Chapter 10: Parametric Equations and Polar Coordinates

10.1: Curves Defined by Parametric Equations

Parametrization of a curve means defining $x$ and $y$ in terms of a third variable $t$ called a parameter. This creates parametric equations:

x = f(t) \qquad \qquad y = g(t)

Each value of $t$ determines a point $(x, y)$ . As $t$ varies over its domain, $(x, y)$ traces a curve $C$ called a parametric curve. (Note that $t$ does not necessarily represent time, and we could use other variable names instead of $t$ ).

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It is entirely possible for two distinct sets of parametric equations to generate the same set of points. It is also possible for two distinct values of $t$ to generate the same $(x, y)$ .

For instance, $x = \cos t, y = \sin t$ and $x = \sin t, y = \cos t$ generate the same set of points (a circle). However, the first set of parametric equations traces the curve in the counterclockwise direction, while the second set traces the curve in the clockwise direction.

example

Describe the parametric curve parametrized by the following: ( $h,k,r$ are constants)

x = h + r \cos t \qquad y = k + r \sin t

We can attempt to eliminate the parameter $t$ to find an equation in terms of $x,y$ .

\begin{gather*} r \cos t = x - h \qquad r \sin t = y - k \\ r^{2} \cos^{2} t + r^{2} \sin ^{2} t = (x - h)^2 + (y - k)^2 \\ r^{2} = (x - h)^2 + (y - k)^2 \\ \end{gather*}

Thus, this set of parametric equations describes a circle centered at $(h, k)$ with radius $r$ .

With parameterization, we can also restrict $t$ to a certain interval, i.e. $a \leq t \leq b$ . For a curve parametrized by the following conditions,

x = f(t) \qquad \qquad y = g(t) \qquad \qquad a \leq t \leq b

it has initial point $(f(a), g(a))$ and terminal point $(f(b), g(b))$ .

Exercise:
Eliminate the parameter

t

for the equations

\qquad x = t^{2} + t \qquad y = t^{2} - t

\begin{gather*} x - y = 2t \qquad x + y = 2t^{2} \\ (x - y)^{2} = 4t^{2} = 2(x + y) \\ x^{2} + y^{2} + 2xy - 2x - 2y = 0 \\ \end{gather*}

This becomes a parabola aligned along $x = y$ .

Example: Cycloid

The cycloid is a curve traced by a point $P$ on the circumference of a circle that rolls along a straight line. (Page 683, Figure 13).

We will skip the derviation of the parametrization of the cycloid. (See Page 684 if interested). It is described by: ( $r$ is constant)

x = r(\theta - \sin \theta) \qquad y = r(1 - \cos \theta) \qquad \theta \in \mathbb{R}

Let's now eliminate the parameter $\theta$

\begin{gather*} r \theta - x = r \sin \theta \qquad r - y = r \cos \theta \\ r^{2} \sin ^{2} \theta + r^{2} \cos ^{2}\theta = (r \theta - x)^{2} + (r - y)^{2} \\ r^{2} = (r \theta - x)^{2} + (r - y)^{2} \\ \sqrt{ r^{2} - (r - y)^{2} } = r\theta - x \\ \theta = \dfrac{\sqrt{ r^{2} - (r - y)^{2} } + x}{r} \\ \end{gather*}

Having solved for $\theta$ in terms of $x,y,r$ , we can effectively substitute for $\theta$ in the original parametric equations to eliminate the parameter. (Although, it is a very, very ugly equation).

10.2 Calculus with Parametric Curves

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In this section, "Cartesian Curve" will refer to a curve $C$ represented by equations with variables $x,y$ , and "Parametric Curve" will refer to a curve $C$ represented by equations $x(t),y(t)$ .

Derivative

Deriving $\frac{dy}{dx}$

\begin{gather*} \frac{dy}{dt} = \frac{d y }{d x } \cdot \frac{d x }{d t } \\ \text{If } \frac{d x }{d t } \neq 0: \frac{d y }{d x } = \dfrac{\frac{d y }{d t } }{\frac{d x }{d t } }. \end{gather*}

Deriving $\frac{d^{2} y }{d x^{2} }$

\begin{align*} \frac{d^{2} y }{d x^{2} } &= \frac{d}{dx} \left( \frac{d y }{d x } \right) && \\ &= \dfrac{\frac{d}{dt} \left( \frac{d y }{d x } \right)}{\frac{d x }{d t } } \end{align*}

The last step above is simply expanding $\frac{d}{dx}$ into $\dfrac{\frac{d}{dt}}{ \frac{d x }{d t }}$ .

Formulas

First and Second Derivative of a Parametric Curve

\begin{gather*} \frac{d y }{d x } = \dfrac{\frac{d y }{d t } }{\frac{d x }{d t } } \\ \frac{d^{2} y }{d x^{2} } = \dfrac{\frac{d}{dt} \left( \frac{d y }{d x } \right)}{ \frac{d x }{d t } } \end{gather*}

Horizontal and Vertical Tangents

The tangent line is...

horizontal when $\frac{d y }{d x } = 0 \implies \frac{d y }{d t } = 0$ .
vertical when $\frac{d y }{d x } = \text{undefined} \implies \frac{d x }{d t } = 0$ .
unknown when $\frac{d x }{d t } = \frac{d y }{d t } = 0$ ; L'Hôpital's Rule is needed.

Area

Suppose that some curve $C$ is traced by $x = f(t)$ and $y = f(t)$ as the parameter increases from $\alpha \to \beta$ . Then, if $a = f(\alpha)$ and $b = f(\beta)$ :

\begin{align*} A &= \int_{a}^{b} y \, dx && \\ &= \int_{\alpha}^{\beta} y \, \frac{d x }{d t } dt \\ \end{align*}

Note that, if instead $b = f(\alpha)$ and $a = f(\beta)$ , we can simply reverse the interval for the integral (the one in respect to $t$ ).

Area under a Parametric Curve

A = \int_{\alpha}^{\beta} y \frac{d x }{d t } \, dt

Arc Length

Deriving for a Cartesian Curve

Let curve $C$ be described by $y=F(x)$ , where $F$ is differentiable. Additionally, let $x = f(t)$ and $y = g(t)$ . Then the arc length $L$ of $C$ between $a\leq x\leq b$ can be obtained by partitioning $[a,b]$ into $n$ subintervals of equal length, deriving an approximation for $L$ based on the partitions, and solving for $\lim_{ n \to \infty } L$ to derive the arc length formula.

First, the partitioning:

a = x_{0} < x_{1} < \dots < x_{n-1} < x_{n} = b

The length of each subinterval is $\Delta x = \frac{b - a}{n}$ .

For each subinterval $x_{i - 1} \to x_{i}$ , we draw a line between the points $(x_{i - 1}, F(x_{i - 1}))$ and $(x_{i}, F(x_{i}))$ . Let $\Delta s_{i}$ be the length of this line.
Additionally, let $\Delta x_{i} = x_{i} - x_{i - 1}$ (yes, this is equivalent to $\Delta x$ ) and let $\Delta y_{i} = y_{i} - y_{i - 1}$ . Then:

\Delta s_{i} = \sqrt{ (\Delta x_{i})^{2} + (\Delta y_{i}^{2}) }

Now, the approximation of $L$ is simply:

L \approx \sum_{i=1}^{n} \Delta s_{i}

Taking the limit:

L = \lim_{ n \to \infty } \sum_{i=1}^{n} \Delta s_{i}

Remember that $\Delta s_{i} = ds$ . Rewritten, we can express this as

L = \int_{a}^b ds

It suffices to then find an expression for $ds$ a.k.a. $\Delta s_{i}$ .

\begin{align*} \Delta s_{i} &= \sqrt{ (\Delta x_{i})^{2} + (\Delta y_{i})^{2} } && \\ &= \sqrt{ 1 + \left( \frac{\Delta x_{i}}{\Delta y_{i}} \right)^{2} } \Delta x_{i} \\ &= \sqrt{ 1 + \left( \frac{\Delta x_{i}}{\Delta y_{i}} \right)^{2} } \Delta x \\ \end{align*}

In other words,

ds = \sqrt{ 1 + \left( \frac{\Delta x_{i}}{\Delta y_{i}} \right)^{2} } dx

And thus,

L = \int_{a}^b \sqrt{ 1 + \left( \frac{\Delta x_{i}}{\Delta y_{i}} \right)^{2} } dx

And we're done!

Formal Interlude

By the Mean Value Theorem (MVT), since $y = F(x)$ is differentiable, $\forall i \in \{1,\; \dots, \; n\}$ , $\exists x_{i}^* \in (x_{i-1}, x_{i})$ such that

F'(x_{i}^*) = \frac{\Delta y_{i}}{\Delta x_{i}}

i.e. there is some $x_{i}^*$ in the interval with the same slope as the slope of the line segment between the points $(x_{i - 1}, F(x_{i - 1}))$ and $(x_{i}, F(x_{i}))$ .

Then,

\begin{align*} L &= \lim_{ n \to \infty } \Delta s_{i} && \\ &= \lim_{ n \to \infty } \sqrt{ 1 + \left( \frac{\Delta y_{i}}{\Delta x_{i}} \right)^{2} } \Delta x \\ &= \lim_{ n \to \infty } \sqrt{ 1 + [F'(x_{i})]^{2} } \Delta x \\ &= \int_{a}^{b} \sqrt{ 1 + [F'(x_{i})]^{2} } \, dx \end{align*}

This is a slightly more formal last step to the proof, but it returns the same answer.

Arc Length for a Cartesian Curve

L = \int_{a}^{b} \sqrt{ 1 + \left( \frac{d y }{d x } \right)^{2} } \, dx

Deriving for a Parametric Curve

The derivation for the arc length formula for a parametric curve follows simply from the formula for a Cartesian curve.

\begin{align*} L &= \int_{a}^{b} \sqrt{ 1 + \left( \frac{d y }{d x } \right)^{2} } \, dx && \\ &= \int_{\alpha}^{\beta} \sqrt{ 1 + \left( \dfrac{\frac{d y }{d t } }{\frac{d x }{d t } } \right)^{2} }\, \frac{d x }{d t } dt \\ &= \int_{\alpha}^{\beta} \sqrt{ \left( \frac{d x }{d t } \right)^{2} + \left( \frac{d y }{d t } \right)^{2} } \, dt \end{align*}

Arc Length for a Parametric Curve

L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{d x }{d t } \right)^{2} + \left( \frac{d y }{d t } \right)^{2} } \, dt

Surface Area

Deriving for a Cartesian Curve

We can revolve a curve $C$ around an axis to obtain a 3-dimensional shape $C'$ . For our purposes, we will consider rotating $C$ around the $x$ -axis on the interval $a \leq x \leq b$ for our derivation.

Just like in the 2-dimensional derivation for arc length, we will approximate $C$ by partitioning it into $n$ subintervals of equal length. Then, by rotating the line segments formed by the subintervals, we obtain an approximation for $C'$ .

cartesian-revolution-surface-area Figure 1. A diagram illustrating the method of approximating the shape of revolution of a Cartesian curve.

We can calculate the surface area of this shape by computing the lateral surface area of a frustum. *Note that the lateral surface area denotes the surface area of the frustum that does not include the circular bases.

Frustum?

A frustum is created when a cross section parallel to the cone's base (circle) splits the cone into two shapes: a cone above and a frustum below.

The lateral surface area of a frustum is

A = 2\pi rl

Where $r=\frac{1}{2}(r_{1}+ r_{2})$ .
*The proof of this will be left out for brevity. Search it up if you're interested!

Thus, we can approximate the surface area of the shape of revolution by summing the surface areas of the frustums. Note that the symbols used here are equivalent to the symbols used [[#Arc Length#Deriving for a Cartesian Curve|here]].

S \approx \sum_{i=1}^n A_{i} = \sum_{i=1}^n 2\pi \frac{y_{i-1} + y_{i}}{2} \Delta s_{i}

$\Delta s_{i}$ is the same as before. So we can write:

\begin{align*} S &= \lim_{ n \to \infty } \sum_{i=1}^{n} 2\pi \frac{y_{i - 1} + y_{i}}{2} \sqrt{ 1 + \left( \frac{\Delta x_{i}}{\Delta y_{i}} \right)^{2} } \Delta x && \\ &= \int_{a}^{b} 2\pi y \sqrt{ 1 + \left( \frac{\textrm{d} y }{\textrm{d} x } \right)^{2} } \, dx \end{align*}

** Note that $y$ is substituted for $\frac{y_{i - 1} + y_{i}}{2}$ for $y$ because $y_{i-1}$ and $y_{i}$ become infinitely close to each other as $n$ tends to $\infty$ . $y$ can be substituted with $F(x)$ when actually solving.

Surface Area of a Cartesian Curve

S = \int_{a}^{b} 2\pi y \sqrt{ 1 + \left( \frac{\textrm{d} y }{\textrm{d} x } \right)^{2} } \, dx

Deriving for a Parametric Curve

The process here is identical to the parametric derivation for arc length.

\begin{align*} S &= \int_{a}^{b} 2\pi y \sqrt{ 1 + \left( \frac{\textrm{d} y }{\textrm{d} x } \right)^{2} } \, dx && \\ &= \int_{\alpha}^{\beta} 2\pi y\sqrt{ 1 + \left( \frac{\textrm{d} y }{\textrm{d} x } \right)^{2} } \, \frac{\textrm{d} x }{\textrm{d} t } dt \\ &= \int_{\alpha}^{\beta} 2\pi y\sqrt{ \left( \frac{\textrm{d} x }{\textrm{d} t } \right)^{2} + \left( \frac{\textrm{d} y }{\textrm{d} t } \right)^{2} } \, dt \end{align*}

Surface Area of a Parametric Curve

S = \int_{\alpha}^{\beta} 2\pi y\sqrt{ \left( \frac{\textrm{d} x }{\textrm{d} t } \right)^{2} + \left( \frac{\textrm{d} y }{\textrm{d} t } \right)^{2} } \, dt

Generalization

You may not necessarily rotate a curve around the $x$ -axis. Generally, you will be assigned to rotate around some line $y=k$ (vertical) or $x = h$ (horizontal). The formula is always similar, however. You simply replace $y$ with a formula for the radius $r$ .

Reference Formulas

Formulas

\begin{gather*} \frac{d y }{d x } = \dfrac{\frac{d y }{d t } }{\frac{d x }{d t } } \qquad\qquad \frac{d^{2} y }{d x^{2} } = \dfrac{\frac{d}{dt} \left( \frac{d y }{d x } \right)}{ \frac{d x }{d t } } \\ A = \int_{\alpha}^{\beta} y \frac{d x }{d t } \, dt \\ L = \int_{a}^{b} \sqrt{ 1 + \left( \frac{d y }{d x } \right)^{2} } \, dx \qquad L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{d x }{d t } \right)^{2} + \left( \frac{d y }{d t } \right)^{2} } \, dt \\ S = \int_{a}^{b} 2\pi y \sqrt{ 1 + \left( \frac{\textrm{d} y }{\textrm{d} x } \right)^{2} } \, dx \qquad S = \int_{\alpha}^{\beta} 2\pi y\sqrt{ \left( \frac{\textrm{d} x }{\textrm{d} t } \right)^{2} + \left( \frac{\textrm{d} y }{\textrm{d} t } \right)^{2} } \, dt \end{gather*}

10.3 Polar Coordinates

Intro

The polar coordinate system is an alternative to the Cartesian coordinate system, where a point $P$ is represented by $(r, \theta)$ . $r$ is the length of the line segment between the pole $O$ , which is typically the origin, and the point $P$ . (We choose the variable $r$ because it means radius). $\theta$ is the angle between the line segment and the polar axis. The polar axis is a ray that is typically equivalent to the positive $x$ -axis.

polar-coords

Figure 2. A graph demonstrating polar coordinates. on the Cartesian plane.

Note that $P(-r, \theta) = P(r, \theta + \pi n)$ for $n \in \mathbb{Z}$ and $n$ odd. Also note that $P(r, \theta) = P(r, \theta + 2\pi n)$ for $n \in \mathbb{Z}$ .

Polar curves are essentially just a special form of parametric curves. Many of the same ideas/formulas for parametric curves will apply here too!

Cartesian to Polar

We can write a couple equations that represent the relationship between polar and Cartesian coordinates.

Based on the definitions of $\cos \theta$ and $\sin\theta$ , we can write

\cos\theta = \frac{x}{r} \qquad\qquad \sin \theta = \frac{y}{r}

It may help to refer to Figure 2 to see why these equations are true.

From this, it's easy to derive the following:

Polar

\to

Cartesian Equations

x = r\cos\theta \qquad\qquad y = r\sin \theta

Subsequently, we may derive the following:

Cartesian

\to

Polar Equations

r^{2} = x^{2} + y^{2} \qquad\qquad \tan\theta = \frac{y}{x}

Finally, note that the equation of a polar curve is typically denoted as $r = f(\theta)$ .

Symmetry

When sketching polar curves, recognizing symmetry can help significantly.

$f(\theta) = f(-\theta) \implies$
The curve is symmetric about the polar axis.
$f(\theta) = -f(\theta) \iff f(\theta) = f(\theta + \pi) \implies$
The curve is symmetric about the pole.
$f(\theta) = f(\pi-\theta) \implies$
The curve is symmetric about $\theta=\frac{\pi}{2}$ (equivalent to $y = x$ )

Derivatives

To derive polar curves, we essentially consider them as parametric. (Because, that basically is what they are!)

As a reminder,

x = r\cos\theta \qquad\qquad y = r\sin\theta

Deriving,

\frac{\textrm{d} x }{\textrm{d} \theta } = \frac{\textrm{d} r }{\textrm{d} \theta } \sin\theta + r\cos\theta \qquad\qquad \frac{\textrm{d} y }{\textrm{d} \theta } = \frac{\textrm{d} r }{\textrm{d} \theta } \cos\theta - r\sin\theta

Therefore,

\frac{\textrm{d} y }{\textrm{d} x } = \dfrac{\frac{\textrm{d} y }{\textrm{d} \theta } }{\frac{\textrm{d} x }{\textrm{d} \theta } } = \frac{\frac{\textrm{d} r }{\textrm{d} \theta } \sin\theta + r\cos\theta}{ \frac{\textrm{d} r }{\textrm{d} \theta } \cos \theta - r\sin\theta }

Derivative for a Polar Curve

\frac{\textrm{d} y }{\textrm{d} x } = \frac{\frac{\textrm{d} r }{\textrm{d} \theta } \sin\theta + r\cos\theta}{ \frac{\textrm{d} r }{\textrm{d} \theta } \cos \theta - r\sin\theta }

For some $\theta=k$ :

$\frac{\textrm{d} y }{\textrm{d} \theta }=0 \; \land \; \frac{\textrm{d} x }{\textrm{d} \theta } \neq 0 \implies$ Vertical Tangent.
$\frac{\textrm{d} y }{\textrm{d} \theta }\neq 0 \; \land \; \frac{\textrm{d} y }{\textrm{d} \theta } = 0 \implies$ Horizontal Tangent.
$\frac{\textrm{d} y }{\textrm{d} \theta } = \frac{\textrm{d} x }{\textrm{d} \theta }= 0 \implies$ Use L'Hôpital's Rule to calculate $\lim_{ \theta \to k } \frac{\textrm{d} y }{\textrm{d} x }$ .

Derivative at

f(\theta)=r=0

(Equivalent to a derivative at the pole)

\frac{\textrm{d} y }{\textrm{d} x } = \tan\theta \qquad \text{ if } \frac{\textrm{d} r }{\textrm{d} \theta } \neq 0

To find the 2nd derivative, we can follow the same steps as for a parametric curve. (Omitted for brevity).

10.4 Areas and Lengths in Polar Coordinates

Area

We define the area of a polar curve as the area swept out by the radius $r=f(\theta)$ between bounds $a$ and $b$ on $\theta$ . For instance,
We can approximate this area by summing several sectors of circles:
The area of a sector of a circle is

A = \frac{1}{2}r^{2}\theta

Therefore, we can approximate the area of the polar curve as

\Delta A_{i} = \sum_{i=1}^{n} \frac{1}{2}[f(\theta_{i})]^{2} \Delta\theta

We take the limit as $\Delta\theta\to \infty \implies n \to \infty$ :

\lim_{ n \to \infty } \sum_{i=1}^{n} \frac{1}{2}[f(\theta_{i})]^{2} \Delta\theta = \int_{a}^{b} \frac{1}{2}[f(\theta)]^{2} \, d\theta

Therefore,

Area of a Polar Curve

A = \int_{a}^{b} \frac{1}{2}r^{2} \, d\theta

Arc Length

For the arc length of a polar curve, it suffices to repurpose the equation for the arc length of a parametric curve. Remember:

L = \int_{a}^{b} \sqrt{ \left( \frac{\textrm{d} x }{\textrm{d} \theta } \right)^{2} + \left( \frac{\textrm{d} y }{\textrm{d} \theta } \right)^{2} } \, d\theta

By definition,

x=r\cos\theta \qquad\qquad y = r\sin\theta

Differentiating,

\frac{\textrm{d} x }{\textrm{d} \theta } = \frac{\textrm{d} r }{\textrm{d} \theta } \cos\theta - r\sin\theta \qquad\qquad \frac{\textrm{d} y }{\textrm{d} \theta } = \frac{\textrm{d} r }{\textrm{d} \theta } \sin\theta + r\cos\theta

So,

\begin{align*} \left( \frac{\textrm{d} x }{\textrm{d} \theta } \right)^{2} + \left( \frac{\textrm{d} y }{\textrm{d} \theta } \right)^{2} &= \left( \frac{\textrm{d} r }{\textrm{d} \theta } \right)^{2}\cos ^{2}\theta - 2r \frac{\textrm{d} r }{\textrm{d} \theta } \cos\theta \sin\theta + r^{2}+\sin ^{2}\theta + \left( \frac{\textrm{d} r }{\textrm{d} \theta } \right)^{2}\sin ^{2}\theta + 2r \frac{\textrm{d} r }{\textrm{d} \theta } \sin\theta \cos\theta + r^{2}\cos ^{2}\theta \\ &= \left( \frac{\textrm{d} r }{\textrm{d} \theta } \right)^{2} + r^{2} \end{align*}

Thus,

Arc Length of a Polar Curve

L = \int_{a}^{b} \sqrt{ r^{2} + \left( \frac{\textrm{d} r }{\textrm{d} \theta } \right)^{2} } \, d\theta

Chapter 10: Parametric Equations and Polar Coordinates

10.1: Curves Defined by Parametric Equations

10.2 Calculus with Parametric Curves

Derivative

Deriving dydx\frac{dy}{dx}dxdy​

Deriving d2ydx2\frac{d^{2} y }{d x^{2} }dx2d2y​

Formulas

Horizontal and Vertical Tangents

Area

Arc Length

Deriving for a Cartesian Curve

Formal Interlude

Deriving for a Parametric Curve

Surface Area

Deriving for a Cartesian Curve

Deriving for a Parametric Curve

Generalization

Reference Formulas

10.3 Polar Coordinates

Intro

Cartesian to Polar

Symmetry

Derivatives

10.4 Areas and Lengths in Polar Coordinates

Area

Arc Length

Deriving $\frac{dy}{dx}$

Deriving $\frac{d^{2} y }{d x^{2} }$