Space curves

The space curve (hodograph) drawn by the point P in space is given by

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

. This is the same as the position vector of the point P with the origin O as $\boldsymbol{r} = \vec{\rm OP} = x \: \boldsymbol{i} + y \: \boldsymbol {j} + z \: \boldsymbol{k}$ , and given as a vector function of

$\displaystyle \boldsymbol{r}(t) = x(t) \: \boldsymbol {i} + y(t) \:\boldsymbol{j} + z(t) \: \boldsymbol{k}$

**Figure 2.1:** Tangent vector
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Next, let's think about what the derivative $\boldsymbol{r}^{\ prime}(t)$ of the vector function $\boldsymbol{r}(t)$ represents geometrically. The derivative of a vector function $\boldsymbol{r}(t)$ , $\boldsymbol{r}^{\ prime}(t)$ is by the definition $\boldsymbol{r}^{\ prime}(t)$ .

$\displaystyle \boldsymbol{r}^{\prime}(t) = \lim_{\Delta t \rightarrow 0}\frac{\boldsymbol{r}(t + \Delta t) - \boldsymbol{r}(t)}{\Delta t}$

Now if $\boldsymbol{r}(t + \Delta t) - \boldsymbol{r}(t)$ is $\boldsymbol{r}^{\prime}(t) \neq 0$ ，then even if $\Delta t$ approaches 0 , it not ${\bf0}$ and it approaches the direction of the tangent vector，Thus, the limit

$\displaystyle \lim_{\Delta t \rightarrow 0} \boldsymbol{r}(t + \Delta t) - \boldsymbol{r}(t)$

might be used as a tangent directional vector.

Unfortunately, this limit cannot be used as a tangent direction vector. Because this limit is ${\bf0}$ and ${\bf0}$ has no direction.

Therefore, to avoid this, consider the following vector that can obtain a large length when $\Delta t$ is small.

$\displaystyle \frac{\boldsymbol{r}(t + \Delta t) - \boldsymbol{r}(t)}{\Delta t}$

This vector is parallel to $\boldsymbol{r}(t + \Delta t) - \boldsymbol{r}(t)$ when $\Delta t$ is non-zero．That is, this vector is parallel to the tangent direction vector. Therefore, when this limit value $\boldsymbol{r}^{\ prime}(t)$ exists, this limit value can be considered as a tangent direction vector, so $\boldsymbol{r}^{\ prime}(t)$ is called tangent vector of curve $\boldsymbol{r} = \boldsymbol{r}(t)$ . Also,

$\displaystyle {\bf t}(t) = \frac{\boldsymbol{r}^{\prime}(t)}{\vert\boldsymbol{r}^{\prime}(t)\vert}$

is called unit tangent vector．

Note that $\vert{\bf t}(t)\vert = 1$ here, from the example 2 ${\bf t}^{\prime}(t)$ and ${\bf t}(t)$ is orthogonal．So, ${\bf t}^{\prime}(t)$ is called normal vector $of the curve \boldsymbol{r}(t)$ . Also， for ${\bf t}^{\prime}(t) \neq 0$ ，

$\displaystyle \boldsymbol{n}(t) = \frac{{\bf t}^{\prime}(t)}{\vert{\bf t}^{\prime}(t)\vert}$

is called principal unit normal vector．

**Figure 2.2:** Normal vector
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Example 2..6

2 Find the equation of straight line goes through the points .

Answer The straight line to be found has a starting point of and a direction of ${}^t [1,4,3]-{}^t [-1,0,2] = {}^t [2,4,1]$ . Let the arbitrary point on the line as . Then，

$\displaystyle (x,y,z) = (-1,0,2) + (2,4,1)t, \ \ t \in R$

Solve this for , we have

$\displaystyle \frac{x+1}{2} = \frac{y}{4} = \frac{z-2}{1}$

Example 2..7

Draw the plane curve $\displaystyle{\boldsymbol{r}(t) = \sin{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j}}$ .

Answer Since $x(t) = \sin{t}, y(t) = \sin(t)$ , we have ．t should be noted here that the value of $\ sin {t}$ is from to , so the curve to be calculated is $y = x, \ -1 \ leq x \ leq 1$ .

Example 2..8

Find the equation of the tangent line to the curve $\displaystyle{\boldsymbol{r}(t) = \cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j}}$ at $\displaystyle{t = \frac{\pi}{4}}$ ．

Answer The tangent vector is give by

$\displaystyle \boldsymbol{r}^{\prime}(\frac{\pi}{4}) = - \sin{t}\:\boldsymbol{i... ...4}} = - \frac{\sqrt{2}}{2}\:\boldsymbol{i} + \frac{\sqrt{2}}{2}\:\boldsymbol{j}$

Thus, the equation of the tangent line is

$\displaystyle (x,y) = \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)t + \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$

$\displaystyle t = \frac{x - \sqrt{2}/{2}}{-{\sqrt{2}}/{2}} = \frac{y - {\sqrt{2}}/{2}}{{\sqrt{2}}/{2}}$

When $\boldsymbol{r} = \boldsymbol{r}(t)$ satisfies the condition $\boldsymbol{r}^{\prime}(t) \neq 0, \boldsymbol{r}^{\prime}(t) \in C[a,b]$ ，we say $\boldsymbol{r}$ is smooth curve．The length of the $a \leq t \leq b$ part of a smooth curve $\boldsymbol{r} = \boldsymbol{r}(t)$ is called arc length an denoted by . How can you find the arc length?

Curve $\boldsymbol{r} = \boldsymbol{r}(t) = (x(t),y(t), z(t))$ corresponding to the small interval $[t, t + \Delta t$ in the interval . The length of the arc PQ is considered to be approximated by the line segment PQ. Then

$\displaystyle {\rm PQ} = \sqrt{(\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}... ...(\frac{\Delta y}{\Delta t})^{2} + (\frac{\Delta z}{\Delta t})^{2} } \ \Delta t$

Thus the length of is expressed by

$\displaystyle s$	$\displaystyle =$	$\displaystyle \lim \sum \sqrt{(\frac{\Delta x}{\Delta t})^{2} + (\frac{\Delta y}{\Delta t})^{2} + (\frac{\Delta z}{\Delta t})^{2} } \ \Delta t$
	$\displaystyle =$	$\displaystyle \int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2} + (\frac{dz}{dt})^{2}} dt$
	$\displaystyle =$	$\displaystyle \int_{a}^{b} \vert\frac{d\boldsymbol{r}}{dt}\vert dt$

Also, the arc lenght from $\boldsymbol{r}(a)$ to $\boldsymbol{r}(t)$ is given by

$\displaystyle s(t) = \int_{a}^{t}\vert\frac{d\boldsymbol{r}}{dt}\vert dt$

Thus,

$\displaystyle \frac{ds}{dt} = \vert\frac{d \boldsymbol{r}}{dt}\vert$

Theorem 2..2

When $\boldsymbol{r} = \boldsymbol{r}(t)$ is smooth curve， the arc length from $\boldsymbol{r}(a)$ to $\boldsymbol{r}(t)$ is given by

$\displaystyle s(t) = \int_{a}^{t}\vert\frac{d\boldsymbol{r}}{dt}\vert dt$

As you may have noticed here, if you think of as time, then $\displaystyle{\vert \frac{d \boldsymbol{r}}{dt}\vert}$ is considered to be a change in position within a minute time. Therefore, it represents the moving speed of the point P. Therefore, the length of the curve can be thought of as the distance that the point P has moved within the time .

Example 2..9

Find the arc length of $\displaystyle{\boldsymbol{r}(t) = \cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j} + t\:\boldsymbol{k}}$ for $0 \leq t \leq 2\pi$ .

Answer Since $x = \cos{t}, y = \sin{t}, z = t$ ，we have . Thus, $\boldsymbol{r}(t)$ is a smooth curve that spirally rotates around a cylinder with a radius of 1.

$\displaystyle \frac{d \boldsymbol{r}}{dt} = -\sin{t}\:\boldsymbol{i} + \cos{t}\:\boldsymbol{j} + \:\boldsymbol{k}$

implies

$\displaystyle \vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{(-\sin{t})^{2} + (\cos{t})^{2} + 1} = \sqrt{1 + 1} = \sqrt{2}$

Therefore,

$\displaystyle s = \int_{0}^{2\pi}\sqrt{2}dt = 2\sqrt{2}\pi$

．

Example 2..10

Express $\displaystyle{\boldsymbol{r}(t) = 5\cos{t}\boldsymbol{i} + 5\sin{t}\:\boldsymbol{j}}$ using the arc length as a parameter.

Answer $\boldsymbol{r}(t) = 5\cos{t}\:\boldsymbol{i} + 5\sin{t}\:\boldsymbol{j}$ implies

$\displaystyle \vert\frac{d\boldsymbol{r}}{dt}\vert = \vert-5\sin{t}\:\boldsymbol{i} + 5\cos{t}\:\boldsymbol{j}\vert = 5$

Thus

$\displaystyle s = \int_{0}^{t} 5 dt = 5t$

or $\vert\boldsymbol{r}(t)\vert^{2} = x^2 + y^2 = 25$ implies $\boldsymbol{r}(t)$ is a circle of the radius 5 with the center at the origin. Thus, the arc length is ．Therefore,

$\displaystyle \boldsymbol{r}(s) = \boldsymbol{r}(t(s)) = 5\cos{\frac{s}{5}}\:\boldsymbol{i} + 5\sin{\frac{s}{5}}\:\boldsymbol{j}$

When the curve is expressed by $\boldsymbol{r}(s)$ using the parameter ，we find the tangent vector.

$\displaystyle \frac{\boldsymbol{r}(s + \Delta s) - \boldsymbol{r}(s)}{\Delta s} = \frac{\mbox{chord length}}{\mbox{arc length}} \longrightarrow 1$

Thus, the unit tangent vector ${\bf t}$ is expressed by

$\displaystyle {\bf t} = \frac{d \boldsymbol{r}}{ds}$

**Figure 2.3:** curvature
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Now let's think about how the curve bends. First, let's think about the curve on the plane．Let $\phi$ be the angle formed by the tangent and the axis at the point P on the curve. The tangents and $\phi$ change as the point P moves．At this time, the rate of change of $\phi$ per unit arc length is called curvature(curvature) .

$\displaystyle \kappa = \vert\frac{d\phi}{ds}\vert$

Where the tangent unit vector ${\bf t}$ can be represented by ${\ bf t} = (\cos {\ phi}, \sin{\ phi})$ ．Therefore, when examining the rate of change of the tangent vector per unit arc length,

$\displaystyle \frac{d{\bf t}}{ds} = (-\sin{\phi}\frac{d\phi}{ds}, \cos{\phi}\frac{d\phi}{ds}) = \frac{d\phi}{ds}(-\sin{\phi},\cos{\phi})$

Thus

$\displaystyle \vert\frac{d{\bf t}}{ds}\vert = \vert\frac{d\phi}{ds}(-\sin{\phi},\cos{\phi})\vert = \vert\frac{d\phi}{ds}\vert = \kappa$

Therefore, the curvature $\kappa$ can also be expressed by the rate of change of the tangent vector per unit arc length.

Example 2..11

Find the curvature of the curve .

Answer First find $\displaystyle{\vert\frac{d\phi}{ds}\vert}$ ． $\tan{\phi}$ is the slope of the tangent line. Thus $\tan{\phi} = y^{\prime}(x)$ Therefore,

$\displaystyle \phi = \tan^{-1}(y^{\prime})$

Differentiate this with respect to .

$\displaystyle \frac{d\phi}{dx} = \frac{1}{1 + (y^{\prime})^{2}} \cdot\frac{d}{dx}(y^{\prime}) = \frac{y^{\prime\prime}}{1 + (y^{\prime})^{2}}$

Note that，

$\displaystyle \frac{d \phi}{dx} = \frac{d \phi}{ds} \frac{ds}{dx} = \frac{d \phi}{ds} \sqrt{1 + (y^{\prime})^2}$

Then

$\displaystyle \frac{d \phi}{ds} \sqrt{1 + (y^{\prime})^2} = \frac{y^{\prime\prime}}{1 + (y^{\prime})^2}$

Therefore,

$\displaystyle \kappa = \vert\frac{d \phi}{ds}\vert = \frac{\vert y^{\prime\prime}\vert}{[1 + (y^{\prime})^2]^{3/2}}$

The curvature of the curve in space is $\kappa = \vert\frac{d {\bf t}}{ds} \vert$ .

$\displaystyle \frac{d {\bf t}}{ds} = \frac{d {\bf t}/dt}{ds/dt}, \ \boldsymbol{... ...c{{\bf t}'}{\vert{\bf t}'\vert} = \frac{d {\bf t}/dt}{\vert d {\bf t}/dt\vert}$

Then

$\displaystyle \frac{d {\bf t}}{ds} = \boldsymbol{n} \frac{\vert d {\bf t}/dt\ve... ...ds/dt} = \boldsymbol{n} \vert\frac{d {\bf t}}{ds} \vert = \boldsymbol{n} \kappa$

(2.1)