Chapter 2: Space Curves

Section 2.3: Tangent Vectors

Example 2.3.6

$$ If $\mathbf{R}\left(p\right)\=x\left(p\right) \mathbf{i}\+y\left(p\right) \mathbf{j}\+z\left(p\right) \mathbf{k}$ is a position vector and $R\left(p\right)\=∥\mathbf{R}\left(p\right)∥$ is its length, show that $\mathbf{R}·\frac{d\mathbf{R}}{\mathrm{dp}}$ = $R \frac{\mathrm{dR}}{\mathrm{dp}}$. $$ Solution $$ $\mathbf{R}·\frac{d\mathbf{R}}{\mathrm{dp}}$ = $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]·\left[\begin{array}{c}x\prime \\ y\prime \\ z\prime \end{array}\right]$ = $x x\prime \+y y\prime \+z z\prime$  $$ $R \frac{\mathrm{dR}}{\mathrm{dp}}$ $\=\sqrt{x^2\+y^2\+z^2} \left(\frac{d}{\mathrm{dp}}\sqrt{x^2\+y^2\+z^2}\right)$ $$ $\=\sqrt{x^2\+y^2\+z^2}$ $\left(\frac{x x\prime \+y y\prime \+z z\prime }{\sqrt{x^2\+y^2\+z^2}}\right)$ $$ $\=x x\prime \+y y\prime \+z z\prime$ $$ Clearly, these two quantities are equal. $$

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