Definimos el vector [b]binormal[/b] [math]B(t)[/math] como:[br][br] [math]B\left(t\right)=T\left(t\right)\times N\left(t\right)[/math][br][br][br][size=150][color=#ff0000]Encuentre el vector binormal [/color][math]B(t)=T(t)×N(t)[/math][color=#ff0000] en [/color][math]t=0[/math][color=#ff0000] y [/color][math]t=1[/math][color=#ff0000]. Además, dibuje la curva trazada por [/color][math]r(t)[/math][color=#ff0000] y los vectores [/color][math]T[/math][color=#ff0000], [/color][math]N[/math][color=#ff0000] y[/color][math]B[/math][color=#ff0000] en estos puntos.[/color][/size]

21.-[math]r\left(t\right)=\left\langle t,2t,t^2\right\rangle[/math][br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle1,2,2t\right\rangle[/math][br][br]Ahora, buscamos el vector [math]T\left(t\right)[/math]:[br][br][math]T\left(t\right)=\frac{r'\left(t\right)}{\parallel r'\left(t\right)\parallel}=\frac{r'\left(t\right)}{\sqrt{1^2+2^2+\left(2t\right)^2}}=\frac{r'\left(t\right)}{\sqrt{5+4t^2}}=\left\langle\frac{1}{\sqrt{5+4t^2}},\frac{2}{\sqrt{5+4t^2}},\frac{2t}{\sqrt{5+4t^2}}\right\rangle[/math][br][br]Ahora, derivamos a [math]T'\left(r\right)[/math]:[br][br][math]T'\left(t\right)=\left\langle-\frac{4t}{\left(5+4t^2\right)^{\frac{3}{2}}},-\frac{8t}{\left(5+4t^2\right)^{\frac{3}{2}}},\frac{10}{\left(5+4t^2\right)\sqrt{5+4t^2}}\right\rangle[/math][br][br]Ahora, buscamos el vector [math]N\left(t\right)[/math]:[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(5+4t^2\right)^{\frac{3}{2}}}\right)^2+\left(-\frac{8t}{\left(5+4t^2\right)^{\frac{3}{2}}}\right)^2+\left(\frac{10}{\left(5+4t^2\right)\sqrt{5+4t^2}}\right)}}=\frac{T'\left(t\right)}{\frac{2\sqrt{5}}{4t^2+5}}=\left\langle\frac{-\frac{4t}{\left(5+4t^2\right)^{\frac{3}{2}}}}{\frac{2\sqrt{5}}{4t^2+5}},\frac{-\frac{8t}{\left(5+4t^2\right)^{\frac{3}{2}}}}{\frac{2\sqrt{5}}{4t^2+5}},\frac{\frac{10}{\left(5+4t^2\right)\sqrt{5+4t^2}}}{\frac{2\sqrt{5}}{4t^2+5}}\right\rangle[/math][br][br]simplificamos:[br][br][math]N\left(t\right)=\left\langle\frac{-2t}{\left(4t^2+5\right)^{\frac{1}{2}}\sqrt{5}},\frac{-4t}{\left(4t^2+5\right)^{\frac{1}{2}}\sqrt{5}},\frac{\sqrt{5}}{\sqrt{5+4t^2}}\right\rangle[/math][br][br]Ahora, buscamos el vector [math]B\left(t\right)[/math]:[br][br][math]B\left(t\right)=T\left(t\right)\times N\left(t\right)=\left\langle\frac{2\sqrt{5}}{5+4t^2}+\frac{8t^2}{\sqrt{5+4t^2}\left(5+4t^2\right)^{\frac{1}{2}}\sqrt{5}},\frac{\sqrt{5}}{5+4t^2}+\frac{4t^2}{\sqrt{5+4t^2}\left(5+4t^2\right)^{\frac{1}{2}}\sqrt{5}},0\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]B\left(t=0\right)=\left\langle\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}},0\right\rangle[/math][br][br]Ahora, cuando [math]t=1[/math]:[br][br][math]B\left(t=1\right)=\left\langle\frac{2\sqrt{5}}{9}+\frac{8}{9\sqrt{5}},\frac{\sqrt{5}}{9}+\frac{4}{9\sqrt{5}},0\right\rangle[/math][br][br]

23.-[math]r\left(t\right)=\left\langle4cos\left(\pi t\right),4sen\left(\pi t\right),t\right\rangle[/math][br][br]Derivamos a [math]r\left(t\right)[/math]:[br][br][math]r'\left(t\right)=\left\langle-4\pi sen\left(\pi t\right),4\pi cos\left(\pi t\right),1\right\rangle[/math][br][br]Ahora, buscamos al vector [math]T\left(t\right)[/math]:[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(-4\pi sen\left(\pi t\right)\right)^2+\left(4\pi cos\left(\pi t\right)\right)^2+1^2}}=\frac{r'\left(t\right)}{\sqrt{16\pi^2+1}}==\left\langle-\frac{4\pi sen\left(\pi t\right)}{\sqrt{16\pi^2+1}},\frac{4\pi cos\left(\pi t\right)}{\sqrt{16\pi^2+1}},\frac{1}{\sqrt{16\pi^2+1}}\right\rangle[/math][br][br]Derivamos a [math]T'\left(t\right)[/math]:[br][br] [math]T'\left(t\right)=\left\langle-\frac{4\pi^2}{\sqrt{16\pi^2+1}}cos\left(\pi t\right),-\frac{4\pi^2}{\sqrt{16\pi^2+1}}sen\left(\pi t\right),0\right\rangle[/math][br][br]Ahora, buscamos al vector [math]N\left(t\right)[/math][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\pi^2}{\sqrt{16\pi^2+1}}cos\left(\pi t\right)\right)^2+\left(-\frac{4\pi^2}{\sqrt{16\pi^2+1}}sen\left(\pi t\right)\right)^2}}=\frac{T'\left(t\right)}{\frac{4\pi\sqrt{16\pi^2+1}}{16\pi^2+116\pi^2+1}}=\left\langle\frac{-\frac{4\pi^2}{\sqrt{16\pi^2+1}}cos\left(\pi t\right)}{\frac{4\pi\sqrt{16\pi^2+1}}{16\pi^2+116\pi^2+1}},\frac{-\frac{4\pi^2}{\sqrt{16\pi^2+1}}sen\left(\pi t\right)}{\frac{4\pi\sqrt{16\pi^2+1}}{16\pi^2+116\pi^2+1}},0\right\rangle[/math][br][br]Simplificamos:[br][br][math]N\left(t\right)=\left\langle\frac{-\pi\left(132\pi^2+1\right)cos\left(\pi t\right)}{16\pi^2+1},\frac{-\pi\left(132\pi^2+1\right)sen\left(\pi t\right)}{16\pi^2+1},0\right\rangle[/math][br][br]Por último, buscamos al vector [math]B\left(t\right)[/math]:[br][br][math]B\left(t\right)=\left\langle\frac{\pi\left(132\pi^2+1\right)\sqrt{16\pi^2+1}sen\left(\pi t\right)}{\left(16\pi^2+1\right)^2},\frac{-\pi\left(132\pi^2+1\right)\sqrt{16\pi^2+1}cos\left(\pi t\right)}{\left(16\pi^2+1\right)^2},\frac{4\pi^2\left(132\pi^2+1\right)\sqrt{16\pi^2+1}}{\left(16\pi^2+1\right)^2}\right\rangle[/math][br][br]Entonces, cuando [math]t=0[/math]:[br][br][math]B\left(t=0\right)=\left\langle0,-\frac{132\pi^3}{\left(16\pi^2+1\right)^{\frac{3}{2}}}-\frac{\pi}{\left(16\pi^2+1\right)^{\frac{3}{2}}},\frac{528\pi^4}{\left(16\pi^2+1\right)^{\frac{3}{2}}}+\frac{4\pi^2}{\left(16\pi^2+1\right)^{\frac{3}{2}}}\right\rangle[/math][br][br]Por último, cuando [math]t=1[/math]:[br][br][math]B\left(t=1\right)=\left\langle0,\frac{132\pi^3}{\left(16\pi^2+1\right)^{\frac{3}{2}}}+\frac{\pi}{\left(16\pi^2+1\right)^{\frac{3}{2}}},\frac{528\pi^4}{\left(16\pi^2+1\right)^{\frac{3}{2}}}+\frac{4\pi^2}{\left(16\pi^2+1\right)^{\frac{3}{2}}}\right\rangle[/math]