calcule \frac{\mathrm{d}}{\mathrm{d} t}\left[\mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}(t)\right] e \frac{\mathrm{d}}{\mathrm{d} t}\left[\mathbf{r}_{1}(t) \times \mathbf{r}_{2}(t)\right].
\mathbf{r}_{1}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}+t^{3} \mathbf{k} ; \mathbf{r}_{2}(t)=t^{4} \mathbf{k}

Passo 1

O enunciado pede para calcular duas derivadas em cálculo vetorial: a derivada do produto escalar \mathbf{r}_{1}(t)\cdot\mathbf{r}_{2}(t) e a derivada do produto vetorial \mathbf{r}_{1}(t)\times\mathbf{r}_{2}(t). Vamos usar as regras de derivação de produtos escalar e vetorial:
\frac{\mathrm{d}}{\mathrm{d}t}\bigl(\mathbf{a}\cdot\mathbf{b}\bigr) =\mathbf{a}'\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{b}',
\frac{\mathrm{d}}{\mathrm{d}t}\bigl(\mathbf{a}\times\mathbf{b}\bigr) =\mathbf{a}'\times\mathbf{b}+\mathbf{a}\times\mathbf{b}'.

Passo 2

Primeiro, calculamos as derivadas dos vetores dados:
\mathbf{r}_{1}(t)=2t\,\mathbf{i}+3t^{2}\,\mathbf{j}+t^{3}\,\mathbf{k} \quad\Longrightarrow\quad \mathbf{r}_{1}'(t)=2\,\mathbf{i}+6t\,\mathbf{j}+3t^{2}\,\mathbf{k},
\mathbf{r}_{2}(t)=t^{4}\,\mathbf{k} \quad\Longrightarrow\quad \mathbf{r}_{2}'(t)=4t^{3}\,\mathbf{k}.

Passo 3

Usando a regra do produto escalar,
\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}\cdot\mathbf{r}_{2}\bigr] =\mathbf{r}_{1}'\cdot\mathbf{r}_{2}+\mathbf{r}_{1}\cdot\mathbf{r}_{2}'.
Calculamos cada termo:
\mathbf{r}_{1}'(t)\cdot\mathbf{r}_{2}(t) =(2\,\mathbf{i}+6t\,\mathbf{j}+3t^{2}\,\mathbf{k})\cdot(0\,\mathbf{i}+0\,\mathbf{j}+t^{4}\,\mathbf{k}) =3t^{2}\cdot t^{4}=3t^{6},
\mathbf{r}_{1}(t)\cdot\mathbf{r}_{2}'(t) =(2t\,\mathbf{i}+3t^{2}\,\mathbf{j}+t^{3}\,\mathbf{k})\cdot(0\,\mathbf{i}+0\,\mathbf{j}+4t^{3}\,\mathbf{k}) =t^{3}\cdot4t^{3}=4t^{6}.
Somando os dois termos,
\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}\cdot\mathbf{r}_{2}\bigr] =3t^{6}+4t^{6}=7t^{6}.

Passo 4

Agora a regra do produto vetorial,
\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}\times\mathbf{r}_{2}\bigr] =\mathbf{r}_{1}'\times\mathbf{r}_{2}+\mathbf{r}_{1}\times\mathbf{r}_{2}'.
Calculamos cada produto:
\mathbf{r}_{1}'(t)\times\mathbf{r}_{2}(t) =(2\,\mathbf{i}+6t\,\mathbf{j}+3t^{2}\,\mathbf{k})\times(t^{4}\,\mathbf{k}) =6t^{5}\,\mathbf{i}-2t^{4}\,\mathbf{j},
\mathbf{r}_{1}(t)\times\mathbf{r}_{2}'(t) =(2t\,\mathbf{i}+3t^{2}\,\mathbf{j}+t^{3}\,\mathbf{k})\times(4t^{3}\,\mathbf{k}) =12t^{5}\,\mathbf{i}-8t^{4}\,\mathbf{j}.
Somando:
\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}\times\mathbf{r}_{2}\bigr] =(6t^{5}+12t^{5})\,\mathbf{i}-(2t^{4}+8t^{4})\,\mathbf{j} =18t^{5}\,\mathbf{i}-10t^{4}\,\mathbf{j}.

Resposta

\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}(t)\cdot\mathbf{r}_{2}(t)\bigr]=7t^{6}. \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\bigl[\mathbf{r}_{1}(t)\times\mathbf{r}_{2}(t)\bigr] =18t^{5}\,\mathbf{i}-10t^{4}\,\mathbf{j}.

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