Seja \mathbf{r}(t)=\cos t \mathbf{i}+\operatorname{sen} t \mathbf{j}+\mathbf{k}. Determine
b) \lim _{t \rightarrow 0}\left[\mathbf{r}(t) \times \mathbf{r}^{\'}(t)\right]

Passo 1

O enunciado dá o vetor
\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+\mathbf{k}
e pede o limite quando t\to0 do produto vetorial \mathbf{r}(t)\times\mathbf{r}'(t). Vamos usar: (a) derivada de \mathbf{r}(t) e (b) definição de produto vetorial via determinante.

Passo 2

Calculamos a derivada
\mathbf{r}'(t)=\frac{d}{dt}\bigl(\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+\mathbf{k}\bigr) =-\sin t\,\mathbf{i}+\cos t\,\mathbf{j}+0\,\mathbf{k}.

Passo 3

Aplicamos a regra do determinante para o produto vetorial
\mathbf{r}(t)\times\mathbf{r}'(t) =\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\\ \cos t & \sin t & 1\\\\ -\sin t & \cos t & 0 \end{vmatrix}.
Expandindo:
\mathbf{i}\bigl(\sin t\cdot0 - 1\cdot\cos t\bigr) -\mathbf{j}\bigl(\cos t\cdot0 - 1\cdot(-\sin t)\bigr) +\mathbf{k}\bigl(\cos t\cdot\cos t - \sin t\cdot(-\sin t)\bigr).
Isso resulta em
\mathbf{r}(t)\times\mathbf{r}'(t) =-\cos t\,\mathbf{i}-\sin t\,\mathbf{j}+(\cos^2 t+\sin^2 t)\,\mathbf{k}.
Como \cos^2 t+\sin^2 t=1, fica
\mathbf{r}(t)\times\mathbf{r}'(t) =-\cos t\,\mathbf{i}-\sin t\,\mathbf{j}+\mathbf{k}.

Passo 4

Tomando o limite quando t\to0,
\lim_{t\to0}\bigl(-\cos t\,\mathbf{i}-\sin t\,\mathbf{j}+\mathbf{k}\bigr) =-1\,\mathbf{i}-0\,\mathbf{j}+\mathbf{k} =-\mathbf{i}+\mathbf{k}.

Resposta

-\mathbf{i}+\mathbf{k}.

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