Mathematical Immaturity

14.13 Exercises

• Set up integrals for the lengths of the curves whose equations are: (a) $y = e^x,$ $0 \leq x \leq 1$ (b) $x = t + \log t,$ $y = t - \log t,$ $1 \leq t \leq e$ Show that the second length is $\sqrt{2}$ times the first one.
• (a) We define the position vector $\mathbf{r}(x)$ and velocity $\mathbf{v}(x)$ as: $$ \begin{align*} \mathbf{r}(x) &= x\,\mathbf{i} + e^{x}\,\mathbf{j} \\ \\ \mathbf{v}(x) &= \mathbf{i} + e^{x}\,\mathbf{j} \end{align*} $$ Thus, the arc length $s_x$ from $x = 0$ to $x = 1$ is: $$ \begin{align*} s_x &= \int_0^1 v(x) \,dx \\ &= \int_0^1 \sqrt{1 + e^{2x}} \,dx \quad \blacksquare \end{align*} $$ (b) We define the position vector $\mathbf{r}(t)$ and velocity vector $\mathbf{v}(t)$ as: $$ \begin{align*} \mathbf{r}(t) &= (t + \log t)\mathbf{i} + (t - \log t)\mathbf{j} \\ \\ \mathbf{v}(t) &= \left(1 + \frac{1}{t}\right)\mathbf{i} + \left(1 - \frac{1}{t}\right)\mathbf{j} \end{align*} $$ Speed $v(t)$ is then defined as: $$ \begin{align*} v(t) &= \sqrt{\left(1 + \frac{2}{t} + \frac{1}{t^2}\right) + \left(1 - \frac{2}{t} + \frac{1}{t^2}\right)} \\ &= \sqrt{2 + \frac{2}{t^2}} \end{align*} $$ With arc length $s_t:$ $$ \begin{align*} s_t &= \int_1^e v(t)\,dt \\ &= \sqrt{2}\int_1^e \sqrt{1 + \frac{1}{t^2}}\,dt \quad \blacksquare \end{align*} $$ To prove that $s_t$ is $\sqrt{2}$ times $s_x,$ we first note that by using the substitution $t = e^x$ and $\frac{dt}{dx} = e^x,$ the value $dt$ becomes $e^x\,dx,$ the limits of integration change from $t = e^{0}$ and $t = e^{1}$ to $x = 0$ and $x = 1,$ and the integral $s_t$ becomes: $$ \begin{align*} s_t &= \sqrt{2}\int_0^1\sqrt{1 + \frac{1}{e^{2x}}}\,e^x\,dx \\ &= \sqrt{2}\int_0^1\sqrt{e^{2x}\left(1 + \frac{1}{e^{2x}}\right)}\,dx \\ &= \sqrt{2}\int_0^1\sqrt{1 + e^{2x}}\,dx \\ &= \sqrt{2}s_x \quad \blacksquare \end{align*} $$