Calculus · Chapter 10 — Parametric Equations & Polar Coordinates
Calculus · Chapter 10

Parametric Equations and Polar Coordinates

Until now, plane curves have been described as a graph of a function or by an implicit relation between x and y. This chapter introduces two alternatives that handle motion and rotational symmetry far more naturally: parametric equations, where both coordinates depend on a third variable, and the polar coordinate system, which locates points by distance and angle. Together they describe trajectories, rolling-circle curves, conic sections, and planetary orbits with equations that stay compact where Cartesian forms would not.

10.1

Curves Defined by Parametric Equations

Instructor note — add your section note here. (Suggested: emphasise that the parameter usually carries physical meaning — time, angle, or arc position — so students read the equations as a description of motion, not just an algebraic trick.)

Parametric Curve

A curve traced when each coordinate is expressed as a function of a common third variable \(t\), called the parameter. Each value of \(t\) fixes one point, and as \(t\) sweeps through its range the point moves, drawing out the curve in a definite direction.

Parametric Equations
\[ x = f(t), \qquad y = g(t) \]

A curve such as the one in Figure 1 can fail the Vertical Line Test, so it cannot be written as \(y=f(x)\). Yet both coordinates of a moving particle are perfectly good functions of time, and the parametric form captures not only where the particle has been but also when and in which direction.

x y P(f(t), g(t))
Figure 1A particle follows a path that no single function \(y=f(x)\) can describe; its position is captured instead by \(x=f(t),\;y=g(t)\). (Original redrawn figure.)
EXAMPLE 1Identifying a Parabola Hidden in Parameters
Sketch and identify the curve given by \(x = t^2 - 2t,\; y = t + 1\).
Solution
Plotting a few values of \(t\) suggests a parabola opening to the right, traced upward as \(t\) increases.
Eliminate the parameter. From the second equation, \(t = y - 1\).
Substitute into the first: \(x = (y-1)^2 - 2(y-1) = y^2 - 4y + 3\).
The curve is the parabola \(x = y^2 - 4y + 3\). The parametric form adds what the Cartesian form omits: the direction and timing of the motion.

Restricting the Parameter

Limiting \(t\) to an interval \(a \le t \le b\) keeps only the arc between the initial point \((f(a),g(a))\) and the terminal point \((f(b),g(b))\). The same curve can have many parametrizations, which may trace it at different speeds, in different directions, or more than once — so a parametric curve carries more information than the point set alone.

EXAMPLE 2A Circle from Trigonometric Parameters
Identify the curve \(x=\cos t,\; y=\sin t,\; 0 \le t \le 2\pi\).
Solution
Use the identity \(x^2 + y^2 = \cos^2 t + \sin^2 t = 1\).
The point lies on the unit circle. Here \(t\) is the angle measured from the positive \(x\)-axis.
As \(t\) runs from \(0\) to \(2\pi\), the point travels once counterclockwise, starting and ending at \((1,0)\).
Circle of Radius r, Center (h, k)
\[ x = h + r\cos t, \qquad y = k + r\sin t, \qquad 0 \le t \le 2\pi \]
The Cycloid

One of the most celebrated parametric curves is the cycloid: the path of a fixed point on the rim of a circle that rolls along a straight line without slipping. It resists a tidy Cartesian description but yields a clean parametric one.

Cycloid

The trajectory of a marked point on a rolling circle. Choosing the angle of rotation \(\theta\) as the parameter, the rolled distance equals the arc length \(r\theta\), which fixes both coordinates of the point.

Cycloid (radius r, one point at the origin)
\[ x = r(\theta - \sin\theta), \qquad y = r(1 - \cos\theta) \]
r one arch
Figure 2One arch of a cycloid (\(0 \le \theta \le 2\pi\)). The same curve solves both the brachistochrone (shortest-time descent) and tautochrone (equal-time descent) problems. (Original redrawn figure.)

Instructor note — add a worked mechanical-engineering tie-in here. (Suggested: cycloidal gear profiles and the cycloidal-drive reducer, where the rolling-circle geometry is exactly this curve.)

10.2

Calculus with Parametric Curves

Instructor note — section note slot. (Suggested: stress that \(dy/dx\) for a parametric curve is a ratio of velocities; the chain rule does all the work and no parameter elimination is needed.)

Tangents

Applying the Chain Rule to \(y=F(x)\) with \(x=f(t)\) gives the slope of a parametric curve directly, without first eliminating the parameter.

Slope of the Tangent
\[ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \quad\text{if}\quad \frac{dx}{dt} \ne 0 \]

Horizontal and Vertical Tangents

The curve has a horizontal tangent where \(dy/dt = 0\) (and \(dx/dt \ne 0\)), and a vertical tangent where \(dx/dt = 0\) (and \(dy/dt \ne 0\)). These points are the key landmarks when sketching the curve by hand.

Second Derivative (Concavity)
\[ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} \]
EXAMPLE 1A Self-Intersecting Curve with Two Tangents
For \(x = t^3 - 3t,\; y = t^2\), show the curve crosses itself and find the two tangent lines there.
Solution
Look for two parameters giving the same point. Setting \(t = \sqrt3\) and \(t = -\sqrt3\): both give \(y = t^2 = 3\) and \(x = t^3 - 3t = 0\). So the curve passes through \((0,3)\) twice and crosses itself there.
Slope: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{3t^2 - 3}\).
At \(t = \sqrt3\), \(\dfrac{dy}{dx} = \dfrac{2\sqrt3}{6} = \dfrac{1}{\sqrt3}\); at \(t = -\sqrt3\), the slope is \(-\dfrac{1}{\sqrt3}\). The two tangents at \((0,3)\) are \(y - 3 = \pm\dfrac{1}{\sqrt3}\,x\).
Areas

The area under one sweep of a parametric curve follows from the Cartesian area formula by the Substitution Rule.

Area Under a Parametric Curve
\[ A = \int_{\alpha}^{\beta} g(t)\,f'(t)\,dt \]
EXAMPLE 3Area Under One Arch of a Cycloid
Find the area under one arch of the cycloid \(x = r(\theta-\sin\theta),\; y = r(1-\cos\theta)\).
Solution
With \(dx = r(1-\cos\theta)\,d\theta\), the area becomes \(\displaystyle\int_0^{2\pi} r^2(1-\cos\theta)^2\,d\theta\).
Expanding and using \(\cos^2\theta = \tfrac12(1+\cos 2\theta)\) gives \(A = 3\pi r^2\).
The area under one arch is exactly three times the area of the rolling circle — a result first proved by Roberval and Torricelli.
Arc Length

When a curve is traversed exactly once as \(t\) increases, its length is obtained from the velocities in each coordinate.

Arc Length of a Parametric Curve

If \(x=f(t),\; y=g(t),\; \alpha \le t \le \beta\), where \(f'\) and \(g'\) are continuous on \([\alpha,\beta]\) and the curve is traversed exactly once as \(t\) increases from \(\alpha\) to \(\beta\), then the length of the curve is

\[ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt \]
EXAMPLE 5Length of One Arch of a Cycloid
Find the length of one arch of the cycloid \(x = r(\theta-\sin\theta),\; y = r(1-\cos\theta)\).
Solution
Here \(\left(\tfrac{dx}{d\theta}\right)^2 + \left(\tfrac{dy}{d\theta}\right)^2 = r^2(2 - 2\cos\theta)\).
Using \(1-\cos\theta = 2\sin^2(\theta/2)\), the integrand becomes \(2r\,\sin(\theta/2)\).
Integrating over \(0 \le \theta \le 2\pi\) gives \(L = 8r\) — the arch is eight radii long (Sir Christopher Wren, 1658).
Surface Area
Surface of Revolution about the x-axis
\[ S = \int_{\alpha}^{\beta} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt \]

Instructor note — note slot. (Suggested: surface-of-revolution integrals are the basis of CNC-turned part surface-area and material estimates.)

10.3

Polar Coordinates

Instructor note — section note slot. (Suggested: warn students early that a point has infinitely many polar names, which is the single biggest source of errors in this section.)

Polar Coordinate System

A point is located by a directed distance \(r\) from a fixed pole \(O\) and an angle \(\theta\) measured from the polar axis. Unlike Cartesian coordinates, each point has many representations: adding any multiple of \(2\pi\) to \(\theta\) names the same point, and a negative \(r\) places the point on the opposite ray.

Polar to Cartesian
\[ x = r\cos\theta, \qquad y = r\sin\theta \]
Cartesian to Polar
\[ r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x} \]
polar axis O P(r, θ) r θ
Figure 3A point \(P\) located by its distance \(r\) from the pole \(O\) and the angle \(\theta\) from the polar axis. (Original redrawn figure.)
EXAMPLE 2Polar to Cartesian Conversion
Convert \((2, \pi/3)\) from polar to Cartesian coordinates.
Solution
\(x = 2\cos(\pi/3) = 2 \cdot \tfrac12 = 1\).
\(y = 2\sin(\pi/3) = 2 \cdot \tfrac{\sqrt3}{2} = \sqrt3\).
The Cartesian point is \((1, \sqrt3)\).
Polar Curves

The graph of \(r = f(\theta)\) consists of all points with at least one polar representation satisfying the equation. Reading \(r\) as a function of \(\theta\) on a Cartesian grid first makes sketching the polar curve far easier.

Cardioid and Rose Curves

\(r = 1 + \sin\theta\) traces a heart-shaped cardioid. Equations of the form \(r = \cos n\theta\) or \(r = \sin n\theta\) produce rose curves: \(n\) petals when \(n\) is odd, and \(2n\) petals when \(n\) is even.

r = 1 + sin θ
Figure 4The cardioid \(r = 1 + \sin\theta\), symmetric about the vertical line \(\theta = \pi/2\). (Original redrawn figure.)
Symmetry

Three Symmetry Tests

A polar equation is symmetric about the polar axis if it is unchanged when \(\theta\) is replaced by \(-\theta\); about the pole if unchanged when \(r \to -r\) (or \(\theta \to \theta + \pi\)); and about the vertical line \(\theta = \pi/2\) if unchanged when \(\theta \to \pi - \theta\). Exploiting symmetry halves the plotting work.

Tangents to Polar Curves

Writing \(x = r\cos\theta,\; y = r\sin\theta\) with \(\theta\) as the parameter and applying the parametric slope formula gives the tangent slope of any polar curve.

Tangent Slope for a Polar Curve
\[ \frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta} \]

Instructor note — note slot. (Suggested: polar plots are the natural language for antenna radiation patterns and microphone pickup patterns — the cardioid mic in Example/Exercise material is a real-world hook.)

10.4

Areas and Lengths in Polar Coordinates

Instructor note — section note slot. (Suggested: remind students the area element is a thin sector, not a rectangle — that is why the integrand is \(\tfrac12 r^2\), not \(r\).)

Area Swept by a Polar Curve

The region bounded by \(r = f(\theta)\) and the rays \(\theta = a,\; \theta = b\) is built from thin circular sectors of area \(\tfrac12 r^2\,d\theta\). Summing these as a Riemann sum and taking the limit gives the polar area integral.

Sector Area (building block)
\[ A = \tfrac{1}{2}r^2\theta \]
Area in Polar Coordinates
\[ A = \int_{a}^{b} \tfrac{1}{2}\,[f(\theta)]^2\,d\theta \;=\; \int_{a}^{b} \tfrac{1}{2}\,r^2\,d\theta \]
O θ=a θ=b r=f(θ)
Figure 5The area is swept out by a ray rotating from \(\theta = a\) to \(\theta = b\); each thin wedge is a circular sector. (Original redrawn figure.)
EXAMPLE 1Area of One Loop of a Four-Leaved Rose
Find the area enclosed by one loop of \(r = \cos 2\theta\).
Solution
The right loop is swept as \(\theta\) goes from \(-\pi/4\) to \(\pi/4\).
\(A = \displaystyle\int_{-\pi/4}^{\pi/4} \tfrac12\cos^2 2\theta\,d\theta = \tfrac12\int_{-\pi/4}^{\pi/4}\tfrac12(1+\cos 4\theta)\,d\theta\).
Evaluating gives \(A = \dfrac{\pi}{8}\).

Finding Intersections

Because a single point has many polar names, solving two polar equations simultaneously can miss intersection points — the pole especially. The reliable method is to graph both curves and inspect where they meet.

Arc Length in Polar Coordinates
Length of a Polar Curve
\[ L = \int_{a}^{b} \sqrt{\,r^2 + \left(\frac{dr}{d\theta}\right)^2\,}\;d\theta \]
EXAMPLE 4Length of a Cardioid
Find the length of the cardioid \(r = 1 + \sin\theta\).
Solution
The full curve corresponds to \(0 \le \theta \le 2\pi\), with \(r^2 + (dr/d\theta)^2 = 2 + 2\sin\theta\).
So \(L = \displaystyle\int_0^{2\pi}\sqrt{2 + 2\sin\theta}\;d\theta\).
Evaluating (the integrand can be simplified algebraically) gives \(L = 8\).

Instructor note — note slot. (Suggested: polar area integrals model the effective coverage area of a directional sensor or sound-pickup field.)

10.5

Conic Sections

Instructor note — section note slot. (Suggested: connect each conic to a real engineering use — parabolic reflectors and satellite dishes, elliptical lithotripsy reflectors, hyperbolic LORAN/cooling-tower profiles.)

Parabolas, ellipses, and hyperbolas are the curves formed when a plane cuts a cone — hence conic sections. Each has a clean standard equation when centered or vertexed at the origin.

Parabola

Parabola

The set of points equidistant from a fixed focus and a fixed line, the directrix. The point midway between them is the vertex; the line through the focus perpendicular to the directrix is the axis.

Parabola, Vertex at Origin (focus (0, p))
\[ x^2 = 4py \qquad\text{(opens up if } p>0,\ \text{down if } p<0) \]
Parabola, Vertex at Origin (focus (p, 0))
\[ y^2 = 4px \qquad\text{(opens right if } p>0,\ \text{left if } p<0) \]
y = −p F(0, p)
Figure 6A parabola \(x^2 = 4py\): every point is equidistant from the focus \(F\) and the directrix. (Original redrawn figure.)
Ellipse

Ellipse

The set of points whose distances to two fixed foci have a constant sum. The longer symmetry segment is the major axis; its endpoints are the vertices. Planetary orbits are ellipses with the sun at one focus.

Ellipse (foci on x-axis, a > b > 0)
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad c^2 = a^2 - b^2,\ \text{foci }(\pm c, 0) \]
Hyperbola

Hyperbola

The set of points whose distances to two foci have a constant difference. It has two separate branches and a pair of asymptotes the branches approach but never reach.

Hyperbola (foci on x-axis)
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad c^2 = a^2 + b^2,\ \text{asymptotes } y = \pm\frac{b}{a}x \]
EXAMPLE 2Sketching an Ellipse and Locating Foci
Sketch \(9x^2 + 16y^2 = 144\) and locate its foci.
Solution
Divide by 144: \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\), so \(a = 4,\ b = 3\).
\(c^2 = a^2 - b^2 = 16 - 9 = 7\), so \(c = \sqrt 7\).
Intercepts at \((\pm 4, 0)\) and \((0, \pm 3)\); foci at \((\pm\sqrt7,\,0)\).

Shifted Conics

Moving a conic so its center (or vertex) sits at \((h,k)\) is done by replacing \(x\) with \(x-h\) and \(y\) with \(y-k\) in the standard equation. Completing the square turns a general second-degree equation back into this recognizable form.

Instructor note — note slot. (Suggested: completing the square to classify a general conic is the same algebra used to find principal axes in stress/inertia problems.)

10.6

Conic Sections in Polar Coordinates

Instructor note — section note slot. (Suggested: the focus-at-origin polar form is the equation of orbital mechanics — it is why this section closes the chapter with Kepler's laws.)

Placing the focus at the origin unifies all three conics under a single number, the eccentricity \(e\), and gives each a compact polar equation — the form used to describe orbits of planets, satellites, and comets.

Focus–Directrix Definition of a Conic

Let \(F\) be a fixed focus, \(l\) a fixed directrix, and \(e > 0\) a fixed eccentricity. The set of points \(P\) with \(\dfrac{|PF|}{|Pl|} = e\) is a conic section:

  • an ellipse if \(e < 1\),
  • a parabola if \(e = 1\),
  • a hyperbola if \(e > 1\).
Polar Equation of a Conic (focus at origin)
\[ r = \frac{ed}{1 \pm e\cos\theta} \qquad\text{or}\qquad r = \frac{ed}{1 \pm e\sin\theta} \]
EXAMPLE 2Identifying a Conic from Its Polar Equation
For \(r = \dfrac{10}{3 - 2\cos\theta}\), find the eccentricity and identify the conic.
Solution
Divide numerator and denominator by 3: \(r = \dfrac{10/3}{1 - \tfrac23\cos\theta}\).
Matching the standard form gives \(e = \tfrac23\).
Since \(e < 1\), the conic is an ellipse.

Orbits: Perihelion and Aphelion

For an elliptical orbit with the sun at a focus, the closest point is the perihelion at distance \(a(1-e)\), and the farthest is the aphelion at distance \(a(1+e)\), where \(a\) is the semimajor axis.

Orbit Equation (semimajor axis a, eccentricity e)
\[ r = \frac{a(1 - e^2)}{1 + e\cos\theta} \]
Kepler's Laws of Planetary Motion

First: each planet moves in an ellipse with the sun at one focus.

Second: the line from the sun to a planet sweeps out equal areas in equal times.

Third: the square of a planet's orbital period is proportional to the cube of the major-axis length of its orbit.

EXAMPLE 5Earth's Orbit and Its Extreme Distances
Given eccentricity \(e \approx 0.017\) and major-axis length \(2a \approx 2.99 \times 10^8\) km, find the perihelion and aphelion distances of Earth's orbit.
Solution
Semimajor axis \(a \approx 1.495 \times 10^8\) km.
Perihelion: \(a(1-e) \approx 1.47 \times 10^8\) km.
Aphelion: \(a(1+e) \approx 1.52 \times 10^8\) km.

Instructor note — closing note slot. (Suggested: this polar-conic + Kepler material is the bridge into orbital mechanics and the vis-viva equation in later dynamics courses.)