Mathematical Immaturity

14.13 Exercises

• Find an integral similar to that in (14.18) for the length of the graph of an equation of the form $x = g(y),$ where $g$ has a continuous derivative on an interval $[c, d].$
• Recall the setup to equation $(14.18):$ Let the graph of a real-valued function $f,$ defined on $[a, b]$ be described by the position vector $\mathbf{r}(t):$ $$ \mathbf{r}(t) = t\,\mathbf{i} + f(t)\,\mathbf{j} $$ Then, its corresponding velocity vector $\mathbf{v}(t)$ and speed $v(t)$ are given by: $$ \begin{align*} \mathbf{v}(t) &= \mathbf{i} + f'(t)\,\mathbf{j} \\ \\ v(t) &= \|\mathbf{v}(t)\| \\ &= \sqrt{1 + \left[f'(t)\right]^2}\,. \end{align*} $$ Therefore, the arc length of the graph of $f$ over a subinterval $[a, x]$ of $[a, b]$ is given by the integral: $$ \begin{align*} s(x) &= \int_a^x v(t)\,dt \\ &= \int_a^x \sqrt{1 + \left[f'(t)\right]^2}\,dt \tag{14.18} \end{align*} $$
• If the graph is of the form $x = g(y),$ where $y$ is a function of $t,$ then the position vector takes the form $\mathbf{r}(t):$ $$ \mathbf{r}(t) = g(y)\,\mathbf{i} + y(t)\,\mathbf{j} $$ Its corresponding velocity $\mathbf{v}(t)$ and speed $v(t)$ are: $$ \begin{align*} \mathbf{v}(t) &= g'(y)\frac{dy}{dt}\,\mathbf{i} + \frac{dy}{dt}\,\mathbf{j} \\ \\ v(t) &= \sqrt{\left(g'(y)\frac{dy}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \end{align*} $$ Its arc length $s,$ given that $c < d,$ is given by: $$ \begin{align*} s &= \int_{t_0}^{t_1}v(t)\,dt \\ &= \int_{t_0}^{t_1}\sqrt{\left(g'(y)\frac{dy}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \\ &= \int_{t_0}^{t_1}\sqrt{1 + \left[g'(y)\right]^2}\frac{dy}{dt}\,dt \\ &= \int_c^d \sqrt{1 + \left[g'(y)\right]^2}\,dy \quad \blacksquare \end{align*} $$ (Note: if $t_0 < t_1,$ then it must be the case that $\frac{dy}{dt} > 0$ since $c < d.$ Thus, $\left|\frac{dy}{dt}\right| = \frac{dy}{dt}$ in the above calculation.)