Questão 2 (3 pontos)

Obtenha os vetores tangente, normal e binormal unitários da curva dada no ponto dado.
\boldsymbol{r}(t)=4 \cos t \boldsymbol{i}+4 \operatorname{sen} t \boldsymbol{j}+3 t \boldsymbol{k} \quad t=\frac{\pi}{6}

Passo 1

O enunciado pede os vetores tangente unitário \mathbf{T}, normal unitário \mathbf{N} e binormal unitário \mathbf{B} para a curva
\mathbf{r}(t)=4\cos t\,\mathbf{i}+4\sin t\,\mathbf{j}+3t\,\mathbf{k}
no ponto t=\pi/6. Vamos usar cálculo de derivadas para achar \mathbf{r}'(t), depois definir \mathbf{T}(t)=\mathbf{r}'(t)/\|\mathbf{r}'(t)\|, em seguida \mathbf{N}(t)=\mathbf{T}'(t)/\|\mathbf{T}'(t)\| e finalmente \mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t).

Passo 2

Derivamos \mathbf{r}(t):
\mathbf{r}'(t)=-4\sin t\,\mathbf{i}+4\cos t\,\mathbf{j}+3\,\mathbf{k}.

Passo 3

O módulo é
\|\mathbf{r}'(t)\|=\sqrt{(-4\sin t)^{2}+(4\cos t)^{2}+3^{2}} =\sqrt{16(\sin^{2}t+\cos^{2}t)+9} =\sqrt{16+9} =5.
Logo o vetor tangente unitário é
\mathbf{T}(t) =\frac{\mathbf{r}'(t)}{5} =\Bigl(-\tfrac{4}{5}\sin t,\;\tfrac{4}{5}\cos t,\;\tfrac{3}{5}\Bigr).

Passo 4

Para achar o vetor normal, derivamos \mathbf{T}(t):
\mathbf{T}'(t) =\Bigl(-\tfrac{4}{5}\cos t,\;-\tfrac{4}{5}\sin t,\;0\Bigr).
O módulo é
\|\mathbf{T}'(t)\| =\sqrt{\Bigl(-\tfrac{4}{5}\cos t\Bigr)^{2}+\Bigl(-\tfrac{4}{5}\sin t\Bigr)^{2}} =\sqrt{\tfrac{16}{25}(\cos^{2}t+\sin^{2}t)} =\tfrac{4}{5}.
Então
\mathbf{N}(t) =\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} =\Bigl(-\tfrac{4}{5}\cos t,\;-\tfrac{4}{5}\sin t,\;0\Bigr)\div\tfrac{4}{5} =\Bigl(-\tfrac{1}{1}\cos t,\;-\tfrac{1}{1}\sin t,\;0\Bigr).

Passo 5

Agora avaliamos em t=\pi/6. Como \sin(\pi/6)=\tfrac12 e \cos(\pi/6)=\tfrac{\sqrt3}{2}, obtemos
\mathbf{T}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(-\tfrac{4}{5}\cdot\tfrac12,\;\tfrac{4}{5}\cdot\tfrac{\sqrt3}{2},\;\tfrac{3}{5}\Bigr) =\Bigl(-\tfrac{2}{5},\;\tfrac{2\sqrt3}{5},\;\tfrac{3}{5}\Bigr),
\mathbf{N}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(-\tfrac{\sqrt3}{2},\;-\tfrac{1}{2},\;0\Bigr).

Passo 6

Finalmente o binormal em t=\pi/6 é
\mathbf{B}\bigl(\tfrac{\pi}{6}\bigr) =\mathbf{T}\bigl(\tfrac{\pi}{6}\bigr)\times\mathbf{N}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(\tfrac{3}{10},\;-\tfrac{3\sqrt3}{10},\;\tfrac{4}{5}\Bigr).

Resposta

\displaystyle \mathbf{T}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(-\tfrac{2}{5},\;\tfrac{2\sqrt3}{5},\;\tfrac{3}{5}\Bigr),\quad \mathbf{N}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(-\tfrac{\sqrt3}{2},\;-\tfrac12,\;0\Bigr),\quad \mathbf{B}\bigl(\tfrac{\pi}{6}\bigr) =\Bigl(\tfrac{3}{10},\;-\tfrac{3\sqrt3}{10},\;\tfrac{4}{5}\Bigr).

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