Item

Fact

1

The derivative $\mathbf{R}\prime \left(p\right)\=\frac{d}{\mathrm{dp}}\mathbf{R}\left(p\right)$ is taken componentwise.

2

The vector $\mathbf{R}\prime \left(p\right)$ is tangent to the curve defined by $\mathbf{R}\left(p\right)$.

3

If $p$ is arc length $s$, $\mathbf{R}\prime \left(s\right)\=\frac{d}{\mathrm{ds}}\mathbf{R}\left(s\right)$ is the unit tangent vector T.

$4$

If T is the unit tangent vector, $\mathbf{T}·\mathbf{T}\prime \=0$, that is, T is orthogonal to its derivative.

$5$

The length of $\mathbf{R}\prime \left(p\right)$ is given by $\mathrm{\ρ}\=\parallel \mathbf{R}\prime \parallel$ = $\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dy}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dz}}{\mathrm{dp}}\right)}^2}$ = $\frac{\mathrm{ds}}{\mathrm{sp}}$

$6$

If the parameter $p$ is the time $t$, then $\mathbf{R}\prime \left(t\right)$ is the velocity vector V; and $\mathrm{\ρ}\=v$, the speed.

Table 2.3.1   Essential tangent-vector facts for a curve defined by the position vector $\mathbf{R}\left(p\right)$.

Example 2.3.1

If $\mathbf{R}\left(p\right)$ is the position-vector representation of the parametric curve $x\=p^2$, $y\=p^3$, $p\ge 0$, show that $\underset{h\to 0}{lim}\frac{\mathbf{R}\left(p\+h\right)-\mathbf{R}\left(p\right)}{h}$, denoted by $\mathbf{R}\prime \left(p\right)\=\frac{d\mathbf{R}}{\mathrm{dp}}$,

is the vector $x\prime \left(p\right) \mathbf{i}\+y\prime \left(p\right) \mathbf{j}$. Thus, the differentiation operator $\frac{d}{\mathrm{dp}}$ is applied to R by applying it to each component of R.

Example 2.3.2

If $\mathbf{R}\left(p\right)$ is the position-vector representation of $C$, the parametric curve $x\=p \cos \left(p\right)$, $y\=p \sin \left(p\right)$, $p\in \left[0\,3 \mathrm{\π}\right]$,

Example 2.3.3

Let $\mathbf{R}\left(p\right)$ be the position-vector representation of the parametric curve $x\=p^2-p\/2\,y\=4\/3 p^{3\/2}$, $p\ge 0$, and let $\mathbf{R}\left(s\right)\=\mathbf{R}\left(p\left(s\right)\right)$ be the reparametrization obtained in Example 2.2.6. (Recall that $s$ is the arc length along the curve.)

Example 2.3.4

If $\mathbf{R}\left(p\right)$ is the position vector for the general parametric curve whose components are $x\left(p\right)\,y\left(p\right)\,z\left(p\right)$, and $s\ge 0$ is arc length, show that $\frac{d}{\mathrm{ds}}\mathbf{R}\left(p\left(s\right)\right)$ is necessarily the unit (tangent) vector T.

Example 2.3.5

If $\mathbf{R}\left(s\right)$ is the position vector for $C$, a curve parametrized by $s$, the arc length, and $\mathbf{T}\left(s\right)\=\mathbf{R}\prime \left(s\right)$ is the unit tangent vector along $C$, show that $\mathbf{T}\mathbf{·}\mathbf{T}\prime \=0$, thereby proving that the unit tangent vector is necessarily orthogonal to its derivative.

Example 2.3.6

If $\mathbf{R}\left(p\right)\=x\left(p\right) \mathbf{i}\+y\left(p\right) \mathbf{j}\+z\left(p\right) \mathbf{k}$ is a position vector and $R\left(p\right)\=∥\mathbf{R}\left(p\right)∥$ is its length, show that $\mathbf{R}·\frac{d\mathbf{R}}{\mathrm{dp}}$ = $R \frac{\mathrm{dR}}{\mathrm{dp}}$.

Example 2.3.7

Given the plane curve $C$ defined by $y\left(x\right)\=\frac{1-x^2}{1\+x^2}$,

Example 2.3.8

If $\mathbf{R}\left(p\right)$ is the position-vector form of the curve $C$ defined parametrically by the equations $x\left(p\right)\=2\+3p- p^2\,y\left(p\right)\=\cos \left(p\right)$, $p\in \left[0\,\mathrm{\π}\right]$,

Example 2.3.9

Given the two plane curves $f\left(x\right)\=x^2\,g\left(x\right)\=8-{\left(\frac{x}{4}\right)}^2\,x\ge 0$,