CALCULO 2
A primeira derivada é:
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt}(e^{3t} \sin t) = \sin t \cdot {d \over dt}(e^{3t}) + e^{3t} {d \over dt}(\sin t) \\ {d \over dt}(3t-2) = {d \over dt}(3t) - {d \over dt}(2) \end{matrix} \right.\) \(\rightarrow \left \{ \begin{matrix} {d \over dt}(e^{3t} \sin t) = \sin t \cdot 3e^{3t} + e^{3t} \cos t \\ {d \over dt}(3t-2) = 3 - 0 \end{matrix} \right.\)
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt}(e^{3t} \sin t) = e^{3t} (3\sin t + \cos t) \\ {d \over dt}(3t-2) = 3 \end{matrix} \right.\)
\(\Longrightarrow \underline {f'(t) = \Big (e^{3t}(3\sin t + \cos t),3 \Big )}\)
Portanto, a segunda derivada é:
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt} \Big ( e^{3t} (3\sin t+ \cos t) \Big ) = (3\sin t+ \cos t) \cdot {d \over dt}(e^{3t}) + e^{3t} \cdot {d \over dt}(3\sin t+ \cos t) \\ {d \over dt}(3) = 0 \end{matrix} \right.\)
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt} \Big ( e^{3t} (3\sin t+ \cos t) \Big ) = (3\sin t+ \cos t) \cdot 3e^{3t} + e^{3t} \cdot (3\cos t - \sin t) \\ {d \over dt}(3) = 0 \end{matrix} \right.\)
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt} \Big ( e^{3t} (3\sin t+ \cos t) \Big ) = e^{3t}(9\sin t+ 3\cos t) + e^{3t} (3\cos t - \sin t) \\ {d \over dt}(3) = 0 \end{matrix} \right.\)
\(\Longrightarrow \left \{ \begin{matrix} {d \over dt} \Big ( e^{3t} (3\sin t+ \cos t) \Big ) = e^{3t}(8\sin t+ 6\cos t) \\ {d \over dt}(3) = 0 \end{matrix} \right.\)
\(\Longrightarrow \fbox { $ f''(t) = \Big (e^{3t}(8\sin t + 6\cos t), 0 \Big ) $}\)
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