# Solución

Apartado a

Esta función describe una circunferencia.

Para encontrar el vector normal principal unitario, primero debemos encontrarel vector tangente unitario $\bold{T}(t)$:

$$\begin{aligned} \bold{T}(t) &= \frac{\bold{r'}(t)}{\|\bold{r'}(t)\|}\\ &= \frac{-4sent\bold{i}-4cost\bold{j}}{\sqrt{(-4sent)^2+(-4cost)^2}}\\ &= \frac{-4sent\bold{i}-4cost\bold{j}}{\sqrt{16sen^2t+16cos^2t}}\\ &= \frac{-4sent\bold{i}-4cost\bold{j}}{\sqrt{16(sen^2t+cos^2t)}}\\ &= \frac{-4sent\bold{i}-4cost\bold{j}}{4}\\ &= -sent\bold{i}-cost\bold{j} \end{aligned}$$

A continuación, usamos la ecuación 3.18:

$$\begin{aligned} \bold{N}(t) &= \frac{\bold{T'}(t)}{\|\bold{T'}(t)\|}\\ &= \frac{-cost\bold{i}+sent\bold{j}}{\sqrt{(-cost)^2+(sent)^2}}\\ &= \frac{-cost\bold{i}+sent\bold{j}}{\sqrt{sen^2t+cos^2t}}\\ &= -cost\bold{i}+sent\bold{j} \end{aligned}$$

Observa que el vector tangente unitario y el vector normal principal unitario son ortogonales entre sí para todos los valores de $t$:

$$\begin{aligned} \bold{T}(t)\cdot\bold{N}(t) &= \lang −sent,−cost\rang\cdot\lang −cost,sent\rang\\ &= sentcost−costsent\\ &= 0 \end{aligned}$$

Además, el vector normal principal unitario apunta hacia el centro del círculo desde cada punto del círculo. Dado que $\bold{r}(t)$ define una curva en dos dimensiones, no podemos calcular el vector binormal.

Apartado b

Esta función se ve así:

Para encontrar el vector normal principal unitario, primero debemos encontrarel vector tangente unitario $\bold{T}(t)$:

$$\begin{aligned} \bold{T}(t) &= \frac{\bold{r'}(t)}{\|\bold{r'}(t)\|}\\ &= \frac{6\bold{i}+10t\bold{j}−8\bold{k}}{\sqrt{6^2+(10t)^2+(-8)^2}}\\ &= \frac{6\bold{i}+10t\bold{j}−8\bold{k}}{\sqrt{36+100t^2+64}}\\ &= \frac{6\bold{i}+10t\bold{j}−8\bold{k}}{\sqrt{100(t^2+1)}}\\ &= \frac{3\bold{i}+5t\bold{j}−4\bold{k}}{5\sqrt{t^2+1}}\\ &= \frac{3}{5}(t^2+1)^{-1/2}\bold{i} - t(t^2+1)^{-1/2}\bold{j} - \frac{4}{5}(t^2+1)^{-1/2}\bold{k} \end{aligned}$$

A continuación, calculamos $\bold{T}'(t)$ y $\|\bold{T}'(t)\|$:

$$\begin{aligned} \bold{T}'(t) &= \frac{3}{5}\big(-\frac{1}{2}\big)(t^2+1)^{-3/2}(2t)\bold{i} &+\bigg((t^2+1)^{-1/2} -t\big(\frac{1}{2}\big)(t^2+1)^{-3/2}(2t)\bold{j}\\ & -\frac{4}{5}\big(-\frac{1}{2}\big)(t^2+1)^{-3/2}(2t)\bold{k}\\ &= -\frac{3t}{5(t^2+1)^{3/2}}\bold{i} - \frac{1}{(t^2+1)^{3/2}}\bold{j} + \frac{4t}{5(t^2+1)^{3/2}}\bold{k}\\ \|\bold{T′}(t)\| &= \sqrt{\Bigg(-\frac{3t}{5(t^2+1)^{3/2}} \Bigg)^2 + \Bigg(- \frac{1}{(t^2+1)^{3/2}} \Bigg)^2 + \Bigg(\frac{4t}{5(t^2+1)^{3/2}} \Bigg)^2}\\ &= \sqrt{\frac{9t^2}{25(t^2+1)^3} + \frac{1}{(t^2+1)^3} + \frac{16t^2}{25(t^2+1)^3}}\\ &= \sqrt{\frac{25t^2+25}{25(t^2+1)^3}}\\ &= \sqrt{\frac{1}{(t^2+1)^2}}\\ &= \frac{1}{t^2+1} \end{aligned}$$

Por lo tanto, de acuerdo con la Ecuación 3.18:

$$\begin{aligned} \bold{N}(t) &= \frac{\bold{T'}(t)}{\|\bold{T'}(t)\|}\\ &= \Bigg(-\frac{3t}{5(t^2+1)^{3/2}}\bold{i} - \frac{1}{(t^2+1)^{3/2}}\bold{j} + \frac{4t}{5(t^2+1)^{3/2}}\bold{k}\Bigg)(t^2+1)\\ &= -\frac{3t}{5(t^2+1)^{1/2}}\bold{i} - \frac{5}{5(t^2+1)^{1/2}}\bold{j} +\frac{4t}{5(t^2+1)^{1/2}}\bold{k}\\ &= -\frac{3t\bold{i}+5\bold{j}−4t\bold{k}}{5\sqrt{t^2+1}} \end{aligned}$$

Una vez más, el vector tangente unitario y el vector normal principal unitario son ortogonales entre sí para todos los valores de $t$:

$$\begin{aligned} \bold{T}(t)\cdot\bold{N}(t) &= \bigg(\frac{3\bold{i}−5t\bold{j}−4\bold{k}}{5\sqrt{t^2+1}}\bigg)\cdot\bigg(-\frac{3t\bold{i}+5\bold{j}−4t\bold{k}}{5\sqrt{t^2+1}}\bigg)\\ &= \frac{3(−3t)−5t(−5)−4(4t)}{25\big(t^2+1\big)}\\ &= \frac{−9t+25t−16t}{25\big(t^2+1\big)}\\ &= 0 \end{aligned}$$

Por último, dado que $\bold{r}(t)$ representa una curva tridimensional, podemos calcular el vector binormal utilizando la Ecuación 3.17:

$$\begin{aligned} \bold{B}(t) &= \bold{T}(t)\times\bold{N}(t)\\ &= \begin{vmatrix} \bold{i} & \bold{j} & \bold{k}\\ \\ \frac{3}{5\sqrt{t^2+1}} & -\frac{5t}{5\sqrt{t^2+1}} & -\frac{4}{5\sqrt{t^2+1}}\\ \\ -\frac{3t}{5\sqrt{t^2+1}} & -\frac{5}{5\sqrt{t^2+1}} & \frac{4t}{5\sqrt{t^2+1}} \end{vmatrix}\\ &= \Bigg(\bigg(-\frac{5t}{5\sqrt{t^2+1}}\bigg)\bigg(\frac{4t}{5\sqrt{t^2+1}}\bigg) - \bigg(-\frac{4}{5\sqrt{t^2+1}}\bigg)\bigg(-\frac{5}{5\sqrt{t^2+1}}\bigg)\Bigg)\bold{i}\\ & \;\;\;\;-\Bigg(\bigg(\frac{3}{5\sqrt{t^2+1}}\bigg)\bigg(\frac{4t}{5\sqrt{t^2+1}}\bigg) - \bigg(-\frac{4}{5\sqrt{t^2+1}}\bigg)\bigg(-\frac{3t}{5\sqrt{t^2+1}}\bigg)\Bigg)\bold{j}\\ & \;\;\;\; +\Bigg(\bigg(\frac{3}{5\sqrt{t^2+1}}\bigg)\bigg(-\frac{5}{5\sqrt{t^2+1}}\bigg) - \bigg(-\frac{5t}{5\sqrt{t^2+1}}\bigg)\bigg(-\frac{3t}{5\sqrt{t^2+1}}\bigg)\Bigg)\bold{k}\\ &= \bigg(\frac{-20t^2-20}{25(t^2+1)}\bigg)\bold{i} +\bigg(\frac{-15-15t^2}{25(t^2+1)}\bigg)\bold{k}\\ &= -20\bigg(\frac{t^2+1}{25(t^2+1)}\bigg)\bold{i} -15\bigg(\frac{t^2+1}{25(t^2+1)}\bigg)\bold{k}\\ &= -\frac{4}{5}\bold{i} - \frac{3}{5}\bold{k} \end{aligned}$$