Calculus/Parametric Integration

Introduction
Because most parametric equations are given in explicit form, they can be integrated like many other equations. Integration has a variety of applications with respect to parametric equations, especially in kinematics and vector calculus.



So, taking a simple example, with respect to t:



Arc length
Consider a function defined by,
 * $$x = f(t)$$
 * $$y = g(t)$$

Say that $$f$$ is increasing on some interval, $$[\alpha, \beta]$$. Recall, as we have derived in a previous chapter, that the length of the arc created by a function over an interval, $$[\alpha, \beta]$$, is given by,


 * $$L = \int_\alpha^\beta \sqrt{1 + (f'(x))^2} \mathrm{d}x$$

It may assist your understanding, here, to write the above using Leibniz's notation,


 * $$L = \int_\alpha^\beta \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x$$

Using the chain rule,


 * $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x}$$

We may then rewrite $$\mathrm{d}x$$,


 * $$\frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t$$

Hence, $$L$$ becomes,


 * $$L = \int_\alpha^\beta \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x}\right)^2} \frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t$$

Extracting a factor of $$\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2$$,


 * $$L = \int_\alpha^\beta \sqrt{\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t$$

As $$f$$ is increasing on $$[\alpha,\beta]$$, $$\sqrt{\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2} = \frac{\mathrm{d}t}{\mathrm{d}x}$$, and hence we may write our final expression for $$L$$ as,


 * $$\int_\alpha^\beta \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \mathrm{d}t$$

Example
Take a circle of radius $$R$$, which may be defined with the parametric equations,


 * $$x = R\sin\theta$$
 * $$y = R\cos\theta$$

As an example, we can take the length of the arc created by the curve over the interval $$[0,R]$$. Writing in terms of $$\theta$$,


 * $$x = 0 \implies \theta = \arcsin\left(\frac{0}{R}\right) = 0$$
 * $$x = R \implies \theta = \arcsin\left(\frac{R}{R}\right) = \arcsin(1) = \frac{\pi}{2}$$

Computing the derivatives of both equations,


 * $$\frac{\mathrm{d}x}{\mathrm{d}\theta} = R\cos\theta$$
 * $$\frac{\mathrm{d}y}{\mathrm{d}\theta} = -R\sin\theta$$

Which means that the arc length is given by,


 * $$L = \int_0^{\frac{\pi}{2}} \sqrt{(-R\sin\theta)^2 + R^2\cos^2\theta}\mathrm{d}\theta$$

By the Pythagorean identity,


 * $$L = R\int_0^{\frac{\pi}{2}} \mathrm{d}\theta = R\frac{\pi}{2}$$

One can use this result to determine the perimeter of a circle of a given radius. As this is the arc length over one "quadrant", one may multiply $$L$$ by 4 to deduce the perimeter of a circle of radius $$R$$ to be $$2\pi R$$.