Theory

Warning - traditional derivative is unnatural

This quantity is unnatural!

• From the point of view of graphs of functions, $\frac{dy}{dx}$ is natural and important. For a graph, space portrays the relation between input and output.
• From the point of view of parametric curves, it is unnatural. Here $x$ and $y$ should be treated on the same footing. Space portrays the output only.

Soon we consider more natural derivative operations in the parametric setting.

Warning - second derivative is not a ratio

$$\frac{d^{2}y}{d{x}^2}\ne \frac{\frac{d^{2}y}{d{t}^2}}{\frac{d^{2}x}{d{t}^2}}$$

Distance traveled

The distance traveled by a particle whose position is given parametrically by $(x(t),y(t))$, as $t$ increases from $a$ to $b$, is computed using this integral:

$$L={\int }_a^b\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}dt.$$

The formula is quickly derived from the usual one for arc length of the graph of a function:

$$L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$$

by using the formula for $\frac{dy}{dx}(t)$ together with the fact that $dx=\left(\frac{dx}{dt}\right)dt$.

It is better to see the formula as the summation of infinitesimal elements of distance $ds$, computing these with the Pythagorean Theorem:

$$ds=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}dt,$$

so the formula is really just:

$$L={\int }_a^{b}ds.$$

Length function

Given a parametric curve $(x(t),y(t))$, the formula for distance traveled can be used to define a length function $s(t)$. It is simply the distance traveled from $t=t_0$ to the time $t$ itself. To write this we need a new dummy variable:

$$s(t)={\int }_{t_0}^t\sqrt{x^{\prime }(u{)}^2+y^{\prime }(u{)}^2}du.$$

Speed

The speed is the rate of distance traveled per unit time: $v(t)=s^{\prime }(t)$. From the Fundamental Theorem of Calculus, we have:

$$v(t)=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.$$

Curvature

The curvature $\kappa (t)$ of a parametric curve in the plane may be defined as the rate of change of the angle of inclination $\varphi$ per unit of arc length.

We can find a formula for this. Let the curve be given by $(x(t),y(t))$. Clearly $\tan (\varphi )=\frac{dy}{dx}$ so $\varphi ={\tan }^{-1}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)$. Using the quotient rule we have:

$$\begin{array}{rl}\frac{d\varphi }{dt}& =\frac{1}{1+{\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)}^2}\cdot \frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }(t{)}^2}\\ & =\frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.\end{array}$$

Then

$$\frac{d\varphi }{ds}=\frac{d\varphi }{dt}\cdot \frac{dt}{ds}=\frac{d\varphi }{dt}\cdot \frac{1}{s^{\prime }(t)}=\frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{{\left(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2\right)}^{3/2}}.$$