Se \mathbf{r}(t)=\left\langle e^{2 t}, e^{-2 t}, t e^{2 t}\right\rangle, encontre \mathbf{T}(0), \mathbf{r}^{\prime \prime}(0) e \mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime \prime}(t).

Passo 1

O enunciado nos dá a curva \mathbf{r}(t)=\langle e^{2t},\,e^{-2t},\,t\,e^{2t}\rangle
e pede:
– o vetor tangente unitário \mathbf{T}(0) no ponto t=0,
– a segunda derivada \mathbf{r}''(0),
– o produto escalar \mathbf{r}'(t)\cdot\mathbf{r}''(t).
Para isso, vamos primeiro obter \mathbf{r}'(t), depois sua norma para achar \mathbf{T}(t), em seguida \mathbf{r}''(t) e finalmente o produto escalar pedido.

Passo 2

Calcular a primeira derivada \mathbf{r}'(t) componente a componente:
\mathbf{r}'(t) = \bigl\langle \tfrac{d}{dt}e^{2t},\,\tfrac{d}{dt}e^{-2t},\,\tfrac{d}{dt}\bigl(t\,e^{2t}\bigr)\bigr\rangle = \langle 2e^{2t},\,-2e^{-2t},\,e^{2t}(1+2t)\rangle.

Passo 3

Para obter o vetor tangente unitário usamos
\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|},
onde
\|\mathbf{r}'(t)\| =\sqrt{(2e^{2t})^2+(-2e^{-2t})^2+\bigl(e^{2t}(1+2t)\bigr)^2}.
Em t=0 isso fica
\|\mathbf{r}'(0)\| =\sqrt{4\cdot1+4\cdot1+1\cdot1^2} =\sqrt{9}=3.
Logo
\mathbf{T}(0) =\frac{1}{3}\langle 2,\,-2,\,1\rangle =\bigl\langle \tfrac{2}{3},\,-\tfrac{2}{3},\,\tfrac{1}{3}\bigr\rangle.

Passo 4

Derivando \mathbf{r}'(t) para obter \mathbf{r}''(t):
\mathbf{r}''(t) = \bigl\langle 4e^{2t},\,4e^{-2t},\,\tfrac{d}{dt}\bigl(e^{2t}(1+2t)\bigr)\bigr\rangle = \langle 4e^{2t},\,4e^{-2t},\,e^{2t}(4+4t)\rangle.
Em t=0 isso dá
\mathbf{r}''(0) =\langle 4,\,4,\,4\rangle.

Passo 5

Finalmente, o produto escalar
\mathbf{r}'(t)\cdot\mathbf{r}''(t) =(2e^{2t})(4e^{2t})+(-2e^{-2t})(4e^{-2t}) +\bigl(e^{2t}(1+2t)\bigr)\bigl(e^{2t}(4+4t)\bigr).
Agrupando termos em e^{4t} e e^{-4t}:
\mathbf{r}'(t)\cdot\mathbf{r}''(t) =(8+4(1+t)(1+2t))\,e^{4t}-8\,e^{-4t} =(12+12t+8t^2)\,e^{4t}-8\,e^{-4t}.

Resposta

\displaystyle \mathbf{T}(0)=\bigl\langle \tfrac{2}{3},-\tfrac{2}{3},\tfrac{1}{3}\bigr\rangle \displaystyle \mathbf{r}''(0)=\langle 4,4,4\rangle \displaystyle \mathbf{r}'(t)\cdot\mathbf{r}''(t)=(12+12t+8t^2)\,e^{4t}-8\,e^{-4t}

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