The Calculus of Parametric Equations

Derivatives of Parametric Systems

Just as we are able to differentiate functions of $x$ , we are able to differentiate $x$ and $y$ , which are functions of $t$ . Consider:

${\begin{aligned}x&=\sin(t)\\y&=t\end{aligned}}$

We would find the derivative of $x$ with respect to $t$ , and the derivative of $y$ with respect to $t$ :

${\begin{aligned}x'&=\cos(t)\\y'&=1\end{aligned}}$

In general, we say that if

${\begin{aligned}x&=x(t)\\y&=y(t)\end{aligned}}$

then:

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

It's that simple.

This process works for any amount of variables.

Slope of Parametric Equations

In the above process, $x'$ has told us only the rate at which $x$ is changing, not the rate for $y$ , and vice versa. Neither is the slope.

In order to find the slope, we need something of the form ${\frac {dy}{dx}}$ .

We can discover a way to do this by simple algebraic manipulation:

${\frac {y'}{x'}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {dy}{dx}}$

So, for the example in section 1, the slope at any time $t$ :

${\frac {1}{\cos(t)}}=\sec(t)$

In order to find a vertical tangent line, set the horizontal change, or $x'$ , equal to 0 and solve.

In order to find a horizontal tangent line, set the vertical change, or $y'$ , equal to 0 and solve.

If there is a time when both $x',y'$ are 0, that point is called a singular point.

Concavity of Parametric Equations

Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.

When you have take the derivative of ${\frac {dy}{dx}}$ in terms of $t$ , you are left with ${\frac {\frac {d^{2}y}{dx}}{dt}}$ :

${\frac {d}{dt}}\left[{\frac {dy}{dx}}\right]={\frac {\frac {d^{2}y}{dx}}{dt}}$ .

By multiplying this expression by ${\frac {dt}{dx}}$ , we are able to solve for the second derivative of the parametric equation:

${\frac {\frac {d^{2}y}{dx}}{dt}}\times {\frac {dt}{dx}}={\frac {d^{2}y}{dx^{2}}}$ .

Thus, the concavity of a parametric equation can be described as:

${\frac {d}{dt}}\left[{\frac {dy}{dx}}\right]\times {\frac {dt}{dx}}$

So for the example in sections 1 and 2, the concavity at any time $t$ :

${\frac {d}{dt}}[\csc(t)]\times \cos(t)=-\csc ^{2}(t)\times \cos(t)$

Parametric Integration

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 $2\pi R$ .