Se \mathbf{r}(t)=\left\langle t, t^{2}, t^{3}\right\rangle, encontre \mathbf{r}^{\prime}(t), \mathbf{T}(1), \mathbf{r}^{\prime \prime}(t) e \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t).

Passo 1

O enunciado dá o vetor posição
\mathbf{r}(t)=\langle t,\;t^{2},\;t^{3}\rangle
e pede:
– a derivada \mathbf{r}'(t),
– o versor tangente \mathbf{T}(1) em t=1,
– a segunda derivada \mathbf{r}''(t),
– o produto vetorial \mathbf{r}'(t)\times\mathbf{r}''(t).
Para isso vamos aplicar derivação componente a componente, usar a definição de versor tangente \mathbf{T}(t)=\mathbf{r}'(t)/\|\mathbf{r}'(t)\|, e a fórmula do produto vetorial em \mathbb{R}^3.

Passo 2

Derivando componente a componente, temos
\mathbf{r}'(t) =\bigl\langle \tfrac{d}{dt}(t),\;\tfrac{d}{dt}(t^{2}),\;\tfrac{d}{dt}(t^{3})\bigr\rangle =\langle 1,\;2t,\;3t^{2}\rangle.

Passo 3

Calculemos o versor tangente \mathbf{T}(t). Primeiro o módulo de \mathbf{r}'(t):
\|\mathbf{r}'(t)\| =\sqrt{1^{2}+(2t)^{2}+(3t^{2})^{2}} =\sqrt{1+4t^{2}+9t^{4}}.
Logo
\mathbf{T}(t) =\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} =\frac{\langle 1,\,2t,\,3t^{2}\rangle}{\sqrt{1+4t^{2}+9t^{4}}}.
Em t=1 temos \mathbf{r}'(1)=\langle1,2,3\rangle e \|\mathbf{r}'(1)\|=\sqrt{1+4+9}=\sqrt{14}, portanto
\mathbf{T}(1) =\bigl\langle\tfrac{1}{\sqrt{14}},\,\tfrac{2}{\sqrt{14}},\,\tfrac{3}{\sqrt{14}}\bigr\rangle.

Passo 4

Derivando novamente \mathbf{r}'(t) componente a componente, obtemos
\mathbf{r}''(t) =\bigl\langle \tfrac{d}{dt}(1),\;\tfrac{d}{dt}(2t),\;\tfrac{d}{dt}(3t^{2})\bigr\rangle =\langle 0,\;2,\;6t\rangle.

Passo 5

Para o produto vetorial, usamos a matriz determinante
\mathbf{r}'(t)\times\mathbf{r}''(t) =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\\\ 1 & 2t & 3t^{2}\\\\ 0 & 2 & 6t \end{vmatrix}.
Calculando,
\mathbf{r}'(t)\times\mathbf{r}''(t) =\langle (2t)(6t)-(3t^{2})(2),\; -\bigl(1\cdot6t-3t^{2}\cdot0\bigr),\; 1\cdot2-(2t)\cdot0 \rangle =\langle 6t^{2},\;-6t,\;2\rangle.

Resposta

\displaystyle \mathbf{r}'(t)=\langle1,\,2t,\,3t^{2}\rangle \displaystyle \mathbf{T}(1)=\bigl\langle\tfrac{1}{\sqrt{14}},\,\tfrac{2}{\sqrt{14}},\,\tfrac{3}{\sqrt{14}}\bigr\rangle \displaystyle \mathbf{r}''(t)=\langle0,\,2,\,6t\rangle \displaystyle \mathbf{r}'(t)\times\mathbf{r}''(t)=\langle6t^{2},\,-6t,\,2\rangle

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