Parametric Equations and Polar Coordinates

In this chapter, we introduce parametric equations on the plane and polar coordinates.

Parametric Equations

Consider the following curve

C

in the plane:

figures/generic-curve.svg — A curve that is not the graph of a function $y = f (x)$

The curve cannot be expressed as the graph of a function

y = f (x)

because there are points

x

associated to multiple values of

y

, that is, the curve does not pass the vertical line test. We may still be interested in describing the points

(x, y)

on the curve. For example, if the curve is the trajectory of a particle moving on a plane then the position

(x, y)

of the particle is a function of time

t

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

This is an example of a set of parametric equations and the variable

t

is called the parameter of the parametrization. In some examples, the parameter could instead be an angle variable

θ

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

The main point is that the points

(x, y)

can be expressed or depend on a third parameter. Parametric equations also come with a domain for the parameter, usually we denote the domain with

I = [a, b]

, and it could be infinite

I = [a, \infty)

, or

I = (- \infty, \infty)

, etc.

Make a sketch of the curve

C

parametrized by

\begin{aligned} x & = x (t) = t^{2} \\ y & = y (t) = t + 1. \end{aligned}

The domain of the parameter is

- 3 \leq t \leq 3

. Eliminate the parameter

t

to find a Cartesian equation of the curve.

Partition the interval

I = [- 3, 3]

into

t_{0} = - 3, t_{1} = - 2, t_{2} = - 1, \dots, t_{7} = 3

and evaluate

(x (t_{i}), y (t_{i}))

and plot points. The resulting curve is given below. The orientation is clockwise.

figures/side-parabola.svg — Curve $x (t) = t^{2}$ , $y (t) = t + 1$ for $- 3 \leq t \leq 3$

The initial point is

(9, - 2)

and final point is

(9, 4)

. It seems as though the curve is a parabola. To find a Cartesian equation, start with

\begin{aligned} x & = t^{2} \\ y & = t + 1 \end{aligned}

and from the

y

-equation we get

t = y - 1

and thus

x = (y - 1)^{2}

Sketch the curve

C

parametrized by the equations below on the interval

I = [0, \frac{3 π}{2}]

. Indicate the orientation of the parametrization with arrows. Eliminate the parameter to find a Cartesian equation of the curve.

\begin{aligned} x & = x (θ) = 2 \cos (θ) \\ y & = y (θ) = 3 \sin (θ) \end{aligned}

Use the estimate

\sqrt{2} \approx 1.4

. Evaluate the parametric equations along convenient

θ

values:

$θ$	0	$π / 4$	$π / 2$	$3 π / 4$	$π$	$5 π / 4$	$3 π / 2$
$x (θ)$	2	$2 \frac{\sqrt{2}}{2}$	0	$- 2 \frac{\sqrt{2}}{2}$	$- 2$	$- 2 \frac{\sqrt{2}}{2}$	0
$y (θ)$	0	$3 \frac{\sqrt{2}}{2}$	3	$3 \frac{\sqrt{2}}{2}$	0	$- 3 \frac{\sqrt{2}}{2}$	$- 3$

figures/part-of-ellipse.svg — Curve $x (θ) = 2 \cos (θ)$ , $y (θ) = 3 \sin (θ)$ for $0 \leq θ \leq 3 π / 2$

The curve seems to be a portion of an ellipse. Recall that an ellipse centered at

with horizontal radius

and vertical radius

has Cartesian equation

Thus, we suspect that the Cartesian equation of the curve is

To eliminate the parameter start with

and then

Now

and thus

In general, a parametrization of a general ellipse

is given by

with interval

depending on how much of the ellipse we want to parametrize. To get a full rotation of the ellipse, we need an interval of length

, and if we take

we start at

and get a counter-clockwise (CCW) orientation with a full rotation.

Draw the ellipse and find a parametrization starting at the point

with a full rotation with CCW orientation.

Sketch the curve parametrized by the equations

where

. Indicate the terminal and final point of the parametrization.

Find a parametrization of the ellipse centered at

, with clockwise orientation, starting at

and passing through the point

, and going around one and a half times (end point is

First determine a CCW orientation and then change the signs accordingly. The ellipse is:

figures/full-ellipse.svg — Ellipse centered at of radii and

The general equations of the parametrization of an ellipse in this case is

We need the interval to be

. This gives a CCW orientation. To change the orientation, we can change the sign in front of the

term:

A familiar type of curve is the graph of a function:

figures/generic-graph.svg — A curve is the graph of a function if whenever is a point on then

Every point

on the curve has

. Therefore, a parametrization is

with

Parameterize the graph of the function

where

with left-to-right orientation. Then find a right-to-left orientation.

A parametrization is

with

. We can find a right-to-left parametrization by changing all

's to

and changing the interval to

. In this case, a right-to-left parametrization is then

with interval

Parameterize the line segment through the points

and

The slope of the line is

The equation of the line is then

. A parametrization of the entire line is

and since we only want the line segment where

then the interval is

For each set of parametric equations, eliminate the parameter to find a Cartesian equation for the curve.

Recall that the equation of the line tangent to the graph of

through the point

where

figures/tangent-line.svg — Equation of tangent line at is

Find equation of line tangent to the graph of

through the point

Find the equation of the line tangent to the given ellipse and passing through the point

We could solve for

in terms of

However, in some cases we may only have a parametric equation for a curve and even if we had a Cartesian equation we may not be able to solve for

(could use implicit differentiation). Use instead a parametrization

We now need a way to find the slope of the tangent line in terms of the parametric equations. We do know that

near

and thus

. Therefore, by the chain rule

and therefore

Because

this is sometimes written as

Hence, in this case

At what

value do we evaluate? It has to correspond to the

value where the parametrization passes through the point

. The value

where the parametrization passes through the point

occurs when

From

either

. In this case, need to take

. Hence, we obtain

Therefore, equation of line is

which simplifies to

Recall that a line is horizontal when its slope is zero and a vertical line could be thought of as a line with infinite slope. Since

the tangent line is horizontal at a

value when

and is vertical when

Consider the parametrized curve

Find the equation of the line tangent to the curve at the point .
Find the points on the curve where the tangent line is horizontal.
Find the points on the curve where the tangent line is vertical.
Find a Cartesian equation for the curve.

The curve is shown in Figure 4.7.

We compute The curve passes through the point when , so but need . At get but at . Hence, value is . Hence, Equation of line is
Tangent line is horizontal when which occurs at values . The points are therefore and .
Tangent line is vertical when which occurs at . The point is .
To find a Cartesian equation notice that and thus is a Cartesian equation.

The arc length of a curve

is the length of the curve. For example, the arc length of a circle of radius

. In general, given a parametrization

, on the interval

of a curve

, the arc length can be computed as

Find the arc length of the given parametric curve.

, ,
Of the graph of for . A parametrization is and . Then
Show that the arc length of a circle of radius is .

Polar Coordinates

Polar coordinates is a coordinate system to represent points in 2D space; it is an alternative to the Cartesian coordinate system. In some problems, it is more natural to use polar coordinates than Cartesian coordinates.

figures/polar-coords.svg — A point with polar coordinates

- distance from origin (directed) and can be negative
- angle measured from a chosen horizontal axis and increasing CCW (can be negative)

Polar coordinates rely on the idea that once an origin is fixed, every point in the 2D plane lies on some circle. It is convention to list polar coordinates with first

and then

, e.g., the polar coordinates

means

and

Draw the points with given polar coordinates.

Given the polar coordinates

of a point

, its Cartesian coordinates

are

On the other hand, given the Cartesian coordinates

of a point

then a set of polar coordinates

are

May need to add

if want

since the range of

. Notice we take

and may need

if we ask that

Find the Cartesian coordinates of the points with given polar coordinates.

Find the polar coordinates of the points with given Cartesian coordinates. (Note:

)

Unless specified otherwise, in this course, we will use the following convention:

In some problems, it is easier to work with mathematical objects expressed in polar coordinates.

Convert the given equations from Cartesian coordinates to polar coordinates.

which can be factored as Now represents only the origin. The other points on therefore satisfy which can be written as
Since then or .
Expanding gives and then or . Use because when get also .

Conversely, we may want to convert an equation from polar coordinates to Cartesian coordinates.

Convert the given equations from polar coordinates to Cartesian coordinates and identify the curve.

or (vertical line)
and completing square gives (circle at of radius )
Write as and then or (line)
Get and thus or and complete square to get .

Regions and curves in polar coordinates

Sketch the region on the 2D plane with given polar coordinates description.

Sketch the curve in the 2D plane with polar coordinates description

. (Note:

)

Create a table of

coordinates by varying

at step-size

	0
	1		2		1		0

pairs for

This curve is called a cardioid. Since

, we can obtain a parametrization for this curve as follows:

In general, if a curve is expressed in polar coordinates as

for

then a parametrization for the curve is:

Find the equation of the line tangent to the cardioid through the point when

Recall that

if given a parametrization

and

. Here we have

We get

and evaluating at

we obtain

. The point on the curve at

and thus the equation of the line is

Arc length in polar coordinates

Recall that the arc length of a curve parametrized by

, for

, is

Given a curve in polar coordinates

, for say

, we have a parametrization

Then the arc length of the curve

This simplifies to

Setup, but do not evaluate, the integral that evaluates to the arc length of the cardioid

, for

A complete revolution of the cardioid requires

. Now

and then

Now

and thus

Find the arc length of the curve

, where

Compute

Areas in polar coordinates

The area of a wedge of radius

and sweeping an angle of

Given say the cardioid

, we can divide the interval

so that we obtain a partition of wedges of the area enclosed by the cardioid:

figures/cardioid-partition.svg — Approximating the area enclosed by a cardioid with wedges

The sum of the area of the wedges approximates the true area

enclosed by the cardioid:

we obtain the true area:

In general, for a curve given in polar coordinates

, the area enclosed by the curve as

ranges from

Find the area enclosed by the cardioid

above the

-axis.

Apply the formula:

Consider the curve

for

Sketch the curve.
Find the area enclosed by the curve.
Setup the integral that evaluates to the arc length of the curve. Simplify the integrand but do not attempt to evaluate the integral.
Use the Trapezoidal rule with to estimate the arc length of the curve.
Find the points on the curve where the tangent line is vertical.

Evaluating and from to at step-size of , we obtain the following graph:

Limacon
Recall that the Trapezoidal rule is for estimating an integral Partitioning into equal subintervals, each subinterval has length . The Trapezoidal rule is then where are the points obtained after subdividing the interval . We wish to apply the Trapezoidal rule to Hence, here , , , and the points to evaluate are , , , , and . Hence,
Recall that given a parametrized curve and , the slope of the line tangent to the curve at is The tangent line is vertical when . Here, Then In the interval , when and . Hence, the points are and . This agrees with the sketch of the curve.

Consider the following polar curve

, for

Sketch the curve.
Find an expression for the arc length of the curve for .
Use Simpson's rule with subintervals to estimate the arc length of the curve on the interval . Label the grid points , , , , , and leave your answer in symbolic form.
Find the area enclosed by the curve.

Evaluating and from to at step-size of , we obtain the curve shown in Figure 4.14.

The polar curve
Since , the arc length of the curve along for is
Let . Applying Simpson's rule, we obtain that and then the grid points are , , , , and . The symbolic form of Simpson's rule is then
The area enclosed by the curve is

Calculus II

4. Parametric Equations and Polar Coordinates