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 $2\pi R$ .

Surface area

Volume

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