Determine o comprimento da curva dada.
4. \mathbf{r}(t)=\cos t \mathbf{i}+\operatorname{sen} t \mathbf{j}+\ln \cos t \mathbf{k}, \quad 0 \leqslant t \leqslant \pi / 4

Passo 1

O enunciado pede o comprimento da curva \mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+\ln(\cos t)\,\mathbf{k}
no intervalo 0\le t\le \tfrac{\pi}{4}. Vamos usar a fórmula do comprimento de arco
L=\int_{a}^{b}\big\|\mathbf{r}'(t)\big\|\,dt
e calcular \mathbf{r}'(t) e sua norma.

Passo 2

Derivando componente a componente,
\mathbf{r}'(t) =\frac{d}{dt}\bigl(\cos t\bigr)\,\mathbf{i} +\frac{d}{dt}\bigl(\sin t\bigr)\,\mathbf{j} +\frac{d}{dt}\bigl(\ln(\cos t)\bigr)\,\mathbf{k} =-\sin t\,\mathbf{i}+\cos t\,\mathbf{j}-\tan t\,\mathbf{k}.

Passo 3

Calculamos o quadrado da norma:
\|\mathbf{r}'(t)\|^2 =(\!-\sin t\!)^2+(\cos t)^2+(\!-\tan t\!)^2 =\sin^2 t+\cos^2 t+\tan^2 t =1+\tan^2 t =\operatorname{sec}^2 t.
Como \cos t>0 em [0,\pi/4],
\|\mathbf{r}'(t)\|=\operatorname{sec} t.

Passo 4

O comprimento é
L=\int_{0}^{\pi/4}\operatorname{sec} t\,dt.
Sabendo que
\int\operatorname{sec} t\,dt=\ln\bigl|\operatorname{sec} t+\tan t\bigr|+C,
avaliamos
L =\Bigl[\ln\bigl(\operatorname{sec} t+\tan t\bigr)\Bigr]_{0}^{\pi/4} =\ln\bigl(\operatorname{sec}(\tfrac{\pi}{4})+\tan(\tfrac{\pi}{4})\bigr) -\ln\bigl(\sec0+\tan0\bigr) =\ln\bigl(\sqrt2+1\bigr)-\ln1 =\ln\bigl(1+\sqrt2\bigr).

Resposta

\ln(1+\sqrt{2})

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