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.

${\begin{aligned}x&=\int x'(t)\mathrm {d} t\\y&=\int y'(t)\mathrm {d} t\end{aligned}}$

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

$y=\int \cos(t)\mathrm {d} t=\sin(t)+C$

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 $2\pi R$.