Questions: Problem C. 6 (3.3-111) Find the length of the curve r(t) = ⟨√2 t, e^t, e^−t⟩ over the interval 0 ≤ t ≤ 1.

Solution Steps

The given vector function is $\vec{r}(t) = \langle \sqrt{2t}, e^t, e^{-t} \rangle$. Its derivative is $\vec{r}'(t) = \langle \frac{\sqrt{2}}{2\sqrt{t}}, e^t, -e^{-t} \rangle$.

The magnitude of the derivative is $||\vec{r}'(t)|| = \sqrt{(\frac{1}{\sqrt{2t}})^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{\frac{1}{2t} + e^{2t} + e^{-2t}}$. Since $e^{2t} + e^{-2t} = 2\cosh(2t)$, we rewrite as $||\vec{r}'(t)|| = \sqrt{\frac{1}{2t} + 2\cosh(2t)}$. However, if $\vec{r}(t) = \langle \sqrt{2}t, e^t, e^{-t} \rangle$, then $\vec{r}'(t) = \langle \sqrt{2}, e^t, -e^{-t} \rangle$ and $||\vec{r}'(t)|| = \sqrt{2 + e^{2t} + e^{-2t}} = \sqrt{2 + 2 \cosh(2t)} = \sqrt{2(1 + \cosh(2t))}$. Using the identity $1 + \cosh(2t) = 2 \cosh^2(t)$, $||\vec{r}'(t)|| = \sqrt{4\cosh^2(t)} = 2\cosh(t)$.

The arc length of the curve $\vec{r}(t)$ over the interval $0 \le t \le 1$ is given by $L = \int_0^1 ||\vec{r}'(t)|| dt = \int_0^1 2 \cosh(t) dt = 2 \sinh(t)|_0^1 = 2(\sinh(1) - \sinh(0)) = 2\sinh(1)$. Since $\sinh(t) = \frac{e^t - e^{-t}}{2}$, $\sinh(1) = \frac{e - e^{-1}}{2}$. Therefore, $L = e - e^{-1}$.

Final Answer