Questão 1. Calcule a curvatura, bem como os vetores \vec{T}, \vec{N} e \vec{B}, da curva
\vec{r}(t)=4 \sin (t) \vec{i}+3 t \vec{j}+4 \cos (t) \vec{k}
Avalie todos esses elementos no ponto que corresponde a t=\frac{\sqrt{2}}{2}.

Passo 1

O enunciado dá a curva no espaço
\vec{r}(t)=4\sin(t)\,\vec{i}+3t\,\vec{j}+4\cos(t)\,\vec{k}
e pede a curvatura \kappa e os vetores de Frenet \vec{T},\vec{N},\vec{B} avaliados em t_{0}=\tfrac{\sqrt{2}}{2}. O conteúdo é geometria diferencial de curvas: vamos precisar de derivadas, produto vetorial e definição de curvatura e do triêdron de Frenet.

Passo 2

Calcular as derivadas primeiras e segundas:
\vec{r}\,'(t)=\frac{d\vec{r}}{dt} =(4\cos t)\,\vec{i} +3\,\vec{j}+(-4\sin t)\,\vec{k},
\vec{r}\,''(t)=\frac{d^{2}\vec{r}}{dt^{2}} =(-4\sin t)\,\vec{i}+0\,\vec{j}+(-4\cos t)\,\vec{k}.

Passo 3

Calcular a velocidade \|\vec{r}\,'(t)\| e o vetor tangente unitário
\|\vec{r}\,'(t)\| =\sqrt{(4\cos t)^{2}+3^{2}+(-4\sin t)^{2}} =\sqrt{16(\cos^{2}t+\sin^{2}t)+9} =\sqrt{25}=5,
\vec{T}(t) =\frac{\vec{r}\,'(t)}{\|\vec{r}\,'(t)\|} =\frac{1}{5}\bigl(4\cos t,\;3,\;-4\sin t\bigr).

Passo 4

Calcular o produto vetorial e a curvatura
\vec{r}\,'(t)\times\vec{r}\,''(t) =\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\\\ 4\cos t&3&-4\sin t\\\\ -4\sin t&0&-4\cos t \end{vmatrix} =(-12\cos t,\,16,\,12\sin t),
\bigl\|\vec{r}\,'\times\vec{r}\,''\bigr\| =\sqrt{(-12\cos t)^{2}+16^{2}+(12\sin t)^{2}} =\sqrt{144+256} =20,
\kappa(t) =\frac{\|\vec{r}\,'\times\vec{r}\,''\|}{\|\vec{r}\,'\|^{3}} =\frac{20}{5^{3}} =\frac{4}{25}.

Passo 5

Calcular o vetor normal principal
\vec{T}\,'(t) =\frac{d}{dt}\biggl(\frac{4\cos t}{5},\frac{3}{5},-\frac{4\sin t}{5}\biggr) =\frac{1}{5}(-4\sin t,0,-4\cos t),
\bigl\|\vec{T}\,'(t)\bigr\| =\frac{1}{5}\sqrt{16(\sin^{2}t+\cos^{2}t)} =\frac{4}{5},
\vec{N}(t) =\frac{\vec{T}\,'(t)}{\|\vec{T}\,'(t)\|} =\frac{1/5(-4\sin t,0,-4\cos t)}{4/5} =(-\sin t,\,0,\,-\cos t).

Passo 6

Calcular o binormal
\vec{B}(t) =\vec{T}(t)\times\vec{N}(t) =\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\\ \frac{4\cos t}{5}&\frac{3}{5}&-\frac{4\sin t}{5}\\\\ -\sin t&0&-\cos t \end{vmatrix} =\Bigl(-\tfrac{3\cos t}{5},\;\tfrac{4}{5},\;\tfrac{3\sin t}{5}\Bigr).

Passo 7

Avaliar em t_{0}=\tfrac{\sqrt{2}}{2}:
\kappa\Bigl(\tfrac{\sqrt{2}}{2}\Bigr)=\frac{4}{25},
\vec{T}\Bigl(\tfrac{\sqrt{2}}{2}\Bigr) =\Bigl(\tfrac{4\cos(\frac{\sqrt{2}}{2})}{5},\;\tfrac{3}{5},\; -\tfrac{4\sin(\frac{\sqrt{2}}{2})}{5}\Bigr),
\vec{N}\Bigl(\tfrac{\sqrt{2}}{2}\Bigr) =\bigl(-\sin(\tfrac{\sqrt{2}}{2}),\,0,\,-\cos(\tfrac{\sqrt{2}}{2})\bigr),
\vec{B}\Bigl(\tfrac{\sqrt{2}}{2}\Bigr) =\Bigl(-\tfrac{3\cos(\frac{\sqrt{2}}{2})}{5},\;\tfrac{4}{5},\; \tfrac{3\sin(\frac{\sqrt{2}}{2})}{5}\Bigr).

Resposta

\displaystyle \kappa=\tfrac{4}{25},\quad \vec{T}(t_{0})=\bigl(\tfrac{4\cos(\tfrac{\sqrt{2}}{2})}{5},\tfrac{3}{5},-\tfrac{4\sin(\tfrac{\sqrt{2}}{2})}{5}\bigr), \displaystyle \vec{N}(t_{0})=\bigl(-\sin(\tfrac{\sqrt{2}}{2}),0,-\cos(\tfrac{\sqrt{2}}{2})\bigr), \quad \vec{B}(t_{0})=\bigl(-\tfrac{3\cos(\tfrac{\sqrt{2}}{2})}{5},\tfrac{4}{5},\tfrac{3\sin(\tfrac{\sqrt{2}}{2})}{5}\bigr).

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