Determine os componentes de \frac{\mathrm{d}^{3} \mathbf{r}}{\mathrm{~d} t^{3}} em coordenadas esféricas.

Passo 1

O enunciado pede a terceira derivada temporal do vetor posição \mathbf{r} em coordenadas esféricas (r,\theta,\varphi). O conteúdo necessário inclui: a expressão de \mathbf{r}=r\,\hat e_{r}, as fórmulas conhecidas para \mathbf{v}= \dot{\mathbf{r}} e \mathbf{a}=\ddot{\mathbf{r}} em bases esféricas, e as derivadas das direções unitárias \hat e_{r},\hat e_{\theta},\hat e_{\varphi}.

Passo 2

Recordamos que
\mathbf{r}=r\,\hat e_{r}
e que as derivadas das bases esféricas são
\frac{d\hat e_{r}}{dt}=\dot\theta\,\hat e_{\theta}+\dot\varphi\sin\theta\,\hat e_{\varphi}, \quad \frac{d\hat e_{\theta}}{dt}=-\dot\theta\,\hat e_{r}+\dot\varphi\cos\theta\,\hat e_{\varphi}, \quad \frac{d\hat e_{\varphi}}{dt}=-\dot\varphi\sin\theta\,\hat e_{r}-\dot\varphi\cos\theta\,\hat e_{\theta}.

Passo 3

A velocidade em esféricas é
\mathbf{v}=\dot r\,\hat e_{r} + r\,\dot\theta\,\hat e_{\theta} + r\,\dot\varphi\sin\theta\,\hat e_{\varphi}.

Passo 4

A aceleração em esféricas é
\mathbf{a}=(\ddot r-r\dot\theta^{2}-r\dot\varphi^{2}\sin^{2}\theta)\,\hat e_{r} +\bigl(r\ddot\theta+2\dot r\dot\theta-r\dot\varphi^{2}\sin\theta\cos\theta\bigr)\,\hat e_{\theta} +\bigl(r\ddot\varphi\sin\theta+2\dot r\dot\varphi\sin\theta+2r\dot\theta\dot\varphi\cos\theta\bigr)\,\hat e_{\varphi}.
Definimos para abreviar
A_{r}=\ddot r-r(\dot\theta^{2}+\dot\varphi^{2}\sin^{2}\theta), \quad A_{\theta}=r\ddot\theta+2\dot r\dot\theta-r\dot\varphi^{2}\sin\theta\cos\theta, \quad A_{\varphi}=r\ddot\varphi\sin\theta+2\dot r\dot\varphi\sin\theta+2r\dot\theta\dot\varphi\cos\theta.

Passo 5

O “jerk” é
\mathbf{J}=\frac{d^{3}\mathbf{r}}{dt^{3}} =\frac{d\mathbf{a}}{dt} =\frac{dA_{r}}{dt}\,\hat e_{r}+A_{r}\frac{d\hat e_{r}}{dt} +\frac{dA_{\theta}}{dt}\,\hat e_{\theta}+A_{\theta}\frac{d\hat e_{\theta}}{dt} +\frac{dA_{\varphi}}{dt}\,\hat e_{\varphi}+A_{\varphi}\frac{d\hat e_{\varphi}}{dt}.
Reagrupando por bases, obtém-se as componentes
\begin{cases} J_{r}=\dfrac{dA_{r}}{dt}-A_{\theta}\,\dot\theta -A_{\varphi}\,\dot\varphi\sin\theta,\\\ J_{\theta}=\dfrac{dA_{\theta}}{dt}+A_{r}\,\dot\theta -A_{\varphi}\,\dot\varphi\cos\theta,\\\ J_{\varphi}=\dfrac{dA_{\varphi}}{dt}+A_{r}\,\dot\varphi\sin\theta +A_{\theta}\,\dot\varphi\cos\theta. \end{cases}

Passo 6

Calculamos \dfrac{dA_{r}}{dt}. Como
A_{r}=\ddot r - r\,C, \quad C=\dot\theta^{2}+\dot\varphi^{2}\sin^{2}\theta,
temos
\frac{dA_{r}}{dt} =\dddot r-\dot r\,C -r\,\frac{dC}{dt}, \quad \frac{dC}{dt} =2\dot\theta\,\ddot\theta+2\dot\varphi\,\ddot\varphi\sin^{2}\theta +2\dot\varphi^{2}\sin\theta\cos\theta\,\dot\theta.

Passo 7

Logo a componente radial é
\begin{aligned} J_{r} &=\dddot r-\dot r(\dot\theta^{2}+\dot\varphi^{2}\sin^{2}\theta) -r\bigl(2\dot\theta\,\ddot\theta+2\dot\varphi\,\ddot\varphi\sin^{2}\theta +2\dot\varphi^{2}\sin\theta\cos\theta\,\dot\theta\bigr)\\\ &\quad -\bigl(r\ddot\theta+2\dot r\dot\theta-r\dot\varphi^{2}\sin\theta\cos\theta\bigr)\dot\theta -\bigl(r\ddot\varphi\sin\theta+2\dot r\dot\varphi\sin\theta+2r\dot\theta\dot\varphi\cos\theta\bigr)\dot\varphi\sin\theta. \end{aligned}

Passo 8

Calculamos \dfrac{dA_{\theta}}{dt}. Com
A_{\theta}=r\ddot\theta+2\dot r\dot\theta-r\dot\varphi^{2}\sin\theta\cos\theta
vem
\begin{aligned} \frac{dA_{\theta}}{dt} &=r\dddot\theta+3\dot r\,\ddot\theta+2\ddot r\,\dot\theta -\dot r\,\dot\varphi^{2}\sin\theta\cos\theta -r\bigl(2\dot\varphi\,\ddot\varphi\sin\theta\cos\theta+\dot\varphi^{2}\cos2\theta\,\dot\theta\bigr). \end{aligned}

Passo 9

A componente polar é
\begin{aligned} J_{\theta} &=\;r\dddot\theta+3\dot r\,\ddot\theta+2\ddot r\,\dot\theta -\dot r\,\dot\varphi^{2}\sin\theta\cos\theta -r\bigl(2\dot\varphi\,\ddot\varphi\sin\theta\cos\theta+\dot\varphi^{2}\cos2\theta\,\dot\theta\bigr)\\\ &\quad +\bigl(\ddot r-r(\dot\theta^{2}+\dot\varphi^{2}\sin^{2}\theta)\bigr)\dot\theta -\bigl(r\ddot\varphi\sin\theta+2\dot r\dot\varphi\sin\theta+2r\dot\theta\dot\varphi\cos\theta\bigr)\dot\varphi\cos\theta. \end{aligned}

Passo 10

Calculamos \dfrac{dA_{\varphi}}{dt}. Como
A_{\varphi}=r\ddot\varphi\sin\theta+2\dot r\dot\varphi\sin\theta+2r\dot\theta\dot\varphi\cos\theta,
obtém-se
\begin{aligned} \frac{dA_{\varphi}}{dt} &=r\dddot\varphi\sin\theta +2\ddot r\,\dot\varphi\sin\theta +3\dot r\,\ddot\varphi\sin\theta +4\dot r\,\dot\theta\,\dot\varphi\cos\theta\\\ &\quad +3r\,\ddot\varphi\,\dot\theta\cos\theta +2r\,\ddot\theta\,\dot\varphi\cos\theta -2r\,\dot\theta^{2}\,\dot\varphi\sin\theta. \end{aligned}

Passo 11

A componente azimutal é
\begin{aligned} J_{\varphi} &=r\dddot\varphi\sin\theta +2\ddot r\,\dot\varphi\sin\theta +3\dot r\,\ddot\varphi\sin\theta +4\dot r\,\dot\theta\,\dot\varphi\cos\theta +3r\,\ddot\varphi\,\dot\theta\cos\theta +2r\,\ddot\theta\,\dot\varphi\cos\theta -2r\,\dot\theta^{2}\,\dot\varphi\sin\theta\\\ &\quad +\bigl(\ddot r-r(\dot\theta^{2}+\dot\varphi^{2}\sin^{2}\theta)\bigr)\dot\varphi\sin\theta +\bigl(r\ddot\theta+2\dot r\dot\theta-r\dot\varphi^{2}\sin\theta\cos\theta\bigr)\dot\varphi\cos\theta. \end{aligned}

Resposta

As componentes de \displaystyle \frac{d^{3}\mathbf{r}}{dt^{3}} em coordenadas esféricas são
\boxed{ \mathbf{J} = J_{r}\,\hat e_{r} + J_{\theta}\,\hat e_{\theta} + J_{\varphi}\,\hat e_{\varphi}, }
onde J_{r},J_{\theta},J_{\varphi} são dadas respectivamente pelas fórmulas acima nos Passos 7, 9 e 11.

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