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

Temos a curva
\vec{r}(t)=4\sin(t)\,\vec{i}+3t\,\vec{j}+4\cos(t)\,\vec{k},
e queremos calcular a curvatura \kappa e o triêdrio de Frenet \{\vec{T},\vec{N},\vec{B}\}.
Conteúdo usado: derivadas de vetor, norma, produto vetorial e fórmulas \vec{T}=\frac{\vec r'}{\|\vec r'\|},\;\kappa=\frac{\|\vec r'\times\vec r''\|}{\|\vec r'\|^3},\;\vec N=\frac{\vec T'}{\|\vec T'\|},\;\vec B=\vec T\times\vec N.

Passo 2

Calculemos primeiro
\vec r'(t) = \bigl(4\cos t,\;3,\;-4\sin t\bigr),
\|\vec r'(t)\| = \sqrt{(4\cos t)^2+3^2+(-4\sin t)^2} =\sqrt{16(\cos^2t+\sin^2t)+9}=\sqrt{25}=5.
Logo
\vec T(t)=\frac{\vec r'(t)}{\|\vec r'(t)\|} =\Bigl(\tfrac{4}{5}\cos t,\;\tfrac{3}{5},\;-\tfrac{4}{5}\sin t\Bigr).

Passo 3

A segunda derivada é
\vec r''(t)=\bigl(-4\sin t,\;0,\;-4\cos t\bigr).
O produto vetorial
\vec r'(t)\times\vec r''(t) =\bigl(-12\cos t,\;16,\;12\sin t\bigr),
tem norma
\|\vec r'\times\vec r''\| =\sqrt{(-12\cos t)^2+16^2+(12\sin t)^2} =\sqrt{144(\cos^2t+\sin^2t)+256} =\sqrt{400}=20.
Portanto a curvatura é
\kappa(t)=\frac{\|\vec r'\times\vec r''\|}{\|\vec r'\|^3} =\frac{20}{5^3}=\frac{20}{125}=\frac{4}{25}.

Passo 4

Para achar \vec N(t) diferenciamos \vec T(t):
\vec T'(t) =\Bigl(-\tfrac{4}{5}\sin t,\;0,\;-\tfrac{4}{5}\cos t\Bigr),
\|\vec T'(t)\| =\sqrt{\Bigl(-\tfrac{4}{5}\sin t\Bigr)^2+\Bigl(-\tfrac{4}{5}\cos t\Bigr)^2} =\frac{4}{5}.
Logo
\vec N(t)=\frac{\vec T'(t)}{\|\vec T'(t)\|} =\bigl(-\sin t,\;0,\;-\cos t\bigr).

Passo 5

Finalmente
\vec B(t)=\vec T(t)\times\vec N(t) =\Bigl(-\tfrac{3}{5}\cos t,\;\tfrac{4}{5},\;\tfrac{3}{5}\sin t\Bigr),
que fecha o sistema ortonormal de Frenet.

Passo 6

Agora avaliamos em
t_0=\frac{\sqrt{2}}{2}.
Denotando \alpha=\cos\!\bigl(\tfrac{\sqrt2}{2}\bigr) e \beta=\sin\!\bigl(\tfrac{\sqrt2}{2}\bigr), temos
\kappa(t_0)=\frac{4}{25},
\vec T(t_0) =\Bigl(\tfrac{4}{5}\,\alpha,\;\tfrac{3}{5},\;-\tfrac{4}{5}\,\beta\Bigr),
\vec N(t_0) =\bigl(-\beta,\;0,\;-\alpha\bigr),
\vec B(t_0) =\Bigl(-\tfrac{3}{5}\,\alpha,\;\tfrac{4}{5},\;\tfrac{3}{5}\,\beta\Bigr).

Resposta

\displaystyle \kappa=\tfrac{4}{25}, \displaystyle \vec T(t_0)=\bigl(\tfrac{4}{5}\cos(\tfrac{\sqrt2}{2}),\;\tfrac{3}{5},\;-\tfrac{4}{5}\sin(\tfrac{\sqrt2}{2})\bigr), \displaystyle \vec N(t_0)=\bigl(-\sin(\tfrac{\sqrt2}{2}),\;0,\;-\cos(\tfrac{\sqrt2}{2})\bigr), \displaystyle \vec B(t_0)=\bigl(-\tfrac{3}{5}\cos(\tfrac{\sqrt2}{2}),\;\tfrac{4}{5},\;\tfrac{3}{5}\sin(\tfrac{\sqrt2}{2})\bigr).

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