \int_ {0} ^ {t} \sqrt {2} d u = \sqrt {2} t \rightarrow \boxed {t = \frac {s}{\sqrt {2}}}$$

Conclusion,

$\boxed{\gamma(s) = \gamma(t(s)) = \left( \cos^2 \left( \frac{s}{\sqrt{2}} \right), \sin^2 \left( \frac{s}{\sqrt{2}} \right) \right)}$ is the unit speed reparametrisation

3)

$$\gamma (t) = (t, \cosh t), \quad f o r - \infty < t < \infty$$ $$x (t) = t \rightarrow x ^ {\prime} (t) = 1 \leftrightarrow \boxed {x ^ {\prime} (t) = 1}$$ $$y (t) = \cosh t \rightarrow y ^ {\prime} (t) = \sinh t \leftrightarrow \boxed {y ^ {\prime} (t) = \sinh t}$$

Then,

$$\gamma^ {\prime} (t) = (1, \sinh t)$$

Since the

$$1 \neq 0,$$

then

$$\gamma^ {\prime} (t) = (1, \sinh t) \neq 0, f o r - \infty < t < \infty \rightarrow$$ $$\boxed {\gamma (t) = (t, \cosh t), \quad f o r - \infty < t < \infty \text { is regular}}$$

Now, we find the unit speed reparametrisation of regular curve $\gamma(t)$

$$s(t) = \int_{0}^{t} |\gamma'(u)| \, du = \int_{0}^{t} \sqrt{1^2 + (\sinh u)^2} \, du = \begin{bmatrix} \text{As we know} \\ \cosh^2 x - \sinh^2 x = 1 \\ \cosh^2 x = 1 + \sinh^2 x \end{bmatrix} = \\ = \int_{0}^{t} \sqrt{\cosh^2 u} \, du = \int_{0}^{t} \cosh u \, du = \sinh u|_0^t = \sinh t$$ $$s(t) = \sinh t \rightarrow \boxed{t(s) = \operatorname{asinh}s}$$

Conclusion,

$$\boxed{\gamma(s) = \gamma(t(s)) = (\operatorname{asinh}s, \cosh(\operatorname{asinh}s)) - \text{the unit speed reparametrisation}}$$

ANSWER:

1)

$$\gamma(t) = (\cos^2 t, \sin^2 t) \quad \text{for} \quad -\infty < t < \infty$$ $$\text{isn't regular}$$

2)

$$\gamma(t) = (\cos^2 t, \sin^2 t), \quad \text{for} \quad 0 < t < \frac{\pi}{2}$$ $$\text{is regular}$$ $$\boxed{\gamma(s)} = \left(\cos^2 \left(\frac{s}{\sqrt{2}}\right), \sin^2 \left(\frac{s}{\sqrt{2}}\right)\right) \text{ is the unit speed reparametrisation}$$

3)

$$\gamma(t) = (t, \cosh t) \quad \text{for} \quad -\infty < t < \infty$$ $$\text{is regular}$$ $$\boxed{\gamma(s)} = (\operatorname{asinh}s, \cosh(\operatorname{asinh}s)) \text{ is the unit speed reparametrisation}$$

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