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.

Template:Math

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

Template:Math

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,

$L=\underset{\alpha }{\overset{\beta }{\int }}\sqrt{1+(f^{\prime }(x){)}^2}\mathrm{d}x$

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

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

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,

$L=\underset{\alpha }{\overset{\beta }{\int }}\sqrt{1+{(\frac{\mathrm{d}y}{\mathrm{d}t}\cdot \frac{\mathrm{d}t}{\mathrm{d}x})}^2}\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t$

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

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

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,

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

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 $\theta$,

$x=0⟹\theta =\arcsin \left(\frac{0}{R}\right)=0$

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

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,

$L=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{(-R\sin \theta {)}^2+R^2{\cos }^2\theta }\mathrm{d}\theta$

By the Pythagorean identity,

$L=R\underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{d}\theta =R\frac{\pi }{2}$

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 $2\pi R$.