Dadas as funções vetoriais F(t)=t i+t^{2} j+5 k e G(t)=t^{3} i+j e a função real h(t)=t^{2}-1, determinar:
F(1 / a)+G(1 / a), a \neq 0

Passo 1

O enunciado dá as funções vetoriais
F(t)=t\,\mathbf{i}+t^{2}\,\mathbf{j}+5\,\mathbf{k}, \quad G(t)=t^{3}\,\mathbf{i}+\mathbf{j}
e pede para calcular
F\bigl(\tfrac1a\bigr)+G\bigl(\tfrac1a\bigr),
com a\neq0. Vamos avaliar cada função em t=\tfrac1a e depois somar componente a componente.

Passo 2

Calcular F\bigl(\tfrac1a\bigr). Basta substituir t=\tfrac1a em cada componente de F:
F_x\bigl(\tfrac1a\bigr)=\tfrac1a,\quad F_y\bigl(\tfrac1a\bigr)=\bigl(\tfrac1a\bigr)^{2}=\tfrac1{a^{2}},\quad F_z\bigl(\tfrac1a\bigr)=5.
Logo
F\bigl(\tfrac1a\bigr)=\tfrac1a\,\mathbf{i}+\tfrac1{a^{2}}\,\mathbf{j}+5\,\mathbf{k}.

Passo 3

Calcular G\bigl(\tfrac1a\bigr). Substituindo t=\tfrac1a em G:
G_x\bigl(\tfrac1a\bigr)=\bigl(\tfrac1a\bigr)^{3}=\tfrac1{a^{3}},\quad G_y\bigl(\tfrac1a\bigr)=1,\quad G_z\bigl(\tfrac1a\bigr)=0.
Logo
G\bigl(\tfrac1a\bigr)=\tfrac1{a^{3}}\,\mathbf{i}+1\,\mathbf{j}.

Passo 4

Somar componente a componente:
\bigl[F\bigl(\tfrac1a\bigr)+G\bigl(\tfrac1a\bigr)\bigr]_x =\tfrac1a+\tfrac1{a^{3}},
\bigl[F\bigl(\tfrac1a\bigr)+G\bigl(\tfrac1a\bigr)\bigr]_y =\tfrac1{a^{2}}+1,
\bigl[F\bigl(\tfrac1a\bigr)+G\bigl(\tfrac1a\bigr)\bigr]_z =5+0=5.

Resposta

F\bigl(\tfrac1a\bigr)+G\bigl(\tfrac1a\bigr) =\Bigl(\tfrac1a+\tfrac1{a^{3}}\Bigr)\mathbf{i} +\Bigl(\tfrac1{a^{2}}+1\Bigr)\mathbf{j} +5\,\mathbf{k}.

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