Note que [br][br] [math]T\left(t\right)=\frac{r'\left(t\right)}{\parallel r'\left(t\right)\parallel}[/math][br][br]también es un vector tangente, pero tiene longitud uno ([math]T(t)=1[/math]). Llamamos a [math]T(t)[/math] la tangente unitaria[br]Unidad vectorial tangente vector a la curva [math]C[/math]. Es decir, para cada [math]t[/math],[math]T(t)[/math] es un vector tangente de longitud uno apuntando en la dirección de la orientación de [math]C[/math].

El [b]vector unitario normal principal[/b] [math]N(t)[/math] es un vector unitario que tiene la misma dirección que [math]T'(t)[/math] y se define por[br][br] [math]N\left(t\right)=\frac{T'\left(t\right)}{\parallel T´\left(t\right)\parallel}[/math][br][br]siempre que [math]T´\left(t\right)\ne0[/math].[br][br][br][color=#0000ff][size=200][size=150]Encontrar la unidad tangente y la unidad principal normal vectores en los puntos dados.[/size][/size][/color]

[b]1.-[/b][math]r\left(t\right)=\left\langle t,t^2\right\rangle,t=0,t=1[/math].[br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle1,2t\right\rangle[/math][br][br]Entonces, usamos la definición de vector tangente:[br][br][math]T\left(t\right)=\frac{r´\left(t\right)}{\parallel r´\left(t\right)\parallel}=\frac{\left\langle1,2t\right\rangle}{\sqrt{\left(1\right)^2+\left(2t\right)^2}}=\frac{\left\langle1,2t\right\rangle}{\sqrt{1+4t^2}}=\frac{1}{\sqrt{1+4t^2}}i+\frac{2t}{\sqrt{1+4t^2}}[/math][br][br]Ahora, derivamos a [math]T\left(t\right)[/math] para encontrar el vector unitario normal.[br][br][math]T'\left(t\right)=\left\langle-\frac{4t}{\left(4t^2+1\right)^{\frac{3}{2}}},\frac{2}{\left(4t^2+1\right)^{\frac{3}{2}}}\right\rangle[/math][br][br]Entonces, usando la definición del vector unitario normal, tenemos que:[br][br][math]N\left(t\right)=\frac{T'\left(t\right)}{\parallel T'\left(t\right)\parallel}=\frac{T´\left(t\right)}{\sqrt{\left(-\frac{4t}{\left(4t^2+1\right)^{\frac{3}{2}}}\right)}^2+\left(\frac{2}{\left(4t^2+1\right)^{\frac{3}{2}}}\right)^2}=\frac{T´\left(t\right)}{\frac{2}{4t^2+1}}=\left\langle\frac{-\frac{4}{\left(4t^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4t^2+1}},\frac{\frac{2}{\left(4t^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4t^2+1}}\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]N\left(t=0\right)=\left\langle\frac{-\frac{4}{\left(4\left(0\right)^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4\left(0\right)^2+1}},\frac{\frac{2}{\left(4\left(0\right)^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4\left(0\right)^2+1}}\right\rangle=\left\langle-2,1\right\rangle[/math] [br][br]Ahora, cuando [math]t=1[/math]:[br][br][math]N\left(t=1\right)=\left\langle\frac{-\frac{4}{\left(4\left(1\right)^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4\left(1\right)^2+1}},\frac{\frac{2}{\left(4\left(1\right)^2+1\right)^{\frac{3}{2}}}}{\frac{2}{4\left(1\right)^2+1}}\right\rangle=\left\langle-\frac{2}{5^{\frac{1}{2}}},\frac{1}{5^{\frac{1}{2}}}\right\rangle[/math][br]

3.-[math]r\left(t\right)=\left\langle cos2t,sen2t\right\rangle ent=0,t=\frac{\pi}{4}[/math][br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle-2sen\left(2t\right),2cos\left(2t\right)\right\rangle[/math][br][br]Entonces, usando la definición del vector tangente:[br][br][math]T\left(t\right)=\frac{r'\left(t\right)}{\sqrt{\left(-2sen\left(2t\right)\right)^2+\left(2cos\left(2t\right)\right)^2}}=\frac{r'\left(t\right)}{\sqrt{4\left(sen^2\left(2t\right)+cos^2\left(2t\right)\right)}}=\left\langle\frac{-2sen\left(2t\right)}{2},\frac{2cos\left(2t\right)}{2}\right\rangle=\left\langle-sen\left(2t\right),cos\left(2t\right)\right\rangle[/math][br][br]Ahora, derivamos a [math]T\left(t\right)[/math] para encontrar el vector unitario normal:[br][br][math]T'\left(t\right)=\left\langle-2cos\left(2t\right),-2sen\left(2t\right)\right\rangle[/math][br][br]Entonces, usando la definición del vector unitario normal, tenemos que:[br][br][math]N\left(t\right)=\frac{T´\left(t\right)}{\parallel T´\left(t\right)\parallel}=\frac{T´\left(t\right)}{\sqrt{\left(-2cos\left(2t\right)\right)^2+\left(-2sen\left(2t\right)\right)^2}}=\frac{T´\left(t\right)}{\sqrt{4\left(sen^2\left(2t\right)+cos^2\left(2t\right)\right)}}=\frac{\left\langle-2cos\left(2t\right),-2sen\left(2t\right)\right\rangle}{2}=\left\langle-cos\left(2t\right),sen\left(2t\right)\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]N\left(t=0\right)=\left\langle-cos\left(2\left(0\right)\right),-sen\left(2\left(0\right)\right)\right\rangle=\left\langle-1,0\right\rangle[/math][br][br]Entonces, cuando [math]t=\frac{\pi}{4}[/math]:[br][br][math]N\left(t=\frac{\pi}{4}\right)=\left\langle-cos\left(2\left(\frac{\pi}{4}\right)\right),-sen\left(2\left(\frac{\pi}{4}\right)\right)\right\rangle=\left\langle0,-1\right\rangle[/math]

5.-[math]r\left(t\right)=\left\langle cos\left(2t\right),t,sen\left(2t\right)\right\rangle,t=0,t=\frac{\pi}{2}[/math][br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle-2sen\left(2t\right),1,2cos\left(2t\right)\right\rangle[/math][br][br]Entonces, usando la definición del vector tangente:[br][br][math]T\left(t\right)=\frac{r'\left(t\right)}{\parallel r'\left(t\right)\parallel}=\frac{r'\left(t\right)}{\sqrt{\left(-2sen\left(2t\right)\right)^2+1^2+\left(2cos\left(2t\right)\right)^2}}=\frac{r'\left(t\right)}{\sqrt{5}}=\left\langle-\frac{2}{\sqrt{5}}sen\left(2t\right),\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}cos\left(2t\right)\right\rangle[/math][br][br]Entonces, derivamos a [math]T\left(t\right)[/math]:[br][br][math]T'\left(t\right)=\left\langle-\frac{4}{\sqrt{5}}cos\left(2t\right),0,-\frac{4}{\sqrt{5}}sen\left(2t\right)\right\rangle[/math][br][br]Entonces, usando la definición del vector unitario normal, tenemos que:[br][br][math]N\left(t\right)=\frac{T'\left(t\right)}{\parallel T'\left(t\right)\parallel}=\frac{T'\left(t\right)}{\sqrt{\left(-\frac{4}{\sqrt{5}}cos\left(2t\right)\right)^2+\left(0\right)^2+\left(-\frac{4}{\sqrt{5}}sen\left(2t\right)\right)^2}}=\frac{T'\left(t\right)}{\sqrt{\frac{16}{5}\left(cos^2\left(2t\right)+sen^2\left(2t\right)\right)}}=\frac{T'\left(t\right)}{\frac{4}{\sqrt{5}}}=\left\langle-cos\left(2t\right),0,-sen\left(2t\right)\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]N\left(t\right)=\left\langle-cos\left(2\left(0\right)\right),0,-sen\left(2\left(0\right)\right)\right\rangle=\left\langle-1,0,0\right\rangle[/math][br][br]Entonces, cuando [br][br][math]N\left(t\right)=\left\langle-cos\left(2\left(\frac{\pi}{2}\right)\right),0,-sen\left(2\left(\frac{\pi}{2}\right)\right)\right\rangle=\left\langle1,0,0\right\rangle[/math]

7.-[math]r\left(t\right)=\left\langle t,t^2-1,t\right\rangle,t=0,t=1[/math][br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle1,2t,1\right\rangle[/math][br][br]Entonces, usando la definición del vector tangente[br][br][math]T'\left(t\right)=\frac{r'\left(t\right)}{\parallel r'\left(t\right)\parallel}=\frac{r'\left(t\right)}{\sqrt{\left(1\right)^2+\left(2t\right)^2+\left(1\right)^2}}=\frac{r'\left(t\right)}{\sqrt{2+4t^2}}=\left\langle\frac{1}{\sqrt{2+4t^2}},\frac{2t}{\sqrt{2+4t^2}},\frac{1}{\sqrt{2+4t^2}}\right\rangle[/math][br][br]Ahora, derivamos a [math]T'\left(t\right)[/math]:[br][br][math]T'\left(t\right)=\left\langle-\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}},\frac{4}{\left(4t^2+2\right)^{\frac{3}{2}}},-\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}}\right\rangle[/math][br][br]Entonces, usando la definición del vector unitario normal, tenemos que:[br][br][math]N\left(t\right)=\frac{T'\left(t\right)}{\parallel T'\left(t\right)\parallel}=\frac{T'\left(t\right)}{\sqrt{\left(-\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}}\right)}^2+\left(\frac{4}{\left(4t^2+2\right)^{\frac{3}{2}}}\right)^2+\left(-\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}}\right)^2}=\frac{T'\left(t\right)}{\frac{\sqrt{2}}{2t^2+1}}=\left\langle-\frac{\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2t^2+1}},\frac{\frac{4}{\left(4t^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2t^2+1}},-\frac{\frac{4t}{\left(4t^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2t^2+1}}\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]N\left(t=0\right)=\left\langle-\frac{\frac{4\left(0\right)}{\left(4\left(0\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(0\right)^2+1}},\frac{\frac{4}{\left(4\left(0\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(0\right)^2+1}}.-\frac{\frac{4\left(0\right)}{\left(4\left(0\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(0\right)^2+1}}\right\rangle=\left\langle0,1,0\right\rangle[/math][br][br]Ahora, cuanto [math]t=1[/math]:[br][br][math]N\left(t=1\right)=\left\langle-\frac{\frac{4\left(1\right)}{\left(4\left(1\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(1\right)^2+1}},\frac{\frac{4}{\left(4\left(1\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(1\right)^2+1}},-\frac{\frac{4\left(1\right)}{\left(4\left(1\right)^2+2\right)^{\frac{3}{2}}}}{\frac{\sqrt{2}}{2\left(1\right)^2+1}}\right\rangle=\left\langle-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.-\frac{1}{\sqrt{3}}\right\rangle[/math]