Questão 4. Determine f\left(\frac{\pi}{2}\right) em que f(t)=\mathscr{L}^{-1}\left[\frac{3 s+5}{s^{2}+9}\right].

Passo 1

O enunciado pede f\bigl(\tfrac{\pi}{2}\bigr), onde
f(t)=\mathscr{L}^{-1}\!\biggl[\frac{3s+5}{s^{2}+9}\biggr].
O conteúdo aqui é transformada inversa de Laplace, usando as fórmulas \mathscr{L}\{\cos(a t)\}=\tfrac{s}{s^{2}+a^{2}} e \mathscr{L}\{\sin(a t)\}=\tfrac{a}{s^{2}+a^{2}}.

Passo 2

Decompomos a função de s:
\frac{3s+5}{s^{2}+9} =\frac{3s}{s^{2}+9}+\frac{5}{s^{2}+9}.

Passo 3

Aplicando a transformada inversa em cada termo, com a=3:
\mathscr{L}^{-1}\!\Bigl[\frac{3s}{s^{2}+9}\Bigr] =3\cos(3t), \qquad \mathscr{L}^{-1}\!\Bigl[\frac{5}{s^{2}+9}\Bigr] =\frac{5}{3}\sin(3t).
Logo
f(t)=3\cos(3t)+\tfrac{5}{3}\sin(3t).

Passo 4

Para t=\tfrac{\pi}{2} temos \cos\bigl(3\cdot\tfrac{\pi}{2}\bigr)=0 e \sin\bigl(3\cdot\tfrac{\pi}{2}\bigr)=-1. Assim,
f\bigl(\tfrac{\pi}{2}\bigr) =3\cdot0+\tfrac{5}{3}\cdot(-1) =-\tfrac{5}{3}.

Resposta

-\tfrac{5}{3}

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