( 2 pts) Considere o cilindro elíptico 2 x^{2}+3 y^{2}=1 e o plano z=2 y. Essas duas superfícies se interceptam e resulta numa elipse C.
(a) Calcule a velocidade, \mathbf{v}(P)=\left.\frac{d \mathbf{r}}{d t}\right|_{P}, avaliada no ponto P
(b) Calcule a aceleração, \mathbf{a}(P)=\left.\frac{d^{2} \mathbf{r}}{d t^{2}}\right|_{P}, avaliada no ponto P
(c) Calcule o curvatura, \kappa, no ponto P, utilizando a equação \kappa(P)=\frac{\|\mathbf{v}(P) \times \mathbf{a}(P)\|}{\|\mathbf{v}(P)\|^{3}} onde P\left(0, \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)

Passo 1

O enunciado nos pede a velocidade \mathbf{v}(P)=\bigl.\tfrac{d\mathbf{r}}{dt}\bigr|_{P}, a aceleração \mathbf{a}(P)=\bigl.\tfrac{d^2\mathbf{r}}{dt^2}\bigr|_{P} e a curvatura \displaystyle \kappa(P)=\frac{\|\mathbf{v}(P)\times\mathbf{a}(P)\|}{\|\mathbf{v}(P)\|^3}
no ponto
P\bigl(0,\tfrac{1}{\sqrt{3}},\tfrac{2}{\sqrt{3}}\bigr)
da curva C obtida pela interseção do cilindro 2x^2+3y^2=1 com o plano z=2y. Vamos usar parametrização de curvas para obter \mathbf{r}(t), depois derivar.

Passo 2

Escolhemos a parametrização do cilindro elíptico
x=\tfrac{1}{\sqrt{2}}\cos t,\quad y=\tfrac{1}{\sqrt{3}}\sin t
e impondo z=2y vem
z=\tfrac{2}{\sqrt{3}}\sin t.
Logo
\mathbf{r}(t)=\Bigl(\tfrac{1}{\sqrt{2}}\cos t,\;\tfrac{1}{\sqrt{3}}\sin t,\;\tfrac{2}{\sqrt{3}}\sin t\Bigr).

Passo 3

Derivando em relação a t obtemos
\mathbf{r}'(t)=\Bigl(-\tfrac{1}{\sqrt{2}}\sin t,\;\tfrac{1}{\sqrt{3}}\cos t,\;\tfrac{2}{\sqrt{3}}\cos t\Bigr),
\mathbf{r}''(t)=\Bigl(-\tfrac{1}{\sqrt{2}}\cos t,\;-\tfrac{1}{\sqrt{3}}\sin t,\;-\tfrac{2}{\sqrt{3}}\sin t\Bigr).

Passo 4

Para encontrar t_0 tal que \mathbf{r}(t_0)=P igualamos
\tfrac{1}{\sqrt{2}}\cos t_0=0,\quad \tfrac{1}{\sqrt{3}}\sin t_0=\tfrac{1}{\sqrt{3}},
logo \sin t_0=1 e \cos t_0=0, ou seja t_0=\tfrac{\pi}{2}. Substituindo em \mathbf{r}' e \mathbf{r}'' temos
\mathbf{v}(P)=\mathbf{r}'\bigl(\tfrac{\pi}{2}\bigr)=\Bigl(-\tfrac{1}{\sqrt{2}},0,0\Bigr),
\mathbf{a}(P)=\mathbf{r}''\bigl(\tfrac{\pi}{2}\bigr)=\Bigl(0,-\tfrac{1}{\sqrt{3}},-\tfrac{2}{\sqrt{3}}\Bigr).

Passo 5

Calculamos o produto vetorial
\mathbf{v}(P)\times\mathbf{a}(P) =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\\\ -\tfrac{1}{\sqrt2}&0&0\\\\ 0&-\tfrac{1}{\sqrt3}&-\tfrac{2}{\sqrt3} \end{vmatrix} =\Bigl(0,\,-\tfrac{2}{\sqrt6},\,\tfrac{1}{\sqrt6}\Bigr).
A norma é
\|\mathbf{v}\times\mathbf{a}\| =\sqrt{\tfrac{4}{6}+\tfrac{1}{6}} =\sqrt{\tfrac{5}{6}} =\tfrac{\sqrt5}{\sqrt6},
enquanto \|\mathbf{v}(P)\|=1/\sqrt2. Assim
\kappa(P) =\frac{\|\mathbf{v}\times\mathbf{a}\|}{\|\mathbf{v}\|^3} =\frac{\tfrac{\sqrt5}{\sqrt6}}{\bigl(\tfrac1{\sqrt2}\bigr)^3} =2\sqrt2\,\frac{\sqrt5}{\sqrt6} =\frac{2\sqrt{15}}{3}.

Resposta

(a) \mathbf{v}(P)=\bigl(-\tfrac{1}{\sqrt2},0,0\bigr)
(b) \mathbf{a}(P)=\bigl(0,-\tfrac{1}{\sqrt3},-\tfrac{2}{\sqrt3}\bigr)
(c) \kappa(P)=\tfrac{2\sqrt{15}}{3}

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