Motion of objects

The curve

$$C : \boldsymbol{r}(t) = (x(t), y(t), z(t)), \ t \in [a,b]$$

can be thought of as the locus drawn by the point P moving in space． Here, the interval $[a, b]$ is considered as the interval of time, and $\boldsymbol{r}(t)$ is considered as the position of the object in time $t$. For the motion $\boldsymbol{r} = \boldsymbol{r}(t)$， $\boldsymbol{r}^{\ prime}(t)$ is called velocity．Also， $\boldsymbol{r}^{\prime\prime}(t)$ is acceleration and expressed by ${\bf v}(t),\boldsymbol{A}(t)$. Thus,，

$${\bf v}(t) = \boldsymbol{r}'(t) , \boldsymbol{A}(t)=\boldsymbol{r}''(t)$$

As we have already learned, the tangent unit vector can be represented by $${{\ bf t} = \frac{d \boldsymbol{r}}{ds}}$$ and the velocity vector is

$${\bf v}(t) = \frac{d\boldsymbol{r}(t)}{dt} = \frac{d\boldsymbol{r}(t)}{ds} \frac{ds}{dt} = \frac{ds}{dt} {\bf t}$$

Thus, velocity vector is a vector tangential to the locus of a point．The magnitude of the velocity vector ${\bf v}$, $${\frac{ds}{dt}}$$ is the rate or speed of change in arc length and denoted by $v$. Thus

$$v = \vert{\bf v}\vert = \vert\boldsymbol{r}'(t)\vert = \frac{ds}{dt}$$

Next, to understand the acceleration a little better, let's consider the velocity vector.　

$${\bf v}(t) = \boldsymbol{r}'(t) = \frac{ds}{dt}{\bf t} = v{\bf t}$$

Differentiate both sides of the above equation,

$$\boldsymbol{A} = \frac{d{\bf v}}{dt} = \frac{dv}{dt}{\bf t} + v\f... ...f t}}{dt} = \frac{dv}{dt}{\bf t} + v\vert\frac{d{\bf t}}{dt}\vert\boldsymbol{n}$$

Note that

$$\frac{d{\bf t}}{dt} = \frac{d{\bf t}}{ds}\frac{ds}{dt} = v\frac{d{\bf t}}{ds}$$

Then

$$\vert\frac{d{\bf t}}{dt}\vert = \vert v \frac{d{\bf t}}{ds} \vert = v \vert\frac{d{\bf t}}{ds}\vert = v\kappa$$

Therefore,

$$\boldsymbol{A} = \frac{dv}{dt}{\bf t} + v^{2}\kappa\boldsymbol{n}$$

$$= a_{{\bf t}} + a_{\boldsymbol{n}}$$

$$a_{{\bf t}} = (\boldsymbol{A}\cdot{\bf t}){\bf t} , a_{\boldsymbol{n}} = (\boldsymbol{A}\cdot\boldsymbol{n})\boldsymbol{n}$$

Example 2..12  

For

$$\boldsymbol{r}(t) = (\cos\pi t, \sin\pi t, t)$$

$t = 1$, find ${\bf v}(t),\boldsymbol{A}(t),v,{\bf t},\boldsymbol{n}$.

Answer

$$\boldsymbol{r}(t) = (\cos\pi t, \sin\pi t, t)$$

$${\bf v}(t) = \boldsymbol{r}'(t) = (-\pi\sin\pi t, \pi\cos\pi t, 1)$$

$$\boldsymbol{A}(t) = \boldsymbol{r}''(t) = (-\pi^{2}\cos\pi t, -\pi^{2}\sin\pi t, 0)$$

Then for $t = 1$,

$${\bf v}(1) = (0, \pi, 1)$$

$$\boldsymbol{A}(1) = (\pi^{2}, 0, 0)$$

Then

$$v = \vert{\bf v}(1)\vert = \sqrt{\pi^{2} + 1}$$

$${\bf t} = \frac{{\bf v}(1)}{\vert{\bf v}(1)\vert} = \frac{(0, \pi, 1)}{\sqrt{\pi^{2} + 1}}$$

Now we have many ways to find $\boldsymbol{n}$. Here we will consider a method that is easy to calculate．

$$\boldsymbol{A} = a_{{\bf t}} + a_{\boldsymbol{n}},a_{{\bf t}} = (\boldsymbol{A}\cdot{\bf t}){\bf t} = 0$$

implies

$$a_{\boldsymbol{n}} = \boldsymbol{A} - \boldsymbol{A}_{{\bf t}} = (\pi^{2}, 0, 0)$$

Thus，

$$\boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1,0,0)$$

Other way is

$$v = \vert{\bf v}(t)\vert = \sqrt{\pi^{2}\sin^{2}\pi t + \pi^{2}\cos^{2}\pi t + 1} = \sqrt{\pi^{2} + 1}$$

$$a_{{\bf t}} = \frac{dv}{dt} = 0$$

Thus, $${\boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1, 0, 0)}$$

The vector

$$\boldsymbol{B} = {\bf t} \times \boldsymbol{n}$$

which is orthogonal to the unit normal vector ${\bf t}$ and the normal vector $\boldsymbol{n}$ is called binormal unit vector. Also, the $\tau$ satisfying

$$\frac{d \boldsymbol{B}}{ds} = - \tau \boldsymbol{n}$$

is called torsion.

Now, let's examine the three unit vectors ${\bf t}, \boldsymbol{n}, \boldsymbol{B}$ that have appeared so far.

Figure 2.4:

The plane made by ${\bf t}$ and $\boldsymbol{n}$ is called osculating plane. the plane made by $\boldsymbol{n}$ and $\boldsymbol{B}$ is called normal plane. the plane made by ${\bf t}$ and $\boldsymbol{B}$is called rectifying plane.

First， ${\bf t}, \boldsymbol{n}, \boldsymbol{B}$ are orthogonal to each other. Also, these vectors satisfies the following relations.

Theorem 2..3  

[Frenet-Serret]

$$\frac{d {\bf t}}{ds} = \kappa \boldsymbol{n}, \frac{d \boldsymbol... ... t} + \tau \boldsymbol{B}, \frac{d \boldsymbol{B}}{ds} = - \tau \boldsymbol{n}$$

Proof By the equation 2.1, $${\frac{d {\bf t}}{ds} = \kappa \boldsymbol{n}}$$. Also, by the definition of torsion, $${\frac{d \boldsymbol{B}}{ds} = - \tau \boldsymbol{n}}$$．Next differentiate $\boldsymbol{n} \times {\bf t}$ with respect to $s$, we have

$$\frac{d \boldsymbol{n}}{ds} = \frac{d \boldsymbol{B}}{ds} \times ... ...ldsymbol{B} \times \frac{d {\bf t}}{ds} = \tau \boldsymbol{B} - \kappa {\bf t}$$

Example 2..13  

For the curve $\boldsymbol{r} = 2a(\sin^{-1}{t} + t\sqrt{1- t^2})\:\boldsymbol{i} + 2at^2\:\boldsymbol{j} + 4at\:\boldsymbol{k}$，find the followings． However, $a$ is an arbitrary positive constant.

$(a)$ the arc length the curve for $t_{1} \leq t \leq t_{2}$

$(b)$ the unit tangent vector ${\bf t}$

$(c)$ the normal vector $\boldsymbol{n}$ and the curvature $\kappa$

$(d)$ the binormal vector $\boldsymbol{B}$ and the torsion $\tau$

Answer (a) $${\frac{d \boldsymbol{r}}{dt} = 4a \sqrt{1 - t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}$$ implies

$$\vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{16a^2 (1- t^2) + 16a^2 t^2 + 16a^2} = \sqrt{32 a^2} = 4\sqrt{2} a$$

Therefore，

$$s = \int_{t_{1}}^{t_{2}} \vert \frac{d \boldsymbol{r}}{dt}\vert dt = 4\sqrt{2} a (t_{2} - t_{1})$$

(b)

$${\bf t} = \frac{d \boldsymbol{r}}{ds} = \frac{d \boldsymbol{r}/dt... ...-t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}{4\sqrt{2} a}$$

(c) $${\frac{d {\bf t}}{dt} = \frac{1}{\sqrt{2}}(\frac{-t}{\sqrt{1 - t^2}}\:\boldsymbol{i} + \boldsymbol{j}), \frac{ds}{dt} = 4\sqrt{2} a}$$ より

$$\kappa = \vert\frac{d {\bf t}}{ds}\vert = \vert\frac{d {\bf t}/dt... ...vert = \frac{1}{8a}\sqrt{\frac{t^2}{1 - t^2} + 1} = \frac{1}{8a \sqrt{1 - t^2}}$$

また， $${\boldsymbol{n} = \frac{1}{\kappa}\frac{d {\bf t}}{ds}}$$ より

$$\boldsymbol{n} = -t\:\boldsymbol{i} + \sqrt{1 - t^2}\:\boldsymbol{j}$$

(d)

$$\boldsymbol{B} = {\bf t} \times \boldsymbol{n} = \frac{1}{\sqrt{2... ...{\sqrt{2}}(-\sqrt{1 - t^2}\:\boldsymbol{i} -t\:\boldsymbol{j} + \boldsymbol{k}$$

Also，

$$\frac{d \boldsymbol{B}}{ds} = \frac{d \boldsymbol{B}/dt}{ds/dt} =... ...frac{t}{\sqrt{1- t^2}}\:\boldsymbol{i} -\boldsymbol{j}) = - \tau \boldsymbol{n}$$

より $${\tau = \frac{1}{8a \sqrt{1 - t^2}}}$$