ache \mathbf{T}(t) e \mathbf{N}(t).
\mathbf{R}(t)=t \mathbf{i}+\operatorname{cosh} t \mathbf{j} ; t \geq 0

Passo 1

Temos a curva
\mathbf{R}(t)=t\,\mathbf{i}+\operatorname{cosh} t\,\mathbf{j},\quad t\ge0
e queremos encontrar o versor tangente \mathbf{T}(t) e o versor normal \mathbf{N}(t).
Por definição
\mathbf{T}(t)=\frac{\mathbf{R}'(t)}{\|\mathbf{R}'(t)\|}, \qquad \mathbf{N}(t)=\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}.

Passo 2

Calcular a derivada \mathbf{R}'(t) e seu módulo.
\mathbf{R}'(t)=\frac{d}{dt}\bigl(t\,\mathbf{i}+\operatorname{cosh} t\,\mathbf{j}\bigr) =\mathbf{i}+\operatorname{sinh} t\,\mathbf{j},
\|\mathbf{R}'(t)\| =\sqrt{1^2+(\operatorname{sinh} t)^2} =\sqrt{1+\operatorname{sinh}^2 t} =\operatorname{cosh} t.

Passo 3

Obter o versor tangente \mathbf{T}(t).
\mathbf{T}(t) =\frac{\mathbf{R}'(t)}{\|\mathbf{R}'(t)\|} =\frac{\mathbf{i}+\operatorname{sinh} t\,\mathbf{j}}{\operatorname{cosh} t} =\sech t\,\mathbf{i}+\operatorname{tanh} t\,\mathbf{j}.

Passo 4

Calcular \mathbf{T}'(t) e seu módulo.
\mathbf{T}'(t) =\frac{d}{dt}\bigl(\sech t\,\mathbf{i}+\operatorname{tanh} t\,\mathbf{j}\bigr) =-\sech t\operatorname{tanh} t\,\mathbf{i}+\sech^2 t\,\mathbf{j},
\|\mathbf{T}'(t)\| =\sqrt{(-\sech t\operatorname{tanh} t)^2+(\sech^2 t)^2} =\sqrt{\sech^2 t\,(\operatorname{tanh}^2 t+\sech^2 t)} =\sech t.

Passo 5

Determinar o versor normal \mathbf{N}(t).
\mathbf{N}(t) =\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} =\frac{-\sech t\operatorname{tanh} t\,\mathbf{i}+\sech^2 t\,\mathbf{j}}{\sech t} =-\operatorname{tanh} t\,\mathbf{i}+\sech t\,\mathbf{j}.

Resposta

\mathbf{T}(t)=\sech t\,\mathbf{i}+\operatorname{tanh} t\,\mathbf{j}, \qquad \mathbf{N}(t)=-\operatorname{tanh} t\,\mathbf{i}+\sech t\,\mathbf{j}.

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