Question

Q. Show that the following curves are unit–speed:
(i) γ(t) = (1/3 〖(1+t)〗^(3/2),1/3 〖(1-t)〗^(3/2),t/√2 )
(ii) γ(t) = (4/5 cos⁡t,1-sin⁡t,-3/5 cos⁡t)

Answer 1

ANSWER on Question #70832 – Math – Geometry

QUESTION

Show that the following curves are unit-speed:

1)

$$s(t) = \left(\frac{1}{3}(1 + t)^{\frac{3}{2}}, \quad \frac{1}{3}(1 - t)^{\frac{3}{2}}, \quad \frac{t}{\sqrt{2}}\right)$$

2)

$$s(t) = \left(\frac{4}{5} \cos t, \; 1 - \sin t, \quad -\frac{3}{5} \sin t\right)$$

SOLUTION

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

The speed at $t$ is a length $|s'(t)|$. The unit-speed is $|s'(t)| = 1, t \geq 0$

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

In our case,

1)

$$s(t) = \left(\frac{1}{3}(1 + t)^{\frac{3}{2}}, \quad \frac{1}{3}(1 - t)^{\frac{3}{2}}, \quad \frac{t}{\sqrt{2}}\right)$$

Then,

$$x(t) = \frac{1}{3}(1 + t)^{\frac{3}{2}} \rightarrow x'(t) = \frac{1}{3} \cdot \frac{3}{2} \cdot (1 + t)^{\frac{3}{2} - 1} \cdot 1 = \frac{1}{2} \cdot (1 + t)^{\frac{1}{2}}$$ $$\boxed{x'(t) = \frac{1}{2} \cdot (1 + t)^{\frac{1}{2}}}$$ $$y(t) = \frac{1}{3}(1 - t)^{\frac{3}{2}} \rightarrow y'(t) = \frac{1}{3} \cdot \frac{3}{2} \cdot (1 - t)^{\frac{3}{2} - 1} \cdot (-1) = -\frac{1}{2} \cdot (1 - t)^{\frac{1}{2}}$$ $$\boxed{y'(t) = -\frac{1}{2} \cdot (1 - t)^{\frac{1}{2}}}$$ $$z(t) = \frac{t}{\sqrt{2}} \rightarrow \boxed{z'(t) = \frac{1}{\sqrt{2}}}$$

Then,

$$\begin{aligned} |s'(t)| &= \sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2 + \left(z'(t)\right)^2} = \\ &= \sqrt{\left(\frac{1}{2} \cdot (1 + t)^{\frac{1}{2}}\right)^2 + \left(-\frac{1}{2} \cdot (1 - t)^{\frac{1}{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \\ &= \sqrt{\frac{(1 + t)^{2\frac{1}{2}}}{4} + \frac{(1 - t)^{2\frac{1}{2}}}{4} + \frac{1}{2}} = \sqrt{\frac{1 + t}{4} + \frac{1 - t}{4} + \frac{1 \cdot 2}{2 \cdot 2}} = \sqrt{\frac{1 + t + 1 - t + 2}{4}} = \sqrt{\frac{4}{4}} = 1 \end{aligned}$$

Conclusion,

$$\boxed{|s'(t)| = 1, \forall t \geq 0}$$

2)

$$s(t) = \left(\frac{4}{5} \cos t, \; 1 - \sin t, \quad -\frac{3}{5} \cos t\right)$$

Then,

$$\begin{aligned} x(t) &= \frac{4}{5} \cos t \rightarrow \boxed{x'(t) = -\frac{4}{5} \sin t} \\ y(t) &= 1 - \sin t \rightarrow \boxed{y'(t) = -\cos t} \\ z(t) &= -\frac{3}{5} \cos t \rightarrow \boxed{z'(t) = \frac{3}{5} \sin t} \end{aligned}$$

Then,

$$\begin{aligned} |s'(t)| &= \sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2 + \left(z'(t)\right)^2} = \\ &= \sqrt{\left(-\frac{4}{5} \sin t\right)^2 + (-\cos t)^2 + \left(\frac{3}{5} \sin t\right)^2} = \sqrt{\frac{16 \sin^2 t}{25} + \frac{9 \sin^2 t}{25} + \cos^2 t} = \sqrt{\frac{16 \sin^2 t}{25} + \frac{9 \sin^2 t}{25} + \cos^2 t} \end{aligned}$$ $$= \sqrt {\frac {(16 + 9) \sin^ {2} t}{25} + \cos^ {2} t} = \sqrt {\frac {25 \sin^ {2} t}{25} + \cos^ {2} t} = \sqrt {\sin^ {2} t + \cos^ {2} t} = \sqrt {1} = 1$$

Conclusion,

$$\boxed{|s'(t)| = 1, \forall t \geq 0}$$

ANSWER:

1)

$$s(t) = \left(\frac{1}{3}(1 + t)^{\frac{3}{2}}, \quad \frac{1}{3}(1 - t)^{\frac{3}{2}}, \quad \frac{t}{\sqrt{2}}\right) \rightarrow |s'(t)| = 1, \forall t \geq 0$$

2)

$$s(t) = \left(\frac{4}{5} \cos t, \; 1 - \sin t, \quad -\frac{3}{5} \sin t\right) \rightarrow |s'(t)| = 1, \forall t \geq 0$$

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