Chapter 13

Medium

Parametric Equations and Polar Coordinates

00 · Symbol Glossary

$t$t — parameter

An independent variable that drives both $x$ and $y$ simultaneously, rather than $y$ depending directly on $x$ . Often (but not always) interpreted as time.

$x=x(t),\ y=y(t)$parametric equations

A pair of functions of the parameter $t$ , together tracing out a curve in the plane as $t$ varies. Distinct from Chapter 01's $y=f(x)$ : here neither coordinate is written directly in terms of the other.

$\dfrac{dy}{dx}$slope of a parametric curve

The same quantity Chapter 02 defined for $y=f(x)$ — but for a parametric curve, it must be recovered from $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ separately, via Section 01's formula.

$r,\ \theta$r, theta — polar coordinates

An alternative way to label a point in the plane: $r$ is the distance from the origin, $\theta$ is the angle measured from the positive $x$ -axis (Chapter 01's angle notation, reused here). $r$ may be taken negative, meaning the point at angle $\theta$ plotted in the opposite direction.

$r=f(\theta)$polar curve

A curve described by giving $r$ as a function of $\theta$ , rather than $y$ as a function of $x$ .

01 · Parametric Equations and Their Slope

Some curves cannot be written as a single function $y=f(x)$ — a circle, for instance, fails the vertical line test. Parametric equations sidestep this by describing both coordinates in terms of a third variable.

Definition — Parametric Curve

A parametric curve is given by $x=x(t)$ , $y=y(t)$ for $t$ ranging over some interval. Each value of $t$ produces one point $(x(t), y(t))$ in the plane.

Example — A circle, parametrized

$x(t)=\cos t$ , $y(t)=\sin t$ for $0\leq t\leq2\pi$ : since $x(t)^2+y(t)^2 = \cos^2t+\sin^2t = 1$ for every $t$ (a standard trigonometric identity), every point $(x(t),y(t))$ lies on the unit circle — and as $t$ ranges over $[0,2\pi]$ , every point of the circle is reached exactly once. This is a curve $y=f(x)$ cannot describe directly, since a single $x$ -value (e.g. $x=0$ ) corresponds to two different $y$ -values ( $y=1$ and $y=-1$ ).

Definition — Slope of a Parametric Curve

If $x(t)$ and $y(t)$ are both differentiable and $\dfrac{dx}{dt}\neq0$ :

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Where this formula comes from

Treating $y$ as a function of $x$ , and $x$ itself as a function of $t$ , the chain rule (Chapter 03) gives $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$ . Solving for $\dfrac{dy}{dx}$ (valid exactly when $\dfrac{dx}{dt}\neq0$ ) gives the formula above — this is Chapter 03's chain rule read in reverse, not a new independent rule.

Step-by-step — Find the slope of $x(t)=t^2$, $y(t)=t^3-3t$ at $t=2$

Differentiate $x(t)$ and $y(t)$ separately with respect to $t$ (Chapter 02, power rule):

\dfrac{dx}{dt}=2t

;

\dfrac{dy}{dt}=3t^2-3

Form the ratio (Section 01's formula):

\dfrac{dy}{dx} = \dfrac{3t^2-3}{2t}

Substitute $t=2$ :

\dfrac{3(4)-3}{2(2)} = \dfrac{12-3}{4} = \dfrac94

Identify the point:

x(2)=4

y(2)=8-6=2

, so the tangent line at

(4,2)

has slope

\dfrac94

❌ Forming dy/dx as y(t)/x(t) instead of the derivative ratio

On the same curve, a common mistake: computing $\dfrac{y(2)}{x(2)} = \dfrac{2}{4}=\dfrac12$ and calling that the slope.

Why it breaks: $\dfrac{y(t)}{x(t)}$ is the ratio of the coordinates themselves (related to the angle from the origin to the point, not the tangent direction of the curve) — completely different from $\dfrac{dy/dt}{dx/dt}$ , the ratio of the rates of change. The slope of a curve is about how $y$ changes relative to $x$ as the curve moves, not about the coordinates' raw ratio at a single instant.

Consequence: $\frac12 \neq \frac94$ (the value correctly computed in the step-block above) — using the coordinate ratio in place of the derivative ratio gives a tangent line with the wrong steepness entirely.

02 · Polar Coordinates

Definition — Polar Coordinates and Conversion

A point is located by $(r,\theta)$ : $r$ is its distance from the origin, $\theta$ is the angle from the positive $x$ -axis (measured counterclockwise, matching Chapter 01's angle convention).

x = r\cos\theta \qquad y = r\sin\theta \qquad r^2 = x^2+y^2 \qquad \tan\theta = \frac{y}{x} \ (x\neq0)

Negative $r$ is permitted: the point $(r,\theta)$ with $r<0$ is plotted at distance $\lvert r\rvert$ in the opposite direction from angle $\theta$ — equivalently, at $(\lvert r\rvert, \theta+\pi)$ .

Example — Converting between coordinate systems

The point $r=4$ , $\theta=\dfrac{\pi}{3}$ : $x = 4\cos\dfrac{\pi}{3} = 4\left(\dfrac12\right)=2$ ; $y = 4\sin\dfrac{\pi}{3} = 4\left(\dfrac{\sqrt3}{2}\right)=2\sqrt3$ . So this polar point is $(2, 2\sqrt3)$ in ordinary coordinates.

03 · Slope of a Polar Curve

A polar curve $r=f(\theta)$ is really a parametric curve in disguise, with $\theta$ as the parameter: $x(\theta) = f(\theta)\cos\theta$ , $y(\theta)=f(\theta)\sin\theta$ . Section 01's formula applies directly.

Step-by-step — Find the slope of $r=1+\cos\theta$ at $\theta=\dfrac{\pi}{2}$

Write $x$ and $y$ in terms of $\theta$ :

x(\theta) = (1+\cos\theta)\cos\theta

;

y(\theta) = (1+\cos\theta)\sin\theta

Differentiate $y(\theta)$ using the product rule (Chapter 02):

\dfrac{dy}{d\theta} = -\sin\theta\cdot\sin\theta + (1+\cos\theta)\cos\theta = -\sin^2\theta+\cos\theta+\cos^2\theta

Differentiate $x(\theta)$ using the product rule:

\dfrac{dx}{d\theta} = -\sin\theta\cdot\cos\theta + (1+\cos\theta)(-\sin\theta) = -\sin\theta\cos\theta - \sin\theta-\sin\theta\cos\theta = -2\sin\theta\cos\theta-\sin\theta

Substitute $\theta=\dfrac{\pi}{2}$ into $\dfrac{dy}{d\theta}$ :

\sin\frac{\pi}{2}=1

\cos\frac{\pi}{2}=0

, so

\dfrac{dy}{d\theta} = -1+0+0=-1

Substitute $\theta=\dfrac{\pi}{2}$ into $\dfrac{dx}{d\theta}$ :

-2(1)(0)-1 = 0-1=-1

Apply Section 01's formula:

\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{-1}{-1}=1

04 · Area in Polar Coordinates

Computing area bounded by a polar curve requires a different building block than the rectangular strips of Chapter 06 — a polar region is naturally built from thin sectors, not thin rectangles.

Definition — Area Element in Polar Coordinates

A circular sector of radius $r$ and angle $d\theta$ (in radians) has area $\dfrac12 r^2\,d\theta$ — this is the familiar sector-area formula $\dfrac{\theta}{2\pi}\times\pi r^2 = \dfrac12 r^2\theta$ , applied to an infinitesimally thin wedge. Summing these wedges (a Riemann-sum idea, Chapter 06, Section 01) and taking the limit:

A = \int_{\alpha}^{\beta} \frac12 \big[f(\theta)\big]^2\,d\theta

for a region traced exactly once as $\theta$ ranges from $\alpha$ to $\beta$ .

Step-by-step — Find the area enclosed by one full loop of $r=2\sin\theta$ (a circle), for $0\leq\theta\leq\pi$

Set up the integral:

A = \displaystyle\int_0^\pi \frac12(2\sin\theta)^2\,d\theta = \int_0^\pi \frac12(4\sin^2\theta)\,d\theta = \int_0^\pi 2\sin^2\theta\,d\theta

Rewrite $\sin^2\theta$ using the identity $\sin^2\theta = \dfrac{1-\cos2\theta}{2}$ (a standard trigonometric identity):

2\sin^2\theta = 2\cdot\dfrac{1-\cos2\theta}{2} = 1-\cos2\theta

Integrate term by term (Chapter 06, 07):

\int(1-\cos2\theta)\,d\theta = \theta - \dfrac{\sin2\theta}{2}+C

(the

\frac12

inside comes from the chain rule reversed on

\cos2\theta

, matching Chapter 07's u-substitution pattern).

Evaluate from $0$ to $\pi$ : at

\theta=\pi

\pi - \dfrac{\sin2\pi}{2} = \pi-0=\pi

. At

\theta=0

0-\dfrac{\sin0}{2}=0

Subtract:

A = \pi - 0 = \pi

. Check:

r=2\sin\theta

traces a circle of diameter

2

(radius

1

), and its area is

\pi(1)^2=\pi

— matches exactly.

❌ Integrating over an interval that traces the curve more than once

On the same curve $r=2\sin\theta$ , integrating over $0\leq\theta\leq2\pi$ instead of $0\leq\theta\leq\pi$ : $A = \displaystyle\int_0^{2\pi} 2\sin^2\theta\,d\theta$ .

Why it breaks: as $\theta$ moves past $\pi$ , $\sin\theta$ becomes negative, so $r=2\sin\theta$ becomes negative too — by the negative- $r$ convention (Section 02), the point is plotted back on the same circle it already traced from $\theta=0$ to $\theta=\pi$ . The full loop from $0$ to $\pi$ already covers the entire circle exactly once; continuing to $2\pi$ retraces it a second time.

Consequence: integrating over $[0,2\pi]$ computes $2\pi$ (double the correct area of $\pi$ found above) — not because the formula is wrong, but because the region is being counted twice. Before applying the area formula, always determine the exact $\theta$ -interval over which the curve traces the desired region exactly once.

05 · Exercises

EXERCISE 13.1

Differentiate $x(t)$ and $y(t)$ separately, form the ratio $\dfrac{dy/dt}{dx/dt}$ , then substitute the given value of $t$ .

$x(t)=t^2-1$ , $y(t)=t^3$ , at $t=1$ . $\dfrac{dx}{dt}=2t$ , $\dfrac{dy}{dt}=3t^2$ . $\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$ . At $t=1$ : $\dfrac{3}{2}$ .

Find the slope of the parametric curve $x(t)=t^2-1$ , $y(t)=t^3$ at $t=1$ .

EXERCISE 13.2

Use $x=r\cos\theta$ and $y=r\sin\theta$ directly, evaluating the trigonometric values at the given angle.

$r=6$ , $\theta=\dfrac{2\pi}{3}$ : $\cos\dfrac{2\pi}{3}=-\dfrac12$ , $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$ . $x=6\left(-\dfrac12\right)=-3$ . $y=6\left(\dfrac{\sqrt3}{2}\right)=3\sqrt3$ . Rectangular coordinates: $(-3, 3\sqrt3)$ .

Convert the polar point $r=6$ , $\theta=\dfrac{2\pi}{3}$ to rectangular coordinates.

EXERCISE 13.3

Write $x$ and $y$ as functions of $\theta$ using the polar-to-rectangular conversion, then apply the parametric slope formula from Section 01.

$r=\theta$ (a spiral) at $\theta=\dfrac{\pi}{2}$ . $x(\theta)=\theta\cos\theta$ , $y(\theta)=\theta\sin\theta$ . $\dfrac{dy}{d\theta}=\sin\theta+\theta\cos\theta$ (product rule); at $\theta=\frac\pi2$ : $1+\frac\pi2(0)=1$ . $\dfrac{dx}{d\theta}=\cos\theta-\theta\sin\theta$ ; at $\theta=\frac\pi2$ : $0-\frac\pi2(1)=-\frac\pi2$ . $\dfrac{dy}{dx}=\dfrac{1}{-\pi/2}=-\dfrac{2}{\pi}$ .

Find the slope of the polar curve $r=\theta$ (a spiral) at $\theta=\dfrac{\pi}{2}$ .

EXERCISE 13.4

Set up the polar area integral over the correct interval, using the trigonometric identity for $\cos^2\theta$ to make the integral solvable with Chapter 06/07 tools.

$r=2\cos\theta$ traces a full circle for $0\leq\theta\leq\pi$ (by the same negative- $r$ retracing logic as Section 04's example). $A=\int_0^\pi \frac12(2\cos\theta)^2\,d\theta = \int_0^\pi 2\cos^2\theta\,d\theta$ . Using $\cos^2\theta=\frac{1+\cos2\theta}{2}$ : $2\cos^2\theta = 1+\cos2\theta$ . Integrate: $\left[\theta+\frac{\sin2\theta}{2}\right]_0^\pi = (\pi+0)-(0+0)=\pi$ . Matches a circle of radius $1$ : area $\pi(1)^2=\pi$ . ✓

Find the area enclosed by $r=2\cos\theta$ , being careful to identify the correct interval of $\theta$ that traces the region exactly once.

EXERCISE 13.5

A point where the curve has a vertical tangent occurs where $dx/dt=0$ but $dy/dt\neq0$ — Section 01's slope formula is undefined there, not zero.

$x(t)=t^3-3t$ , $y(t)=t^2$ . $\dfrac{dx}{dt}=3t^2-3=3(t-1)(t+1)$ , zero at $t=\pm1$ . $\dfrac{dy}{dt}=2t$ , which at $t=1$ is $2\neq0$ and at $t=-1$ is $-2\neq0$ . At both $t=1$ and $t=-1$ , $\dfrac{dx}{dt}=0$ while $\dfrac{dy}{dt}\neq0$ , so $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ is undefined (division by zero) — the curve has a vertical tangent at both points, at $(x(1),y(1))=(-2,1)$ and $(x(-1),y(-1))=(2,1)$ .

Find the points on the parametric curve $x(t)=t^3-3t$ , $y(t)=t^2$ where the tangent line is vertical.

06 · Chapter Summary

Concept	Formula / Rule
Parametric curve	$x=x(t)$ , $y=y(t)$
Parametric slope	$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , undefined where $dx/dt=0$
Polar conversion	$x=r\cos\theta$ , $y=r\sin\theta$ , $r^2=x^2+y^2$
Negative $r$	Plots at $(\lvert r\rvert, \theta+\pi)$
Polar curve as parametric	$x(\theta)=f(\theta)\cos\theta$ , $y(\theta)=f(\theta)\sin\theta$
Polar area	$A=\int_\alpha^\beta \frac12[f(\theta)]^2\,d\theta$ , over an interval tracing the region exactly once

Next: Chapter 14 — Vectors in 2D/3D, which introduces directed quantities with magnitude and direction to describe curves and motion more directly than a pair of coordinate functions.