Mathematical Immaturity

14.13 Exercises

• A curve has the equation $y^2 = x^3.$ Find the length of the arc joining $(1, -1)$ to $(1, 1).$
• Let position be a function of $t,$ what happens to $\mathbf{v}(t)$ and $v(t)$ when we differentiate $y^2 = x^3$ with respect to $t\,?$
• Define the position vector $\mathbf{r}(t) = (x, y)$ as a function of $t.$ In other words: $$ \begin{align*} \mathbf{r}(t) &= x(t)\,\mathbf{i} + y(t)\,\mathbf{j} \\ \\ \mathbf{v}(t) &= \frac{dx}{dt}\,\mathbf{i} + \frac{dy}{dt}\,\mathbf{j} \end{align*} $$ But we know that the graph of $\mathbf{r}(t)$ is confined by the equation $y^2 = x^3,$ so its velocity $\mathbf{v}(t)$ must be confined by: $$ 2y\frac{dy}{dt} = 3x^2\frac{dx}{dt} $$ We can rewrite velocity as $$ \mathbf{v}(t) = \frac{dx}{dt}\,\mathbf{i} + \frac{3x^2}{2y}\frac{dx}{dt}\,\mathbf{j} $$ Recalling that $y^2 = x^3,$ speed $v(t)$ is: $$ \begin{align*} v(t) &= \sqrt{\left(\frac{dx}{dt}\right)^2\left(1 + \frac{9x^4}{4y^2}\right)} \\ &= \left|\frac{dx}{dt}\right|\sqrt{1 + \frac{9}{4}x} \end{align*} $$ If we assume that $x$ and $y$ are real-valued functions, and $x$ increases monotonically, then $x \geq 0$ and $\frac{dx}{dt} \geq 0$ for all $t.$ Thus, $\left|\frac{dx}{dt}\right| = \frac{dx}{dt}$ for all $t.$ To find the arc length from $(1, -1)$ to $(1, 1),$ we first note that $y^2 = x^3,$ equivalent to $|y| = x^{3/2}$ $(x \geq 0),$ is in fact two functions of $x,$ symmetric about the $x$-axis. Thus, assuming $x(t_0) = 0$ and $x(t_1) = 1,$ the arc length $s$ can be expressed as: $$ \begin{align*} s &= 2\int_{t_0}^{t_1}v(t)\,dt \\ &= 2\int_{t_0}^{t_1}\sqrt{1 + \frac{9}{4}x}\, \frac{dx}{dt}\,dt \\ &=2\int_0^1\sqrt{1 + \frac{9}{4}x}\,dx \\ &=\frac{16}{27}\left(1 + \frac{9}{4}x\right)^{3/2}\,\Biggr|_0^1 \\ &= \frac{26\sqrt{13} - 16}{27} \quad \blacksquare \end{align*} $$