Chapter 2: Space Curves

Section 2.3: Tangent Vectors

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. Solution Recall that $\frac{\mathrm{ds}}{\mathrm{dp}}\=\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dy}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dz}}{\mathrm{dp}}\right)}^2}$, and that $\frac{\mathrm{dp}}{\mathrm{ds}}\=1\/\left(\frac{\mathrm{ds}}{\mathrm{dp}}\right)$.   Apply the chain rule of differentiation to obtain   $\frac{d}{\mathrm{ds}}\mathbf{R}\left(p\left(s\right)\right)\=\frac{d\mathbf{R}}{\mathrm{dp}} \frac{\mathrm{dp}}{\mathrm{ds}}$    Compute the norm, obtaining   $∥\frac{d}{\mathrm{ds}}\mathbf{R}\left(p\left(s\right)\right)∥$ $\=\frac{\mathrm{dp}}{\mathrm{ds}}\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dy}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dz}}{\mathrm{dp}}\right)}^2}$ $\=\frac{\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dy}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dz}}{\mathrm{dp}}\right)}^2}}{\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dy}}{\mathrm{dp}}\right)}^2\+{\left(\frac{\mathrm{dz}}{\mathrm{dp}}\right)}^2}}$ $\=1$

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