Ejer_vectT01

El vector tangente unitario de una función vectorial $\textbf{r}(t)$ es el vector \[\textbf{T}(t)=\frac{\textbf{r}'(t)}{||\textbf{r}'(t)||}\]

Por tanto,

(%i2)

define( r( t),[ t ^ 3 + 3 * t, t ^ 2 + 1, 3 * t + 4]) $
define( T( t), diff( r( t), t) / sqrt( diff( r( t), t). diff( r( t), t))) ;

\begin{matrix} (%o2) & T (t) := [\frac{3 t^{2} + 3}{\sqrt{{(3 t^{2} + 3)}^{2} + 4 t^{2} + 9}}, \frac{2 t}{\sqrt{{(3 t^{2} + 3)}^{2} + 4 t^{2} + 9}}, \frac{3}{\sqrt{{(3 t^{2} + 3)}^{2} + 4 t^{2} + 9}}] \end{matrix}

Ahora solo nos resta sustituir en $t=1$:

(%i3)

T( 1) ;

\begin{matrix} (%o3) & [\frac{6}{7}, \frac{2}{7}, \frac{3}{7}] \end{matrix}

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