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.
![{\displaystyle {\begin{aligned}x&=\int x'(t)\mathrm {d} t\\y&=\int y'(t)\mathrm {d} t\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39235030af5076f3cae19dc617f8b048e603a100)
So, taking a simple example, with respect to t:
![{\displaystyle y=\int \cos(t)\mathrm {d} t=\sin(t)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78f2e54eb7260426b50624eff1ecccafbda8ac3f)
Consider a function defined by,
![{\displaystyle x=f(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/013620f52abdd8b9c8a658ef0c5ef5e99aef1b43)
![{\displaystyle y=g(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8bf432cea843238c7b86c954aa2028614cb31ec)
Say that
is increasing on some interval,
. Recall, as we have derived in a previous chapter, that the length of the arc created by a function over an interval,
, is given by,
![{\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {1+(f'(x))^{2}}}\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbc1c2bbcb9dec492d11171120875258edc8f0f5)
It may assist your understanding, here, to write the above using Leibniz's notation,
![{\displaystyle L=\int _{\alpha }^{\beta }{\sqrt {1+\left({\frac {\mathrm {d} y}{\mathrm {d} x}}\right)^{2}}}\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00d6654e1a73c2727d260b2244deeff43bb79188)
Using the chain rule,
![{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {\mathrm {d} y}{\mathrm {d} t}}\cdot {\frac {\mathrm {d} t}{\mathrm {d} x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7676a6b31743ea53901f56fce2143fe9f546148)
We may then rewrite
,
![{\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} t}}\mathrm {d} t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12217d589695bd8113f1e176762c1b6a93dea99f)
Hence,
becomes,
![{\displaystyle 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84d2dad7fd3bc5e3dc926180dacacc5f6324e7f5)
Extracting a factor of
,
![{\displaystyle 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8737e1a2a42a3155e81dd96bffa373be52b5df3f)
As
is increasing on
,
, and hence we may write our final expression for
as,
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9581fb788883c40873feee4514ca7684559971a)
Take a circle of radius
, which may be defined with the parametric equations,
![{\displaystyle x=R\sin \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/05606c0ae0b3cece3d2cc3d71cacba641a138f2e)
![{\displaystyle y=R\cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4a817584691731cd4bb942bb8ce20577d21a47a)
As an example, we can take the length of the arc created by the curve over the interval
. Writing in terms of
,
![{\displaystyle x=0\implies \theta =\arcsin \left({\frac {0}{R}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68ed860bcb938c9e72adaf719b1374bd463a04af)
![{\displaystyle x=R\implies \theta =\arcsin \left({\frac {R}{R}}\right)=\arcsin(1)={\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7e91cb2dd5eef32b917f72fc81b813b22957bea)
Computing the derivatives of both equations,
![{\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} \theta }}=R\cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/38f4d69bd666cbff8e1076a28debf9b970137d0a)
![{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} \theta }}=-R\sin \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/22919379385ca9624b05661e9a30749256ca125e)
Which means that the arc length is given by,
![{\displaystyle L=\int _{0}^{\frac {\pi }{2}}{\sqrt {(-R\sin \theta )^{2}+R^{2}\cos ^{2}\theta }}\mathrm {d} \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/343de2b3c8bf142bf616a391726e949551805fa4)
By the Pythagorean identity,
![{\displaystyle L=R\int _{0}^{\frac {\pi }{2}}\mathrm {d} \theta =R{\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f714e1199cd1382b9e6cdaa214c3fce4e63ebc6)
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
by 4 to deduce the perimeter of a circle of radius
to be
.