Contents / Calculus / Parametric Equations and Polar Coordinates

Chapter 8

Parametric Equations and Polar Coordinates

Parametric curves, polar coordinates, and conic sections.

Introduction

Most curves you meet in calculus are written as $y = f (x)$ . Many of the most important paths in physics, engineering and geometry are not: a planet's orbit, a point on a rolling wheel, the petals of a rose curve. None of them is the graph of a function, and forcing them into the form $y = f (x)$ either fails outright or produces an equation nobody can work with.

This chapter supplies two other languages. Parametric equations give $x$ and $y$ separately as functions of a third variable $t$ , so that the curve is traced as $t$ runs — the language of motion, where $t$ is time and $(x (t), y (t))$ is position. Polar coordinates replace horizontal and vertical distance with distance from the origin and direction, so that anything circular, spiralling or rotationally symmetric becomes simple.

Every tool of single-variable calculus — tangents, concavity, arc length, area, surface area — is rebuilt in each language, and the chapter ends with the conic sections, first in Cartesian form and then, with a single parameter deciding ellipse, parabola or hyperbola, in polar form.

8.1Curves defined by parametric equations

Definition 8.1 (Parametric curve). Let $x = f (t)$ and $y = g (t)$ be functions defined on an interval $I$ . The parametric curve with these parametric equations is the set of points $(f (t), g (t))$ for $t$ in $I$ ; the variable $t$ is the parameter. If $I = [a, b]$ , the curve has initial point $(f (a), g (a))$ and terminal point $(f (b), g (b))$ .

The curve is a set of points, but the equations say more than the set does. They say in which direction the curve is traced as $t$ increases — its orientation — and how fast, and whether some part of it is covered more than once. Two different pairs of equations can trace the same set of points, and when they do it is the equations, not the picture, that decide the answers to every calculus question in this chapter.

The classic example is the unit circle: $x = cos t$ , $y = sin t$ for $0 \leq t \leq 2 π$ . Eliminating the parameter, which here means using $cos^{2} t + sin^{2} t = 1$ , recovers $x^{2} + y^{2} = 1$ , but the parametric form also tells you that the circle is traced counterclockwise, starting and ending at $(1, 0)$ , exactly once. A projectile's position is another: $x = (v_{0} cos α) t$ and $y = (v_{0} sin α) t - \frac{1}{2} g t^{2}$ , where the parameter is literally time.

To eliminate the parameter, solve one equation for $t$ and substitute into the other, or find an identity that combines the two equations without $t$ . When that is impossible, or produces an unreadable equation, the parametric form is the only sensible description — which is precisely why it exists.

Intuition. A hiker's GPS logs a position every second: $(x, y)$ at $t = 0$ , at $t = 1$ , at $t = 2$ and so on. The trail on the map is the curve; the log is the parametrization. Two hikers can walk the same trail and produce completely different logs — one faster, one in the opposite direction, one who walks a loop twice — and the trail alone cannot tell you which log was which. Everything in this chapter that depends on speed or direction depends on the log.

Example 8.2 (Sketching from a table of values). Sketch and identify the curve $x = t^{2} - 2 t$ , $y = t + 1$ .

Solution. Tabulate a few parameter values. At $t = - 2, - 1, 0, 1, 2, 3, 4$ the points are $(8, - 1)$ , $(3, 0)$ , $(0, 1)$ , $(- 1, 2)$ , $(0, 3)$ , $(3, 4)$ and $(8, 5)$ . Plotting them and joining in order of increasing $t$ gives a curve that comes in from the far right, turns at $(- 1, 2)$ and leaves to the right again, moving upward throughout.

To identify it, eliminate the parameter. From the second equation $t = y - 1$ , so

x = (y - 1)^{2} - 2 (y - 1) = y^{2} - 4 y + 3 = (y - 2)^{2} - 1.

This is a parabola with vertex $(- 1, 2)$ opening to the right, which matches the table: the vertex is the point at $t = 1$ , and the orientation is upward because $y = t + 1$ increases with $t$ .□

Example 8.3 (Eliminating the parameter). Given $x = 2 cos t$ and $y = 3 sin t$ , $0 \leq t \leq 2 π$ , eliminate $t$ to find the Cartesian equation, and describe how the curve is traced.

Solution.

Solve for the trigonometric functions: $cos t = x /2$ and $sin t = y /3$ .
Use $cos^{2} t + sin^{2} t = 1$ :

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1.

This is an ellipse centered at the origin with semi-axes $2$ (along $x$ ) and $3$ (along $y$ ).
At $t = 0$ the point is $(2, 0)$ ; at $t = π /2$ it is $(0, 3)$ . The ellipse is traced counterclockwise, once, starting and ending at $(2, 0)$ .

□

Example 8.4 (Same set, different parametrization). Describe the curve $x = sin 2 t$ , $y = cos 2 t$ , $0 \leq t \leq 2 π$ .

Solution. Since $x^{2} + y^{2} = sin^{2} 2 t + cos^{2} 2 t = 1$ , every point lies on the unit circle. But the details differ from the standard parametrization in three ways.

The initial point is $(sin 0, cos 0) = (0, 1)$ , the top of the circle. At $t = π /4$ the point is $(1, 0)$ : the motion is clockwise. And $sin 2 t$ , $cos 2 t$ have period $π$ , so as $t$ runs from $0$ to $2 π$ the circle is traced twice. Any arc length computed from these equations over $[0, 2 π]$ will be $4 π$ , not $2 π$ .□

Pitfall. Eliminating the parameter can enlarge the curve. For $x = sin t$ , $y = sin^{2} t$ the relation is $y = x^{2}$ , but the parametric curve is only the piece of that parabola with $- 1 \leq x \leq 1$ , traced back and forth forever. After eliminating, always ask which part of the Cartesian curve the parameter actually reaches.

Method 8.5 (Standard parametrizations). The line segment from $P_{0} = (x_{0}, y_{0})$ to $P_{1} = (x_{1}, y_{1})$ :

x = x_{0} + (x_{1} - x_{0}) t, y = y_{0} + (y_{1} - y_{0}) t, 0 \leq t \leq 1.

The circle of radius $R$ centered at $(h, k)$ , counterclockwise from the rightmost point:

x = h + R cos t, y = k + R sin t, 0 \leq t \leq 2 π .

The graph of $y = F (x)$ for $a \leq x \leq b$ : $x = t$ , $y = F (t)$ , $a \leq t \leq b$ . Replacing $t$ by $- t$ in any parametrization reverses the orientation.

For instance, the segment from $(1, 2)$ to $(4, - 3)$ is $x = 1 + 3 t$ , $y = 2 - 5 t$ for $0 \leq t \leq 1$ : at $t = 0$ it sits at $(1, 2)$ and at $t = 1$ at $(4, - 3)$ , as required.

Definition 8.6 (Cycloid). A cycloid is the curve traced by a fixed point $P$ on the rim of a circle as the circle rolls without slipping along a straight line.

Example 8.7 (Parametrizing the cycloid). A circle of radius $a$ rolls along the $x$ -axis, and $P$ is the point of the rim that starts at the origin. Find parametric equations for the path of $P$ .

Solution. Use as parameter the angle $θ$ through which the circle has rotated. Because the circle rolls without slipping, the distance it has travelled along the axis equals the arc length that has touched the axis, which is $a θ$ . So after rotating through $θ$ the center is at $C = (a θ, a)$ .

The point $P$ started at the bottom of the circle and has rotated clockwise through $θ$ about the center, so it sits at horizontal displacement $- a sin θ$ and vertical displacement $- a cos θ$ from $C$ . Therefore

x = a (θ - sin θ), y = a (1 - cos θ), θ \in R .

Check: at $θ = 0$ the point is $(0, 0)$ ; at $θ = π$ it is $(πa, 2 a)$ , the top of the first arch; at $θ = 2 π$ it is $(2 πa, 0)$ , back on the axis one circumference to the right. One arch corresponds to $0 \leq θ \leq 2 π$ .□

Eliminating $θ$ from the cycloid is possible but produces an equation involving $cos^{- 1}$ and a square root that nobody uses. The cycloid is famous for two properties that were discovered in the seventeenth century: it is the brachistochrone, the curve along which a bead slides under gravity from one point to a lower one in the least time, and the tautochrone, along which the bead reaches the bottom in the same time wherever it starts. Both were found by working with the parametric equations directly.

Exercises for §8.1 →

8.2Tangents and concavity of parametric curves

Theorem 8.8 (Slope of a parametric curve). Let $x = f (t)$ and $y = g (t)$ have continuous derivatives, with $f^{'} (t) \neq = 0$ . Then the curve has a tangent line at the point with parameter $t$ , with slope

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{g ^{'} ( t )}{f ^{'} ( t )} .

Proof. Because $f^{'} (t) \neq = 0$ and $f^{'}$ is continuous, $f^{'}$ keeps its sign on an interval around $t$ , so $f$ is monotone there and has a differentiable inverse: near the point, $t$ is a function of $x$ with $d t / d x = 1/ f^{'} (t)$ by the formula for the derivative of an inverse. Along the curve $y = g (t (x))$ , so by the chain rule

\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x} = g^{'} (t) \cdot \frac{1}{f ^{'} ( t )} .

∎

The formula is a ratio of speeds: the vertical speed $d y / d t$ divided by the horizontal speed $d x / d t$ . It says nothing about the parameter itself, which is why reparametrizing a curve (walking it faster or in the other direction) never changes the tangent line.

Horizontal tangents occur where $d y / d t = 0$ and $d x / d t \neq = 0$ ; vertical tangents where $d x / d t = 0$ and $d y / d t \neq = 0$ . When both vanish at the same $t$ the formula says nothing, and the curve may have a cusp, a corner or a perfectly ordinary tangent; the only way to know is to examine the limit of $d y / d x$ as the parameter approaches that value.

Intuition. A car's dashboard shows an eastward speedometer reading $3$ m/s and a northward one reading $4$ m/s. The road under the car is climbing at slope $4/3$ , whatever the clock says. If the driver had been going twice as fast, both readings would double and the slope would be unchanged; if the eastward reading drops to zero while the northward one does not, the car is heading straight north and the road is vertical.

Theorem 8.9 (Second derivative of a parametric curve). Under the hypotheses of the slope theorem, if $f$ and $g$ are also twice differentiable then

\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{\frac{d x}{d t}} = \frac{f ^{'} ( t ) g ^{''} ( t ) - g ^{'} ( t ) f ^{''} ( t )}{( f ^{'} ( t ) ) ^{3}} .

Proof. The slope $d y / d x$ is itself a function of $t$ , and so is $x$ ; the pair $(x (t), \frac{d y}{d x} (t))$ is a parametric curve, and the second derivative is the slope of that curve. Applying the slope theorem to it gives the first expression. For the second, differentiate the quotient $g^{'} / f^{'}$ with the quotient rule and divide by $f^{'}$ .∎

Pitfall. The second derivative is not $g^{''} (t) / f^{''} (t)$ . You must differentiate the slope with respect to $t$ and then divide by $d x / d t$ again. Getting this wrong is the single most common error in the whole chapter, and it produces answers with the wrong sign, so concavity comes out backwards.

Example 8.10 (Tangent line and a cusp). Find the tangent line to $x = t^{2}$ , $y = t^{3}$ at $t = 1$ , and describe the curve at $t = 0$ .

Solution.

Derivatives: $d x / d t = 2 t$ and $d y / d t = 3 t^{2}$ , so for $t \neq = 0$ the slope is $3 t^{2} / (2 t) = 3 t /2$ .
At $t = 1$ the slope is $3/2$ and the point is $(1, 1)$ , so the tangent line is $y - 1 = \frac{3}{2} (x - 1)$ , that is, $y = \frac{3}{2} x - \frac{1}{2}$ .
At $t = 0$ both derivatives vanish, so the slope theorem is silent. But $d y / d x = 3 t /2 \to 0$ as $t \to 0$ from either side, so both branches arrive at the origin tangent to the $x$ -axis. Eliminating the parameter gives $y^{2} = x^{3}$ : the curve lives in $x \geq 0$ and has a cusp at the origin, with the two branches ( $y = \pm x^{3/2}$ ) meeting at a sharp point along a horizontal direction.

□

Example 8.11 (A curve that crosses itself). The curve $x = t^{2}$ , $y = t^{3} - 3 t$ passes through $(3, 0)$ twice. Find both tangent lines there, find every point with a horizontal or vertical tangent, and determine where the curve is concave up.

Solution.

The point $(3, 0)$ needs $t^{3} - 3 t = t (t^{2} - 3) = 0$ and $t^{2} = 3$ , so $t = \pm 3$ . The slope is

\frac{d y}{d x} = \frac{3 t ^{2} - 3}{2 t} .

At $t = 3$ this is $6/ (23) = 3$ and at $t = - 3$ it is $- 3$ . The two tangent lines are $y = 3 (x - 3)$ and $y = - 3 (x - 3)$ .

Horizontal tangents: $3 t^{2} - 3 = 0$ gives $t = \pm 1$ , where $d x / d t = \pm 2 \neq = 0$ . The points are $(1, - 2)$ at $t = 1$ and $(1, 2)$ at $t = - 1$ .
Vertical tangents: $2 t = 0$ gives $t = 0$ , where $d y / d t = - 3 \neq = 0$ . The point is $(0, 0)$ .
Concavity: write the slope as $\frac{3}{2} (t - t^{- 1})$ . Then

\frac{d}{d t} (\frac{d y}{d x}) = \frac{3}{2} (1 + \frac{1}{t ^{2}}) = \frac{3 ( t ^{2} + 1 )}{2 t ^{2}}, \frac{d ^{2} y}{d x ^{2}} = \frac{3 ( t ^{2} + 1 )}{2 t ^{2}} \cdot \frac{1}{2 t} = \frac{3 ( t ^{2} + 1 )}{4 t ^{3}} .

The sign is the sign of $t$ : the curve is concave up for $t > 0$ (the lower half of the loop and the lower branch) and concave down for $t < 0$ . Sanity check: at $t = 1$ the tangent is horizontal at $(1, - 2)$ and the curve is concave up there, so $(1, - 2)$ is the lowest point of the loop — which it is.□

Example 8.12 (Concavity of a circle). Compute the second derivative for $x = cos t$ , $y = sin t$ , and check it against the shape of the circle.

Solution. Here $d x / d t = - sin t$ and $d y / d t = cos t$ , so $d y / d x = - cot t$ for $sin t \neq = 0$ . Differentiating in $t$ gives $csc^{2} t$ , and dividing by $d x / d t = - sin t$ ,

\frac{d ^{2} y}{d x ^{2}} = \frac{csc ^{2} t}{- sin t} = - \frac{1}{sin ^{3} t} .

For $0 < t < π$ (the upper semicircle) this is negative, and for $π < t < 2 π$ it is positive: the top of a circle is concave down and the bottom concave up, as it should be. Had we wrongly computed $y^{''} (t) / x^{''} (t) = (- sin t) / (- cos t) = tan t$ , the sign would change at $t = π /2$ , which is nonsense for a circle.□

Example 8.13 (Tangents to the cycloid). For the cycloid $x = a (θ - sin θ)$ , $y = a (1 - cos θ)$ , find the slope at $θ = π /3$ and locate every horizontal and vertical tangent.

Solution.

$d x / d θ = a (1 - cos θ)$ and $d y / d θ = a sin θ$ , so

\frac{d y}{d x} = \frac{sin θ}{1 - cos θ} .

At $θ = π /3$ : $sin θ = 3 /2$ and $1 - cos θ = 1/2$ , so the slope is $3$ . The point is $(a (π /3 - 3 /2), a /2)$ .
Horizontal tangents: $sin θ = 0$ with $1 - cos θ \neq = 0$ gives $θ = π, 3 π, \dots$ , the tops of the arches at $((2 n + 1) πa, 2 a)$ .
Vertical tangents: $1 - cos θ = 0$ at $θ = 2 nπ$ , but $sin θ = 0$ there too. Examine the limit with L'Hôpital's Rule:

θ \to 2 n π^{+} lim \frac{sin θ}{1 - cos θ} = θ \to 2 n π^{+} lim \frac{cos θ}{sin θ} = + \infty,

and the left-hand limit is $- \infty$ . So the cycloid has a vertical tangent at each point $(2 nπa, 0)$ where it touches the axis. These points are cusps: the curve comes down vertically and leaves vertically, with $P$ momentarily at rest.□

Exercises for §8.2 →

8.3Arc length of parametric curves

Theorem 8.14 (Arc length of a parametric curve). Suppose $x = f (t)$ and $y = g (t)$ have continuous derivatives on $[a, b]$ and the curve is traversed exactly once as $t$ increases from $a$ to $b$ . Then its length is

L = \int_{a}^{b} (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} d t .

Proof. Partition $[a, b]$ into subintervals of width $Δ t$ with endpoints $t_{i}$ , and join the corresponding points of the curve by a polygon. The $i$ -th side has length $(Δ x_{i})^{2} + (Δ y_{i})^{2}$ , where $Δ x_{i} = f (t_{i}) - f (t_{i - 1})$ and similarly for $y$ . By the Mean Value Theorem there are points $t_{i}^{*}$ and $t_{i}^{**}$ in the subinterval with $Δ x_{i} = f^{'} (t_{i}^{*}) Δ t$ and $Δ y_{i} = g^{'} (t_{i}^{**}) Δ t$ , so the polygon has length

i \sum f^{'} (t_{i}^{*})^{2} + g^{'} (t_{i}^{**})^{2} Δ t .

This is almost a Riemann sum for the integrand $f^{'2} + g^{'2}$ — almost, because the two sample points differ. Continuity of $f^{'}$ and $g^{'}$ makes that difference vanish as $Δ t \to 0$ (the real-analysis detail is the uniform continuity of the integrand on $[a, b]$ ), and the polygon lengths converge to the integral.∎

The integrand is the speed $∣ v (t) ∣ = x^{'} (t)^{2} + y^{'} (t)^{2}$ of a particle moving along the curve, so the theorem says that distance travelled is the integral of speed — the odometer integrates the speedometer. Taking $x = t$ , $y = F (t)$ recovers the familiar formula $L = \int_{a}^{b} 1 + F^{'} (x)^{2} d x$ for a graph.

The hypothesis that the curve is traversed once matters. Speed is never negative, so a particle that retraces its path accumulates length on every pass; the integral measures distance travelled, which exceeds the length of the curve whenever any part is covered twice.

Intuition. Two runners complete the same $400$ m track. One runs it once in $60$ seconds, the other twice in $120$ seconds; their logs are different parametrizations of the same curve. Integrating speed over each log gives $400$ and $800$ : the arc-length integral reports what the runner did, not what the track looks like. If you want the track, integrate over one lap.

Example 8.15 (A circle traced once and twice). Find the length of $x = cos t$ , $y = sin t$ for $0 \leq t \leq 2 π$ , and the distance travelled along $x = sin 2 t$ , $y = cos 2 t$ for $0 \leq t \leq 2 π$ .

Solution. For the first curve the speed is $sin^{2} t + cos^{2} t = 1$ , so $L = \int_{0}^{2 π} 1 d t = 2 π$ , the circumference of the unit circle.

For the second, $d x / d t = 2 cos 2 t$ and $d y / d t = - 2 sin 2 t$ , so the speed is $2$ and the integral is $\int_{0}^{2 π} 2 d t = 4 π$ . This is twice the circumference, because the circle is traced twice; the length of the curve itself is still $2 π$ , obtained by integrating over $0 \leq t \leq π$ .□

Example 8.16 (Arc length by substitution). Find the length of $x = t^{2}$ , $y = t^{3}$ for $0 \leq t \leq 1$ .

Solution.

$d x / d t = 2 t$ and $d y / d t = 3 t^{2}$ , so the speed is $4 t^{2} + 9 t^{4} = t 4 + 9 t^{2}$ (using $t \geq 0$ ).
Substitute $u = 4 + 9 t^{2}$ , $d u = 18 t d t$ ; when $t = 0$ , $u = 4$ , and when $t = 1$ , $u = 13$ :

L = \int_{0}^{1} t 4 + 9 t^{2} d t = \frac{1}{18} \int_{4}^{13} u d u = \frac{1}{18} \cdot \frac{2}{3} [u^{3/2}]_{4}^{13} = \frac{13 13 - 8}{27} .

Sanity check: $L \approx 1.44$ . The straight line from $(0, 0)$ to $(1, 1)$ has length $2 \approx 1.41$ , and the curve bows slightly below it, so a length just above $1.41$ is right.

□

Example 8.17 (Length of one arch of the cycloid). Find the length of one arch of the cycloid $x = a (θ - sin θ)$ , $y = a (1 - cos θ)$ .

Solution.

One arch is $0 \leq θ \leq 2 π$ , with $d x / d θ = a (1 - cos θ)$ and $d y / d θ = a sin θ$ .
The speed squared is $a^{2} [(1 - cos θ)^{2} + sin^{2} θ] = a^{2} (2 - 2 cos θ)$ .
Use the half-angle identity $1 - cos θ = 2 sin^{2} (θ /2)$ : the speed is $a 4 sin^{2} (θ /2) = 2 a ∣ sin (θ /2) ∣ = 2 a sin (θ /2)$ , since $sin (θ /2) \geq 0$ on $[0, 2 π]$ .
Integrate:

L = \int_{0}^{2 π} 2 a sin \frac{θ}{2} d θ = 2 a [- 2 cos \frac{θ}{2}]_{0}^{2 π} = 2 a (2 + 2) = 8 a .

One arch of a cycloid is exactly eight times the radius of the rolling circle: a clean number, and the first result of this kind ever found (by Wren, in 1658). Sanity check: the arch spans $2 πa \approx 6.28 a$ along the axis and rises to $2 a$ , so a length a little over $6.28 a$ is reasonable.□

Example 8.18 (Length of a logarithmic spiral). Find the length of $x = e^{t} cos t$ , $y = e^{t} sin t$ for $0 \leq t \leq π$ .

Solution. By the product rule $d x / d t = e^{t} (cos t - sin t)$ and $d y / d t = e^{t} (sin t + cos t)$ . Squaring and adding, the cross terms $\mp 2 sin t cos t$ cancel and

(\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} = e^{2 t} [(cos t - sin t)^{2} + (sin t + cos t)^{2}] = 2 e^{2 t} .

So the speed is $2 e^{t}$ and $L = 2 \int_{0}^{π} e^{t} d t = 2 (e^{π} - 1) \approx 31.3$ . The spiral's distance from the origin is $e^{t}$ , which grows from $1$ to $e^{π} \approx 23$ , so a length of about $31$ for half a turn is sensible.□

Exercises for §8.3 →

8.4Surface area of revolution

Theorem 8.19 (Surface area of a parametric curve rotated about an axis). Let $x = f (t)$ , $y = g (t)$ have continuous derivatives on $[a, b]$ , traverse the curve once, and satisfy $g (t) \geq 0$ . The area of the surface obtained by rotating the curve about the $x$ -axis is

S = \int_{a}^{b} 2 π y (\frac{d x}{d t})^{2} + (\frac{d y}{d t})^{2} d t .

Rotating about the $y$ -axis (with $f (t) \geq 0$ ) replaces $2 π y$ by $2 π x$ .

Proof. Cut the curve into small arcs. Rotating the arc between parameters $t$ and $t + Δ t$ produces a thin band that is nearly a frustum of a cone with slant length $Δ s$ , the arc length, and radius $y$ ; its area is $2 π y Δ s$ up to an error that vanishes faster than $Δ t$ . Summing and passing to the limit gives $S = \int 2 π y d s$ , and the arc length theorem supplies $d s = x^{'2} + y^{'2} d t$ .∎

The formula is the Cartesian surface-area formula $S = \int 2 π y d s$ with the parametric expression for $d s$ substituted; all that changes is the bookkeeping. The factor $2 π y$ is the circumference of the circle swept out by the point, so the integrand is circumference times a sliver of arc.

Example 8.20 (The sphere). Rotate the semicircle $x = R cos t$ , $y = R sin t$ , $0 \leq t \leq π$ , about the $x$ -axis.

Solution. The speed is $R$ and $y = R sin t \geq 0$ on $[0, π]$ , so

S = \int_{0}^{π} 2 π R sin t \cdot R d t = 2 π R^{2} [- cos t]_{0}^{π} = 4 π R^{2},

the surface area of a sphere of radius $R$ .□

Example 8.21 (A parabolic arc). Find the area of the surface obtained by rotating $x = t^{2}$ , $y = 2 t$ , $0 \leq t \leq 1$ , about the $x$ -axis.

Solution.

$d x / d t = 2 t$ and $d y / d t = 2$ , so $d s = 4 t^{2} + 4 d t = 2 t^{2} + 1 d t$ .
With $y = 2 t \geq 0$ ,

S = \int_{0}^{1} 2 π (2 t) \cdot 2 t^{2} + 1 d t = 8 π \int_{0}^{1} t t^{2} + 1 d t .

Substitute $u = t^{2} + 1$ , $d u = 2 t d t$ : the integral is $\frac{1}{2} \int_{1}^{2} u d u = \frac{1}{3} [u^{3/2}]_{1}^{2} = \frac{1}{3} (22 - 1)$ .
So $S = \frac{8 π}{3} (22 - 1) \approx 15.3$ . As a check, the curve runs from $(0, 0)$ to $(1, 2)$ ; a cone with that slant edge would have area $π \cdot 2 \cdot 5 \approx 14.0$ , and the curve bulges outward, so a slightly larger area is right.

□

The polar version of the formula, and an example, appear in the section on arc length in polar coordinates.

Exercises for §8.4 →

8.5Areas with parametric curves

Proposition 8.22 (Area under a parametric curve). Suppose the curve $x = f (t)$ , $y = g (t)$ , $α \leq t \leq β$ , is the graph of a function $y = F (x) \geq 0$ traversed once from left to right, so that $f (α) = a$ , $f (β) = b$ and $f^{'} (t) \geq 0$ . Then the area between the curve and the $x$ -axis is

A = \int_{a}^{b} F (x) d x = \int_{α}^{β} g (t) f^{'} (t) d t .

Proof. This is the substitution $x = f (t)$ , $d x = f^{'} (t) d t$ in the ordinary area integral; the integrand $F (x)$ becomes $F (f (t)) = g (t)$ and the limits $x = a, b$ become $t = α, β$ .∎

If the curve is traversed from right to left the same integral gives $- A$ ; either take the absolute value or reverse the limits. Nothing else changes.

Proposition 8.23 (Area enclosed by a closed curve). If the closed curve $x = f (t)$ , $y = g (t)$ , $α \leq t \leq β$ , is traversed exactly once counterclockwise and does not cross itself, the area it encloses is

A = \int_{α}^{β} x \frac{d y}{d t} d t = - \int_{α}^{β} y \frac{d x}{d t} d t = \frac{1}{2} \int_{α}^{β} (x \frac{d y}{d t} - y \frac{d x}{d t}) d t .

Proof. Suppose first that the region is bounded above by a curve $y = T (x)$ and below by $y = B (x)$ for $a \leq x \leq b$ . A counterclockwise traversal runs along the bottom from left to right and along the top from right to left. By the previous proposition the bottom contributes $\int_{a}^{b} B (x) d x$ to $\int y d x$ and the top contributes $- \int_{a}^{b} T (x) d x$ , so

\int_{α}^{β} y \frac{d x}{d t} d t = \int_{a}^{b} B (x) d x - \int_{a}^{b} T (x) d x = - A .

The same argument with the roles of $x$ and $y$ exchanged gives $\int x d y = A$ (a counterclockwise traversal runs up the right side and down the left). Averaging the two expressions gives the third. A region that is not of this simple shape is cut into pieces that are; the contributions along the cuts cancel because each cut is traversed twice in opposite directions. This is a special case of Green's Theorem.∎

The symmetric form $\frac{1}{2} \int (x d y - y d x)$ is usually the easiest to compute for closed curves, because $x d y - y d x$ often simplifies dramatically — for a circle or ellipse it is a constant.

Intuition. A surveyor walks the boundary of a field counterclockwise, noting position at every step. The running total of $x Δ y$ — each step's eastward position times its northward displacement — adds strips of area while walking up the right-hand side and subtracts them while walking down the left, leaving exactly the area of the field. The planimeter, a nineteenth-century brass instrument that measures the area of any shape when its arm is traced around the boundary, is this formula built in metal.

Example 8.24 (Area under one arch of the cycloid). Find the area between one arch of the cycloid $x = a (θ - sin θ)$ , $y = a (1 - cos θ)$ and the $x$ -axis.

Solution.

On $0 \leq θ \leq 2 π$ the arch runs left to right, since $d x / d θ = a (1 - cos θ) \geq 0$ , and $y \geq 0$ . The first proposition applies:

A = \int_{0}^{2 π} a (1 - cos θ) \cdot a (1 - cos θ) d θ = a^{2} \int_{0}^{2 π} (1 - 2 cos θ + cos^{2} θ) d θ .

Use $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ : the integrand becomes $\frac{3}{2} - 2 cos θ + \frac{1}{2} cos 2 θ$ .
The cosine terms integrate to zero over a full period, leaving

A = a^{2} [\frac{3}{2} θ]_{0}^{2 π} = 3 π a^{2} .

The area under one arch is exactly three times the area of the rolling circle — Galileo guessed this by weighing cut-outs, and Roberval proved it in 1634. Sanity check: the arch fits inside a $2 πa \times 2 a$ rectangle of area $4 π a^{2}$ , and fills three quarters of it.□

Example 8.25 (Area of an ellipse). Find the area enclosed by $x = a cos t$ , $y = b sin t$ , $0 \leq t \leq 2 π$ .

Solution. The ellipse is traversed once counterclockwise. With $d x / d t = - a sin t$ and $d y / d t = b cos t$ ,

x \frac{d y}{d t} - y \frac{d x}{d t} = ab cos^{2} t + ab sin^{2} t = ab,

so $A = \frac{1}{2} \int_{0}^{2 π} ab d t = πab$ . With $a = b = R$ this is $π R^{2}$ .□

Example 8.26 (Area of the astroid). Find the area enclosed by the astroid $x = cos^{3} t$ , $y = sin^{3} t$ , $0 \leq t \leq 2 π$ .

Solution.

The astroid is a four-cusped star with cusps at $(\pm 1, 0)$ and $(0, \pm 1)$ , traversed counterclockwise. Its derivatives are $d x / d t = - 3 cos^{2} t sin t$ and $d y / d t = 3 sin^{2} t cos t$ .
Then

x \frac{d y}{d t} - y \frac{d x}{d t} = 3 sin^{2} t cos^{4} t + 3 sin^{4} t cos^{2} t = 3 sin^{2} t cos^{2} t (cos^{2} t + sin^{2} t) = \frac{3}{4} sin^{2} 2 t .

A = \frac{1}{2} \int_{0}^{2 π} \frac{3}{4} sin^{2} 2 t d t = \frac{3}{8} \int_{0}^{2 π} \frac{1 - cos 4 t}{2} d t = \frac{3}{8} \cdot π = \frac{3 π}{8} .

Sanity check: the astroid is inscribed in the unit square rotated by $4 5^{\circ}$ (the diamond with vertices at the cusps), which has area $2$ , and its concave sides leave it well under half of that; $3 π /8 \approx 1.18$ fits.□

Example 8.27 (Area of a loop). Find the area of the loop of $x = t^{2}$ , $y = t^{3} - 3 t$ .

Solution.

The loop is the part of the curve between the two visits to $(3, 0)$ , that is, $- 3 \leq t \leq 3$ . At $t = - 3, - 1, 0, 1, 3$ the points are $(3, 0)$ , $(1, 2)$ , $(0, 0)$ , $(1, - 2)$ , $(3, 0)$ : the loop is traversed counterclockwise.
Compute $x y^{'} - y x^{'} = t^{2} (3 t^{2} - 3) - (t^{3} - 3 t) (2 t) = 3 t^{4} - 3 t^{2} - 2 t^{4} + 6 t^{2} = t^{4} + 3 t^{2}$ .
The integrand is even, so

A = \frac{1}{2} \int_{- 3}^{3} (t^{4} + 3 t^{2}) d t = \int_{0}^{3} (t^{4} + 3 t^{2}) d t = [\frac{t ^{5}}{5} + t^{3}]_{0}^{3} = \frac{9 3}{5} + 33 = \frac{24 3}{5} .

Sanity check: the loop spans $0 \leq x \leq 3$ and $- 2 \leq y \leq 2$ , a box of area $12$ , and $243 /5 \approx 8.3$ is a plausible fraction of it.□

Exercises for §8.5 →

8.6Polar coordinates

Definition 8.28 (Polar coordinates). Fix a point $O$ , the pole, and a ray from $O$ , the polar axis (drawn as the positive $x$ -axis). A point $P$ has polar coordinates $(r, θ)$ if $r$ is the distance $∣ OP ∣$ and $θ$ is the angle, measured counterclockwise in radians, from the polar axis to the ray $OP$ . By convention $(- r, θ)$ with $r > 0$ denotes the point at distance $r$ on the ray opposite to $θ$ , so that $(- r, θ)$ and $(r, θ + π)$ are the same point; the pole is $(0, θ)$ for every $θ$ .

A point has one pair of Cartesian coordinates but infinitely many polar pairs: $(r, θ)$ , $(r, θ + 2 πn)$ and $(- r, θ + (2 n + 1) π)$ all name the same point. The negative- $r$ convention looks strange, but it is what allows a single equation $r = f (θ)$ to describe curves such as limaçons and roses without splitting them into cases, and it is the source of several genuine traps later in the chapter.

Theorem 8.29 (Converting between polar and Cartesian coordinates). If a point has polar coordinates $(r, θ)$ and Cartesian coordinates $(x, y)$ , then

x = r cos θ, y = r sin θ, r^{2} = x^{2} + y^{2}, tan θ = \frac{y}{x} (x \neq = 0) .

Proof. For $r > 0$ the first two are the definitions of cosine and sine in the right triangle with hypotenuse $OP$ ; for $r < 0$ both sides change sign together. Squaring and adding gives the third, dividing gives the fourth.∎

The first pair converts polar to Cartesian without any thought. The second pair converts Cartesian to polar only with thought, because $tan θ = y / x$ has two solutions in $[0, 2 π)$ and only one of them points at the given quadrant. Always sketch the point, or check the signs of $x$ and $y$ , before trusting an arctangent.

Pitfall. $θ = tan^{- 1} (y / x)$ is correct only when $x > 0$ . The calculator's arctangent lies in $(- π /2, π /2)$ , which covers quadrants I and IV; for a point in quadrant II or III, add $π$ . The point $(- 1, - 1)$ has $tan θ = 1$ , but $θ = π /4$ would put it in the first quadrant; the correct angle is $5 π /4$ .

Intuition. A radar operator does not say "the aircraft is $30$ km east and $40$ km north". She says "range $50$ km, bearing $5 3^{\circ}$ " — because the antenna sweeps in a circle and reports distance and direction. Both descriptions locate the same aircraft. Anything that spins, radiates from a center or is measured by a rotating beam is a polar problem.

Example 8.30 (Plotting polar points). Locate the points with polar coordinates $(1, 5 π /4)$ , $(2, 3 π)$ , $(2, - 2 π /3)$ and $(- 3, 3 π /4)$ .

Solution. $(1, 5 π /4)$ is one unit from the pole in the direction $22 5^{\circ}$ , in the third quadrant. $(2, 3 π)$ is at angle $3 π$ , which is the same direction as $π$ , so it is the point two units along the negative $x$ -axis, $(- 2, 0)$ . $(2, - 2 π /3)$ is measured clockwise by $12 0^{\circ}$ , landing in the third quadrant at Cartesian $(- 1, - 3)$ . Finally $(- 3, 3 π /4)$ has negative $r$ : face the direction $13 5^{\circ}$ (second quadrant) and walk three units backwards, into the fourth quadrant, arriving at the same point as $(3, 7 π /4)$ , which is $(32 /2, - 32 /2)$ .□

Example 8.31 (Converting points). Convert $(2, π /3)$ to Cartesian coordinates, and find all polar representations of the Cartesian point $(1, - 1)$ .

Solution.

$x = 2 cos (π /3) = 1$ and $y = 2 sin (π /3) = 3$ , so the point is $(1, 3)$ .
For $(1, - 1)$ : $r = 1 + 1 = 2$ and $tan θ = - 1$ . The point is in the fourth quadrant, so $θ = - π /4$ (equivalently $7 π /4$ ). One representation is $(2, - π /4)$ ; all of them are $(2, - π /4 + 2 πn)$ and $(- 2, 3 π /4 + 2 πn)$ for integers $n$ .

□

Example 8.32 (Converting equations). Write the polar curve $r = 2 cos θ$ in Cartesian form, and write the Cartesian curves $x = 4$ and $x^{2} - y^{2} = 1$ in polar form.

Solution.

Multiply $r = 2 cos θ$ by $r$ to make the conversions available: $r^{2} = 2 r cos θ$ , that is, $x^{2} + y^{2} = 2 x$ . Completing the square, $(x - 1)^{2} + y^{2} = 1$ : the circle of radius $1$ centered at $(1, 0)$ . (Multiplying by $r$ adds the pole as a solution, but the pole was already on the curve at $θ = π /2$ .)
$x = 4$ becomes $r cos θ = 4$ , that is, $r = 4 sec θ$ .
$x^{2} - y^{2} = 1$ becomes $r^{2} cos^{2} θ - r^{2} sin^{2} θ = 1$ , that is, $r^{2} cos 2 θ = 1$ .

The direction of the traffic is telling. Circles through the origin and spirals are simpler in polar form; lines not through the origin are simpler in Cartesian form.□

Exercises for §8.6 →

8.7Polar curves: sketching and the standard catalogue

A polar curve $r = f (θ)$ is sketched by letting $θ$ increase and following the radius as it grows, shrinks, passes through zero and possibly turns negative. Each sketch below is described precisely rather than shown; read the description with a pencil in hand and reproduce it, then check it against the rose drawn at the end of the section.

Method 8.33 (Sketching a polar curve).

Find the range of $θ$ needed to trace the curve once. Trigonometric curves usually need $[0, 2 π]$ , sometimes only $[0, π]$ .
Apply the symmetry tests below; each symmetry halves the work.
Tabulate $r$ at the standard angles, and find every $θ$ where $r = 0$ (the curve passes through the pole) and where $∣ r ∣$ is largest.
Plot and join the points in order of increasing $θ$ , remembering that a negative $r$ is plotted on the opposite ray. Where $r$ changes sign the curve passes through the pole.

Proposition 8.34 (Symmetry tests for polar curves). Let $C$ be the curve $r = f (θ)$ .

If the equation is unchanged when $θ$ is replaced by $- θ$ , then $C$ is symmetric about the polar axis (the $x$ -axis).
If the equation is unchanged when $r$ is replaced by $- r$ , or when $θ$ is replaced by $θ + π$ , then $C$ is symmetric about the pole.
If the equation is unchanged when $θ$ is replaced by $π - θ$ , then $C$ is symmetric about the line $θ = π /2$ (the $y$ -axis).

Proof. The point $(r, - θ)$ has Cartesian coordinates $(r cos θ, - r sin θ)$ , the reflection of $(r, θ)$ in the $x$ -axis; $(- r, θ)$ is the reflection through the pole; and $(r, π - θ)$ has coordinates $(- r cos θ, r sin θ)$ , the reflection in the $y$ -axis. If the equation is unchanged under the substitution, the reflected point lies on the curve whenever the original does.∎

These are sufficient conditions, not necessary ones: a curve can have a symmetry that none of the three substitutions detects, because of the multiplicity of polar representations. In practice they catch what you need.

The standard curves come in a small number of families, and recognising the family from the equation is most of the skill.

Summary. Lines: $θ = α$ is the line through the pole at angle $α$ ; $r = a sec θ$ is the vertical line $x = a$ ; $r = a csc θ$ is the horizontal line $y = a$ .

Circles: $r = a$ is the circle of radius $a$ about the pole; $r = 2 a cos θ$ is the circle of radius $a$ centered at $(a, 0)$ ; $r = 2 a sin θ$ is the circle of radius $a$ centered at $(0, a)$ . Both of the latter pass through the pole.

Limaçons $r = a \pm b cos θ$ and $r = a \pm b sin θ$ ( $a, b > 0$ ): an inner loop if $a < b$ ; a cardioid (heart shape with a cusp at the pole) if $a = b$ ; a dimple if $b < a < 2 b$ ; convex, oval-like, if $a \geq 2 b$ . The cosine versions are symmetric about the $x$ -axis and the sine versions about the $y$ -axis.

Roses $r = a cos n θ$ and $r = a sin n θ$ for an integer $n \geq 2$ : $n$ petals of length $a$ if $n$ is odd, $2 n$ petals if $n$ is even.

Lemniscates $r^{2} = a^{2} cos 2 θ$ (figure eight along the $x$ -axis) and $r^{2} = a^{2} sin 2 θ$ (along the line $y = x$ ), each loop reaching distance $a$ from the pole.

Spirals: the Archimedean spiral $r = a θ$ , with successive turns a constant distance $2 πa$ apart, and the logarithmic spiral $r = a e^{b θ}$ , which crosses every ray at the same angle.

Theorem 8.35 (Petal rule for roses). For an integer $n \geq 2$ the rose $r = cos n θ$ has $n$ petals if $n$ is odd, traced once as $θ$ runs over $[0, π]$ , and $2 n$ petals if $n$ is even, traced once over $[0, 2 π]$ . The same holds for $r = sin n θ$ .

Proof. The function $cos n θ$ vanishes at $θ = π / (2 n) + kπ / n$ ; between consecutive zeros it keeps one sign and its absolute value rises from $0$ to $1$ and falls back to $0$ . So each interval of length $π / n$ between zeros traces one petal: a loop that leaves the pole, reaches distance $1$ at the midpoint of the interval and returns. There are $2 n$ such intervals in $[0, 2 π]$ .

Compare the petals from $θ$ and from $θ + π$ . If $n$ is odd, $cos n (θ + π) = cos (n θ + nπ) = - cos n θ$ , so the point at parameter $θ + π$ is $(- r, θ + π)$ , which is the same point as $(r, θ)$ . The second half-turn retraces the first: only $n$ distinct petals, all traced on $[0, π]$ . If $n$ is even, $cos n (θ + π) = cos n θ$ , so the point at $θ + π$ is $(r, θ + π)$ , diametrically opposite to $(r, θ)$ . It lies on a different petal, and all $2 n$ petals are distinct.∎

Intuition. A petal is one breath of the radius: out from the pole to full length and back, which takes $θ$ through an interval of width $π / n$ . In one full turn there are $2 n$ breaths, but for odd $n$ the negative breaths land exactly on top of positive ones half a turn later, so only half of them are visible. Three-petal roses really do have three petals; four-petal roses would have had two if the negative breaths had not gone to new places.

Example 8.36 (Sketching a cardioid). Sketch $r = 1 + cos θ$ .

Solution. Replacing $θ$ by $- θ$ leaves the equation unchanged, so the curve is symmetric about the $x$ -axis and it suffices to trace $0 \leq θ \leq π$ . The values are $r = 2$ at $θ = 0$ , $r = 3/2$ at $π /3$ , $r = 1$ at $π /2$ , $r = 1/2$ at $2 π /3$ , and $r = 0$ at $π$ . The radius never goes negative.

Start at $(2, 0)$ on the positive $x$ -axis, the point farthest from the pole. As $θ$ increases the curve swings up and to the left, passing through $(0, 1)$ on the $y$ -axis at $θ = π /2$ , then curls inward, reaching the pole from the direction $θ = π$ : it arrives at the origin travelling along the negative $x$ -axis. The reflection in the $x$ -axis completes the lower half. The result is heart-shaped, widest at $x = 0$ where it spans $- 1 \leq y \leq 1$ , with a cusp at the origin pointing left and a rounded end at $(2, 0)$ .□

Example 8.37 (Sketching a limaçon with an inner loop). Sketch $r = 1 + 2 cos θ$ .

Solution. Again the curve is symmetric about the $x$ -axis. Now $r = 3$ at $θ = 0$ , $r = 2$ at $π /3$ , $r = 1$ at $π /2$ , $r = 0$ at $2 π /3$ , and $r = - 1$ at $π$ . The radius is negative for $2 π /3 < θ < 4 π /3$ , which is what creates the inner loop.

Trace it. From $(3, 0)$ the outer loop swings up through $(0, 1)$ at $θ = π /2$ and reaches the pole at $θ = 2 π /3$ , arriving along the ray at $12 0^{\circ}$ . For $θ$ between $2 π /3$ and $π$ the radius is negative, so the points are plotted on the opposite rays, at angles between $- 6 0^{\circ}$ and $0^{\circ}$ : the curve is now in the fourth quadrant, close to the pole, and at $θ = π$ it reaches $(- 1, π)$ , which is the Cartesian point $(1, 0)$ . That is the far end of the inner loop, on the positive $x$ -axis, one unit from the pole. Reflecting in the $x$ -axis completes both loops. The outer loop is a large closed curve from $(3, 0)$ that returns to the pole from the directions $\pm 12 0^{\circ}$ ; its leftmost point is the pole itself. The inner loop is a small closed curve traced entirely to the right of the pole, spanning $0 \leq x \leq 1$ , tucked inside the outer one.□

Example 8.38 (Sketching a four-petal rose). Sketch $r = cos 2 θ$ .

Solution. The petal rule with $n = 2$ promises four petals, traced on $[0, 2 π]$ . The radius vanishes at $θ = π /4, 3 π /4, 5 π /4, 7 π /4$ and has $∣ r ∣ = 1$ at $θ = 0, π /2, π, 3 π /2$ .

On $- π /4 \leq θ \leq π /4$ the radius is positive, rising to $1$ at $θ = 0$ : one petal along the positive $x$ -axis with tip at $(1, 0)$ , symmetric about that axis. On $π /4 \leq θ \leq 3 π /4$ the radius is negative, reaching $- 1$ at $θ = π /2$ ; the point $(- 1, π /2)$ is $(0, - 1)$ , so this petal points down the negative $y$ -axis. The next interval, $3 π /4 \leq θ \leq 5 π /4$ , gives the petal along the negative $x$ -axis with tip at $(- 1, 0)$ , and the last gives the petal along the positive $y$ -axis. Each petal is a teardrop of length $1$ , and the lines $θ = \pm π /4$ , where the curve passes through the pole, separate consecutive petals. The pen visits the petals in the order right, down, left, up.□

Example 8.39 (Sketching a three-petal rose). Sketch $r = sin 3 θ$ .

Solution. Here $n = 3$ is odd, so there are three petals and $[0, π]$ suffices. The radius vanishes at $θ = 0, π /3, 2 π /3, π$ .

On $[0, π /3]$ the radius is positive with maximum $1$ at $θ = π /6$ : a petal pointing in the direction $3 0^{\circ}$ . On $[π /3, 2 π /3]$ the radius is negative with $r = - 1$ at $θ = π /2$ ; the point $(- 1, π /2)$ is $(0, - 1)$ , so this petal points straight down. On $[2 π /3, π]$ the radius is positive again, maximal at $5 π /6$ : a petal in the direction $15 0^{\circ}$ . The three petals point at $3 0^{\circ}$ , $15 0^{\circ}$ and $27 0^{\circ}$ , equally spaced, and the whole figure is symmetric about the $y$ -axis. As $θ$ continues from $π$ to $2 π$ the pen retraces the same three petals.□

Figure 8.40. The rose

r = sin 3 θ

, traced once as

θ

runs from

0

π

: three petals of length

1

, pointing at

3 0^{\circ}

15 0^{\circ}

and

27 0^{\circ}

Example 8.41 (Sketching a lemniscate and a spiral). Sketch $r^{2} = 4 cos 2 θ$ and $r = θ$ for $θ \geq 0$ .

Solution. For the lemniscate, $r^{2} \geq 0$ forces $cos 2 θ \geq 0$ , so the curve exists only for $- π /4 \leq θ \leq π /4$ and $3 π /4 \leq θ \leq 5 π /4$ ; there are no points at all in the directions between $4 5^{\circ}$ and $13 5^{\circ}$ . On the first interval $r = \pm 2 cos 2 θ$ runs from $0$ at $θ = - π /4$ up to $\pm 2$ at $θ = 0$ and back to $0$ : a loop through the pole with its far point at $(2, 0)$ , and the $\pm$ automatically supplies the mirror loop through $(- 2, 0)$ . The result is a figure eight lying along the $x$ -axis, crossing itself at the pole, where the two loops leave along the lines $θ = \pm π /4$ .

For the spiral $r = θ$ , the radius grows steadily with the angle. Starting at the pole, the curve crosses the positive $y$ -axis at distance $π /2$ , the negative $x$ -axis at $π$ , the negative $y$ -axis at $3 π /2$ , and returns to the positive $x$ -axis at distance $2 π$ ; each further turn crosses the positive $x$ -axis $2 π$ farther out. The gap between successive turns, measured along any ray, is always $2 π$ — the spiral of a coiled rope or a roll of paper.□

Pitfall. A curve that passes through negative $r$ may be traced more than once in $[0, 2 π]$ . The circle $r = 2 cos θ$ is traced completely on $[0, π]$ : for $π /2 < θ < π$ the negative radius plots the lower semicircle in the fourth quadrant, and $[π, 2 π]$ repeats the whole circle. Integrating an area or arc length for this curve over $[0, 2 π]$ counts everything twice.

Exercises for §8.7 →

8.8Tangents to polar curves

A polar curve $r = f (θ)$ is a parametric curve in disguise: with parameter $θ$ ,

x = r cos θ = f (θ) cos θ, y = r sin θ = f (θ) sin θ,

and every result about parametric curves applies. In particular the slope of the tangent is $d y / d x = (d y / d θ) / (d x / d θ)$ .

Theorem 8.42 (Slope of a polar curve). If $r = f (θ)$ has a continuous derivative then, wherever $d x / d θ \neq = 0$ ,

\frac{d y}{d x} = \frac{\frac{d r}{d θ} sin θ + r cos θ}{\frac{d r}{d θ} cos θ - r sin θ} .

Proof. By the product rule, $d x / d θ = f^{'} (θ) cos θ - f (θ) sin θ$ and $d y / d θ = f^{'} (θ) sin θ + f (θ) cos θ$ ; divide.∎

Each derivative has two terms with a clear meaning: the first, with $d r / d θ$ , is the contribution of moving outward along the ray, and the second, with $r$ , is the contribution of the ray swinging around the pole. Horizontal tangents occur where $d y / d θ = 0$ and $d x / d θ \neq = 0$ ; vertical tangents where $d x / d θ = 0$ and $d y / d θ \neq = 0$ . When both vanish, take the limit of $d y / d x$ , exactly as for parametric curves.

Pitfall. $d r / d θ$ is not a slope. It measures how fast the radius changes, and it is zero at the points farthest from and nearest to the pole, which are usually not the points with horizontal tangents. The cardioid $r = 1 + cos θ$ has $d r / d θ = 0$ at $θ = 0$ , where the tangent is vertical.

Theorem 8.43 (Tangents at the pole). If $f (α) = 0$ and $f^{'} (α) \neq = 0$ , then the line $θ = α$ is tangent to the curve $r = f (θ)$ at the pole.

Proof. At $θ = α$ the slope formula has $r = 0$ , so it collapses to

\frac{d y}{d x} = \frac{f ^{'} ( α ) sin α}{f ^{'} ( α ) cos α} = tan α,

the slope of the line $θ = α$ (with a vertical tangent when $cos α = 0$ ). Since the curve passes through the pole at $θ = α$ , this line is its tangent there.∎

This theorem is what makes rose curves and limaçons easy to draw near the origin: the curve enters and leaves the pole along the rays where $r = 0$ , and those rays are found by solving one trigonometric equation.

Intuition. Stand at the pole and watch a rose petal being drawn. The pen leaves your feet heading in the direction $θ = α$ , the angle at which $r$ became zero, exactly as a thrown ball leaves the hand in the direction it is moving at the instant of release. The petal curves away afterwards, but the initial direction is the tangent at the pole.

Example 8.44 (Tangents to a cardioid). For the cardioid $r = 1 + cos θ$ : find the slope of the tangent at $θ = π /6$ , and find every point where the tangent is horizontal or vertical.

Solution.

With $r = 1 + cos θ$ and $d r / d θ = - sin θ$ ,

\frac{d y}{d θ} = - sin θ sin θ + (1 + cos θ) cos θ = 2 cos^{2} θ + cos θ - 1 = (2 cos θ - 1) (cos θ + 1),

\frac{d x}{d θ} = - sin θ cos θ - (1 + cos θ) sin θ = - sin θ (1 + 2 cos θ) .

At $θ = π /6$ : $cos θ = 3 /2$ and $sin θ = 1/2$ , so $d y / d θ = (3 - 1) (3 /2 + 1) = \frac{1}{2} (3 - 1) (3 + 2) = \frac{1}{2} (1 + 3)$ and $d x / d θ = - \frac{1}{2} (1 + 3)$ . The slope is $- 1$ .
Horizontal tangents: $d y / d θ = 0$ when $cos θ = 1/2$ or $cos θ = - 1$ , that is, $θ = π /3, 5 π /3, π$ . At $π /3$ and $5 π /3$ , $d x / d θ \neq = 0$ ; the points are $(3/2, π /3)$ and $(3/2, 5 π /3)$ .
Vertical tangents: $d x / d θ = 0$ when $sin θ = 0$ or $cos θ = - 1/2$ , that is, $θ = 0, π, 2 π /3, 4 π /3$ . At $θ = 0$ , $2 π /3$ and $4 π /3$ the numerator is nonzero; the points are $(2, 0)$ , $(1/2, 2 π /3)$ and $(1/2, 4 π /3)$ .
At $θ = π$ both derivatives vanish, so examine the limit. Using $1 + cos θ = 2 cos^{2} (θ /2)$ and $sin θ = 2 sin (θ /2) cos (θ /2)$ ,

\frac{d y}{d x} = \frac{( 2 cos θ - 1 ) ( 1 + cos θ )}{- sin θ ( 1 + 2 cos θ )} = \frac{( 2 cos θ - 1 ) cos ( θ /2 )}{- sin ( θ /2 ) ( 1 + 2 cos θ )} ⟶ \frac{( - 3 ) \cdot 0}{- 1 \cdot ( - 1 )} = 0

as $θ \to π$ . The cardioid has a horizontal tangent at the pole: the cusp points along the $x$ -axis, as the sketch in the previous section showed.

Altogether: horizontal tangents at $(3/2, π /3)$ , $(3/2, 5 π /3)$ and the pole; vertical tangents at $(2, 0)$ , $(1/2, 2 π /3)$ and $(1/2, 4 π /3)$ . Sanity check: the vertical tangent at $(2, 0)$ is the rightmost point of the heart and the horizontal ones at $θ = \pm π /3$ are its highest and lowest points, at height $\pm \frac{3}{2} sin (π /3) = \pm 33 /4 \approx \pm 1.3$ .□

Example 8.45 (Tangent lines at the pole). Find the tangent lines at the pole to the rose $r = sin 3 θ$ and to the lemniscate $r^{2} = 4 cos 2 θ$ .

Solution. For the rose, $r = 0$ at $θ = 0, π /3, 2 π /3$ (and their repeats), and $d r / d θ = 3 cos 3 θ = \pm 3 \neq = 0$ at each. The tangent lines at the pole are $θ = 0$ , $θ = π /3$ and $θ = 2 π /3$ — the three lines that separate the petals, and along which each petal leaves and returns.

For the lemniscate, $r = 0$ where $cos 2 θ = 0$ , at $θ = \pm π /4$ in the right-hand range. The theorem does not apply as stated: writing the loop as $r = 2 cos 2 θ$ , the derivative $d r / d θ$ is unbounded as $θ \to π / 4^{-}$ rather than finite and nonzero. Differentiate $r^{2} = 4 cos 2 θ$ implicitly instead, $2 r r^{'} = - 8 sin 2 θ$ , which gives the ratio that matters:

\frac{r}{r ^{'}} = \frac{r ^{2}}{r r ^{'}} = \frac{4 cos 2 θ}{- 4 sin 2 θ} = - cot 2 θ ⟶ 0 as θ \to \frac{π}{4} .

Now divide the numerator and denominator of the slope formula by $r^{'}$ , which is nonzero for $0 < θ < π /4$ :

\frac{d y}{d x} = \frac{sin θ + ( r / r ^{'} ) cos θ}{cos θ - ( r / r ^{'} ) sin θ} ⟶ \frac{sin ( π /4 )}{cos ( π /4 )} = 1.

The same computation at $θ = - π /4$ gives $- 1$ . So the tangent lines at the pole are $θ = π /4$ and $θ = - π /4$ , and the figure eight crosses itself at right angles. The argument shows more than the theorem: the line $θ = α$ is tangent at the pole whenever $r \to 0$ and $r / r^{'} \to 0$ as $θ \to α$ , whether $r^{'}$ tends to a nonzero number or to infinity.□

Example 8.46 (Slope of a spiral). Find the slope of the Archimedean spiral $r = θ$ at $θ = π /2$ .

Solution. With $r = π /2$ and $d r / d θ = 1$ at $θ = π /2$ , where $sin θ = 1$ and $cos θ = 0$ ,

\frac{d y}{d x} = \frac{1 \cdot 1 + ( π /2 ) \cdot 0}{1 \cdot 0 - ( π /2 ) \cdot 1} = - \frac{2}{π} .

Sanity check: at that moment the point is at $(0, π /2)$ on the positive $y$ -axis and moving mainly leftward (the ray is swinging counterclockwise) while drifting slowly upward (the radius grows), so a small negative slope is right.□

Exercises for §8.8 →

8.9Area in polar coordinates

Theorem 8.47 (Area in polar coordinates). Let $f$ be continuous and non-negative on $[α, β]$ with $0 < β - α \leq 2 π$ . The area of the region bounded by the curve $r = f (θ)$ and the rays $θ = α$ and $θ = β$ is

A = \frac{1}{2} \int_{α}^{β} [f (θ)]^{2} d θ = \frac{1}{2} \int_{α}^{β} r^{2} d θ .

Proof. Partition $[α, β]$ into subintervals of width $Δ θ$ and pick a sample angle $θ_{i}^{*}$ in each. The piece of the region between the rays $θ_{i - 1}$ and $θ_{i}$ is approximately a circular sector of radius $f (θ_{i}^{*})$ and angle $Δ θ$ , whose area is $\frac{1}{2} f (θ_{i}^{*})^{2} Δ θ$ (a sector is the fraction $Δ θ /2 π$ of a disc of area $π r^{2}$ ). The sum $\sum_{i} \frac{1}{2} f (θ_{i}^{*})^{2} Δ θ$ is a Riemann sum for $\frac{1}{2} \int_{α}^{β} f (θ)^{2} d θ$ , and since $f$ is continuous the sums converge to the integral as $Δ θ \to 0$ .∎

The formula replaces the thin rectangles of Cartesian area with thin sectors; that is where the $\frac{1}{2}$ and the square come from, and both are essential. A common mental check: with $f (θ) = a$ the formula gives $\frac{1}{2} a^{2} \cdot 2 π = π a^{2}$ , the area of a disc.

Figure 8.48. The region measured by

\frac{1}{2} \int r^{2} d θ

is swept from the pole outward: here the cardioid

r = 1 + cos θ

between the rays

θ = 0

and

θ = π /3

Two things need care. The limits $α$ and $β$ must sweep the region exactly once — a curve traced twice on $[0, 2 π]$ must be integrated over half that range. And the formula does not mind negative $r$ , because $r$ is squared; but negative $r$ is often the sign that the curve is being retraced, so it should prompt a look at the sketch.

Pitfall. The integrand is $\frac{1}{2} r^{2}$ , not $r$ and not $\frac{1}{2} r$ . And the region is measured from the pole outward: the formula gives the area between the curve and the pole, not between the curve and the $x$ -axis.

Intuition. A lighthouse beam of length $r$ sweeps through a small angle $Δ θ$ and paints a thin sector of the sea. Doubling the beam length quadruples the painted area, because the sector is twice as long and twice as wide at its far end — that is the $r^{2}$ . Sweeping from $α$ to $β$ and adding up the sectors is the integral. If the beam length changes with direction, as it does for every curve but a circle, each sector uses its own $r$ .

Example 8.49 (Area of one petal of a rose). Find the area enclosed by one petal of $r = cos 2 θ$ .

Solution.

The petal along the positive $x$ -axis is traced for $- π /4 \leq θ \leq π /4$ , between two consecutive zeros of $r$ .
Then

A = \frac{1}{2} \int_{- π /4}^{π /4} cos^{2} 2 θ d θ = \frac{1}{2} \int_{- π /4}^{π /4} \frac{1 + cos 4 θ}{2} d θ = \frac{1}{4} [θ + \frac{sin 4 θ}{4}]_{- π /4}^{π /4} = \frac{1}{4} \cdot \frac{π}{2} = \frac{π}{8} .

The four petals together enclose $π /2$ , exactly half of the unit disc that contains them. Sanity check: a petal of length $1$ and maximum width well under $1$ should have area well under $1$ , and $π /8 \approx 0.39$ does.

□

Example 8.50 (Area of a cardioid). Find the area enclosed by $r = a (1 + cos θ)$ .

Solution.

The cardioid is traced once on $[0, 2 π]$ with $r \geq 0$ throughout.
Expand and use $cos^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ :

A = \frac{a ^{2}}{2} \int_{0}^{2 π} (1 + cos θ)^{2} d θ = \frac{a ^{2}}{2} \int_{0}^{2 π} (\frac{3}{2} + 2 cos θ + \frac{1}{2} cos 2 θ) d θ .

The cosine terms integrate to zero over a full period, so $A = \frac{a ^{2}}{2} \cdot \frac{3}{2} \cdot 2 π = \frac{3}{2} π a^{2}$ .

Sanity check: the cardioid spans $- a \leq y \leq a$ at its widest and $0 \leq x \leq 2 a$ , roughly a $2 a \times 2 a$ box of area $4 a^{2}$ , and $1.5 π a^{2} \approx 4.7 a^{2}$ is too big — until you recall that the widest vertical extent is not at $x = 0$ but at $θ = \pm π /3$ , where $y = \pm 33 a /4 \approx \pm 1.3 a$ . The bounding box is closer to $2 a \times 2.6 a = 5.2 a^{2}$ , and the cardioid fills most of it.□

Example 8.51 (The two loops of a limaçon). Find the area inside the outer loop but outside the inner loop of $r = 1 + 2 cos θ$ .

Solution.

From the sketch in the previous section, $r = 0$ at $θ = 2 π /3$ and $4 π /3$ . The outer loop is traced for $- 2 π /3 \leq θ \leq 2 π /3$ (where $r \geq 0$ ) and the inner loop for $2 π /3 \leq θ \leq 4 π /3$ (where $r \leq 0$ ).
The integrand is $(1 + 2 cos θ)^{2} = 1 + 4 cos θ + 4 cos^{2} θ = 3 + 4 cos θ + 2 cos 2 θ$ , with antiderivative $3 θ + 4 sin θ + sin 2 θ$ .
Outer loop:

A_{outer} = \frac{1}{2} [3 θ + 4 sin θ + sin 2 θ]_{- 2 π /3}^{2 π /3} = \frac{1}{2} (4 π + 8 \cdot \frac{3}{2} + 2 sin \frac{4 π}{3}) = \frac{1}{2} (4 π + 43 - 3) = 2 π + \frac{3 3}{2} .

Inner loop:

A_{inner} = \frac{1}{2} [3 θ + 4 sin θ + sin 2 θ]_{2 π /3}^{4 π /3} = \frac{1}{2} (2 π - 43 + 3) = π - \frac{3 3}{2} .

The required area is the difference, $A_{outer} - A_{inner} = π + 33 \approx 8.34$ .

Sanity check: the inner loop is small (about $0.54$ ), and the outer loop, which reaches $x = 3$ and $y = \pm 2$ or so, has area a little over $8$ ; the numbers fit. Note that integrating $\frac{1}{2} r^{2}$ over all of $[0, 2 π]$ would give $A_{outer} + A_{inner}$ , counting the inner loop as part of the region rather than a hole.□

Example 8.52 (A circle traced twice). Find the area inside the circle $r = 2 cos θ$ .

Solution. The curve is the circle of radius $1$ centered at $(1, 0)$ , so the answer must be $π$ . It is traced once for $- π /2 \leq θ \leq π /2$ :

A = \frac{1}{2} \int_{- π /2}^{π /2} 4 cos^{2} θ d θ = \int_{- π /2}^{π /2} (1 + cos 2 θ) d θ = π .

Integrating over $[0, 2 π]$ instead gives $2 π$ , twice the true area, because the circle is traced twice. The formula cannot know this; only the sketch can.□

Exercises for §8.9 →

8.10Area between polar curves

Proposition 8.53 (Area between polar curves). If $f (θ) \geq g (θ) \geq 0$ on $[α, β]$ , the area of the region between $r = g (θ)$ (inner) and $r = f (θ)$ (outer) for $α \leq θ \leq β$ is

A = \frac{1}{2} \int_{α}^{β} ([f (θ)]^{2} - [g (θ)]^{2}) d θ .

Proof. Subtract the area inside the inner curve from the area inside the outer curve over the same range of angles; both are given by the area theorem.∎

The formula is the easy part. The hard part is finding $α$ and $β$ , which means finding where the curves intersect, and here polar coordinates set a trap that Cartesian coordinates do not. Solving $f (θ) = g (θ)$ finds only those intersection points at which the two curves arrive with the same polar coordinates. A point can lie on both curves and still be missed, in two ways.

The pole: if $f (θ_{1}) = 0$ and $g (θ_{2}) = 0$ for different angles, both curves pass through the origin but $f (θ) = g (θ) = 0$ may have no solution.

Different representations: a point on the first curve as $(r, θ)$ may lie on the second as $(- r, θ + π)$ , so it is also necessary to solve $f (θ) = - g (θ + π)$ .

Method 8.54 (Finding all intersections of two polar curves).

Solve $f (θ) = g (θ)$ for $θ$ in $[0, 2 π)$ .
Solve $f (θ) = - g (θ + π)$ , which finds the points reached with opposite signs of $r$ .
Check separately whether each curve passes through the pole.
Sketch both curves and confirm that the count of intersection points matches the picture.

Intuition. Two cars drive on two roads that cross. Solving $f (θ) = g (θ)$ finds the points where both cars are at the same place at the same clock time $θ$ . But the roads can cross at a point the cars pass at different times — one at $θ = π /6$ and the other at $θ = 7 π /6$ with negative $r$ , say — and the crossroads is still a crossroads. Area problems care about the roads, not the timetable.

Example 8.55 (Inside one curve and outside another). Find the area of the region that lies inside the circle $r = 3 sin θ$ and outside the cardioid $r = 1 + sin θ$ .

Solution.

The curves meet where $3 sin θ = 1 + sin θ$ , so $sin θ = 1/2$ and $θ = π /6$ or $5 π /6$ . Both curves also pass through the pole (the circle at $θ = 0$ , the cardioid at $θ = 3 π /2$ ), but that point is not on the boundary of the required region, which lies between the two crossing points in the upper half-plane.
For $π /6 \leq θ \leq 5 π /6$ the circle is outside: at $θ = π /2$ it has $r = 3$ against the cardioid's $r = 2$ .
Set up and simplify using $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ :

A = \frac{1}{2} \int_{π /6}^{5 π /6} (9 sin^{2} θ - (1 + sin θ)^{2}) d θ = \frac{1}{2} \int_{π /6}^{5 π /6} (8 sin^{2} θ - 2 sin θ - 1) d θ = \frac{1}{2} \int_{π /6}^{5 π /6} (3 - 4 cos 2 θ - 2 sin θ) d θ .

Integrate:

A = \frac{1}{2} [3 θ - 2 sin 2 θ + 2 cos θ]_{π /6}^{5 π /6} = \frac{1}{2} [(\frac{5 π}{2} + 3 - 3) - (\frac{π}{2} - 3 + 3)] = π .

Sanity check: the circle has area $9 π /4 \approx 7.07$ and the cardioid $3 π /2 \approx 4.71$ ; the region is the part of the circle not covered by the cardioid, and $π \approx 3.14$ is a believable slice of it.□

Example 8.56 (Intersections that solving misses). Find all points of intersection of the rose $r = cos 2 θ$ and the circle $r = 1/2$ .

Solution.

Solving $cos 2 θ = 1/2$ gives $2 θ = π /3, 5 π /3, 7 π /3, 11 π /3$ , so $θ = π /6, 5 π /6, 7 π /6, 11 π /6$ : four points.
But the rose also has points with $r = - 1/2$ , wherever $cos 2 θ = - 1/2$ , at $θ = π /3, 2 π /3, 4 π /3, 5 π /3$ . The point $(- 1/2, π /3)$ is the same as $(1/2, 4 π /3)$ , which lies on the circle — but the circle reaches it at $θ = 4 π /3$ and the rose at $θ = π /3$ , so the equation $cos 2 θ = 1/2$ never sees it. Equivalently, these are the solutions of $f (θ) = - g (θ + π)$ , that is, $cos 2 θ = - 1/2$ .
The sketch confirms it: the circle of radius $1/2$ cuts each of the four petals twice, once on the way out and once on the way back, so there are $8$ intersection points, at $θ = π /6, π /3, 2 π /3, 5 π /6, 7 π /6, 4 π /3, 5 π /3, 11 π /6$ .

□

Example 8.57 (Area common to a rose and a circle). Find the area of the region inside both $r = cos 2 θ$ and $r = 1/2$ .

Solution.

By the symmetry of the rose, the region consists of four congruent pieces, one per petal. Work with the petal along the positive $x$ -axis, $- π /4 \leq θ \leq π /4$ , and multiply by $4$ . Within that petal the relevant intersections are $θ = \pm π /6$ (from the previous example); the negative- $r$ intersections belong to other petals.
For $∣ θ ∣ \leq π /6$ the rose is outside the circle ( $cos 2 θ \geq 1/2$ ), so the region is bounded by the circle; for $π /6 \leq ∣ θ ∣ \leq π /4$ the rose is inside, and bounds the region.
Using the symmetry in $θ \mapsto - θ$ ,

A_{petal} = \frac{1}{2} \int_{- π /6}^{π /6} \frac{1}{4} d θ + 2 \cdot \frac{1}{2} \int_{π /6}^{π /4} cos^{2} 2 θ d θ = \frac{π}{24} + [\frac{θ}{2} + \frac{sin 4 θ}{8}]_{π /6}^{π /4} = \frac{π}{24} + \frac{π}{8} - \frac{π}{12} - \frac{3}{16} = \frac{π}{12} - \frac{3}{16} .

Multiply by four: $A = \frac{π}{3} - \frac{3}{4} \approx 0.61$ .

Sanity check: the disc of radius $1/2$ has area $π /4 \approx 0.79$ , and the rose covers most of it but leaves four gaps near the lines $θ = \pm π /4, \pm 3 π /4$ ; a common area of $0.61$ is right.□

Example 8.58 (Two circles through the pole). Find the area of the region inside both $r = sin θ$ and $r = cos θ$ .

Solution.

Solving $sin θ = cos θ$ gives only $θ = π /4$ , the point $(2 /2, π /4)$ . But both circles pass through the pole ( $r = sin θ$ at $θ = 0$ , $r = cos θ$ at $θ = π /2$ ), so the pole is a second intersection point, and the region is the lens between them.
Sweeping from $θ = 0$ to $π /2$ , the boundary of the lens is the circle nearer the pole: $r = sin θ$ for $0 \leq θ \leq π /4$ and $r = cos θ$ for $π /4 \leq θ \leq π /2$ . The two pieces are mirror images in the line $θ = π /4$ , so

A = 2 \cdot \frac{1}{2} \int_{0}^{π /4} sin^{2} θ d θ = \int_{0}^{π /4} \frac{1 - cos 2 θ}{2} d θ = [\frac{θ}{2} - \frac{sin 2 θ}{4}]_{0}^{π /4} = \frac{π}{8} - \frac{1}{4} .

Sanity check: the circles have radius $1/2$ and area $π /4 \approx 0.79$ each, and their centers $(1/2, 0)$ and $(0, 1/2)$ are $2 /2 \approx 0.71$ apart, so they overlap in a modest lens; $π /8 - 1/4 \approx 0.14$ is plausible. Had we integrated with $r = sin θ$ as the boundary all the way from $0$ to $π /2$ we would have found $π /8$ , the area of half a circle, which is too much.□

Pitfall. Never set up an area-between integral without a sketch. Which curve is outer can change from one interval to the next (the rose-and-circle example), the pole can be an intersection that no equation reports (the two-circles example), and a curve with negative $r$ can be inside the other on the sketch while its equation says otherwise. The integral is the last step, not the first.

Exercises for §8.10 →

8.11Arc length and surface area in polar coordinates

Theorem 8.59 (Arc length in polar coordinates). If $r = f (θ)$ has a continuous derivative on $[α, β]$ and the curve is traced exactly once as $θ$ runs from $α$ to $β$ , its length is

L = \int_{α}^{β} r^{2} + (\frac{d r}{d θ})^{2} d θ .

Proof. Treat the curve as parametric with parameter $θ$ : $x = f (θ) cos θ$ , $y = f (θ) sin θ$ . Then

(\frac{d x}{d θ})^{2} + (\frac{d y}{d θ})^{2} = (f^{'} cos θ - f sin θ)^{2} + (f^{'} sin θ + f cos θ)^{2} = (f^{'})^{2} + f^{2},

because the cross terms $\mp 2 f f^{'} sin θ cos θ$ cancel and $cos^{2} θ + sin^{2} θ = 1$ . Substitute into the parametric arc length formula.∎

The integrand has a geometric reading. In a small change $d θ$ the point moves outward by $d r$ and sideways by $r d θ$ ; these two displacements are perpendicular, and the integrand is the hypotenuse $(d r)^{2} + (r d θ)^{2}$ divided by $d θ$ . As with parametric curves, the range of $θ$ must trace the curve once; a circle $r = 2 a cos θ$ is integrated over $[0, π]$ , not $[0, 2 π]$ .

Example 8.60 (Circumference check). Find the length of $r = 2 cos θ$ , $0 \leq θ \leq π$ .

Solution. $r^{2} + (d r / d θ)^{2} = 4 cos^{2} θ + 4 sin^{2} θ = 4$ , so $L = \int_{0}^{π} 2 d θ = 2 π$ , the circumference of a circle of radius $1$ . Over $[0, 2 π]$ the integral would report $4 π$ : the circle traced twice.□

Example 8.61 (Length of a cardioid). Find the length of $r = 1 + cos θ$ .

Solution.

With $d r / d θ = - sin θ$ ,

r^{2} + (\frac{d r}{d θ})^{2} = 1 + 2 cos θ + cos^{2} θ + sin^{2} θ = 2 + 2 cos θ = 4 cos^{2} \frac{θ}{2},

by the half-angle identity $1 + cos θ = 2 cos^{2} (θ /2)$ .

So the integrand is $2∣ cos (θ /2) ∣$ . The absolute value matters: $cos (θ /2)$ is negative for $π < θ < 2 π$ . Use the symmetry of the cardioid about the $x$ -axis to integrate over $[0, π]$ , where the cosine is positive, and double:

L = 2 \int_{0}^{π} 2 cos \frac{θ}{2} d θ = 4 [2 sin \frac{θ}{2}]_{0}^{π} = 8.

Sanity check: the cardioid is contained in a box roughly $2 \times 2.6$ whose perimeter is about $9$ , and hugs most of it; $8$ is right. Ignoring the absolute value and integrating $2 cos (θ /2)$ straight over $[0, 2 π]$ gives $0$ , an impossible length.□

Example 8.62 (Length of a logarithmic spiral). Find the length of $r = e^{θ}$ for $0 \leq θ \leq 2 π$ .

Solution. $d r / d θ = e^{θ}$ , so $r^{2} + (d r / d θ)^{2} = 2 e^{2 θ}$ and

L = \int_{0}^{2 π} 2 e^{θ} d θ = 2 (e^{2 π} - 1) \approx 756.

The spiral ends at distance $e^{2 π} \approx 535$ from the pole after one turn, so a length of about $756$ — a bit more than $2$ times the final radius — is consistent with the logarithmic spiral's constant $4 5^{\circ}$ angle to every ray.□

Example 8.63 (Length of an Archimedean spiral). Find the length of $r = θ$ for $0 \leq θ \leq 2 π$ .

Solution. Here the integrand is $θ^{2} + 1$ , whose antiderivative (by the substitution $θ = tan u$ , or from a table) is $\frac{1}{2} [θ θ^{2} + 1 + ln (θ + θ^{2} + 1)]$ . So

L = \frac{1}{2} [2 π 4 π^{2} + 1 + ln (2 π + 4 π^{2} + 1)] = π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1) \approx 21.26.

Sanity check: the first turn of the spiral is roughly a circle of average radius $π$ , with circumference $2 π^{2} \approx 19.7$ ; the spiral is a little longer because it also moves outward.□

Proposition 8.64 (Surface area of revolution for a polar curve). Under the hypotheses of the polar arc length theorem, and with $r sin θ \geq 0$ , the area of the surface obtained by rotating $r = f (θ)$ , $α \leq θ \leq β$ , about the polar axis is

S = \int_{α}^{β} 2 π r sin θ r^{2} + (\frac{d r}{d θ})^{2} d θ .

Rotating about the line $θ = π /2$ (the $y$ -axis) replaces $r sin θ$ by $r cos θ$ .

Proof. This is $S = \int 2 π y d s$ with $y = r sin θ$ and $d s = r^{2} + (d r / d θ)^{2} d θ$ .∎

Example 8.65 (Surface generated by a cardioid). Find the area of the surface obtained by rotating the cardioid $r = 1 + cos θ$ about the polar axis.

Solution.

The surface is generated by the upper half of the cardioid, $0 \leq θ \leq π$ , where $y \geq 0$ . From the cardioid length example, $d s = 2 cos (θ /2) d θ$ on this range.
Write everything in the half angle: $r = 2 cos^{2} (θ /2)$ and $sin θ = 2 sin (θ /2) cos (θ /2)$ , so $y = r sin θ = 4 cos^{3} (θ /2) sin (θ /2)$ .
Then

S = \int_{0}^{π} 2 π \cdot 4 cos^{3} \frac{θ}{2} sin \frac{θ}{2} \cdot 2 cos \frac{θ}{2} d θ = 16 π \int_{0}^{π} cos^{4} \frac{θ}{2} sin \frac{θ}{2} d θ .

Substitute $u = cos (θ /2)$ , $d u = - \frac{1}{2} sin (θ /2) d θ$ ; as $θ$ goes from $0$ to $π$ , $u$ goes from $1$ to $0$ :

S = 16 π \int_{1}^{0} u^{4} (- 2 d u) = 32 π \int_{0}^{1} u^{4} d u = \frac{32 π}{5} .

Sanity check: $32 π /5 \approx 20.1$ . The cardioid rotated about its axis gives a shape like an apple, with maximum radius about $1.3$ and length $2$ ; a sphere of radius $1.2$ would have area about $18$ , so $20$ is reasonable.□

Exercises for §8.11 →

8.12Conic sections in Cartesian form

The conic sections — parabolas, ellipses and hyperbolas — are the curves obtained by cutting a double cone with a plane, and the Greeks studied them that way. The definitions below, in terms of distances to fixed points and lines, are equivalent and are the ones calculus uses, because each leads in a few lines to a standard equation.

Definition 8.66 (Parabola). A parabola is the set of points in the plane equidistant from a fixed point $F$ (the focus) and a fixed line $l$ (the directrix) not through $F$ . The line through $F$ perpendicular to $l$ is the axis of the parabola, and the point where the parabola crosses its axis, midway between focus and directrix, is the vertex.

Theorem 8.67 (Standard equation of a parabola). The parabola with focus $(0, p)$ and directrix $y = - p$ has equation

x^{2} = 4 p y .

It opens upward if $p > 0$ and downward if $p < 0$ . Interchanging the roles of $x$ and $y$ , the parabola with focus $(p, 0)$ and directrix $x = - p$ is $y^{2} = 4 p x$ , opening to the right if $p > 0$ and to the left if $p < 0$ .

Proof. A point $(x, y)$ is at distance $x^{2} + (y - p)^{2}$ from the focus and $∣ y + p ∣$ from the directrix. Setting them equal and squaring,

x^{2} + y^{2} - 2 p y + p^{2} = y^{2} + 2 p y + p^{2},

which simplifies to $x^{2} = 4 p y$ . Squaring introduces no spurious solutions, because both distances are non-negative.∎

Example 8.68 (Focus and directrix of a parabola). Find the focus and directrix of $y^{2} = 10 x$ .

Solution. Match with $y^{2} = 4 p x$ : $4 p = 10$ , so $p = 5/2$ . The focus is $(5/2, 0)$ and the directrix is the line $x = - 5/2$ . The parabola opens to the right, and its vertex is the origin, midway between them.□

Definition 8.69 (Ellipse). An ellipse is the set of points whose distances from two fixed points $F_{1}$ and $F_{2}$ (the foci) have a constant sum. The midpoint of $F_{1} F_{2}$ is the center; the points of the ellipse on the line through the foci are the vertices, and the segment joining them is the major axis. The segment through the center perpendicular to the major axis, with endpoints on the ellipse, is the minor axis.

Theorem 8.70 (Standard equation of an ellipse). The ellipse with foci $(\pm c, 0)$ and constant sum of distances $2 a$ , where $a > c \geq 0$ , has equation

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, b^{2} = a^{2} - c^{2} .

Its vertices are $(\pm a, 0)$ and the minor axis has endpoints $(0, \pm b)$ . If the foci are $(0, \pm c)$ instead, the equation is $x^{2} / b^{2} + y^{2} / a^{2} = 1$ with the same relation, and the vertices are $(0, \pm a)$ .

Proof. The condition is $(x + c)^{2} + y^{2} + (x - c)^{2} + y^{2} = 2 a$ . Isolate the second root, square, and cancel the common terms $x^{2} + c^{2} + y^{2}$ :

(x + c)^{2} + y^{2} = 4 a^{2} - 4 a (x - c)^{2} + y^{2} + (x - c)^{2} + y^{2} ⟹ a (x - c)^{2} + y^{2} = a^{2} - c x .

Square again and collect:

a^{2} (x^{2} - 2 c x + c^{2} + y^{2}) = a^{4} - 2 a^{2} c x + c^{2} x^{2} ⟹ (a^{2} - c^{2}) x^{2} + a^{2} y^{2} = a^{2} (a^{2} - c^{2}) .

The triangle inequality forces $2 a > 2 c$ , so $a^{2} - c^{2} > 0$ ; call it $b^{2}$ and divide by $a^{2} b^{2}$ . (The two squarings can be checked not to add solutions, since $a^{2} - c x > 0$ throughout the ellipse.) The vertices and minor-axis endpoints are read off by setting $y = 0$ and $x = 0$ .∎

The number $b^{2} = a^{2} - c^{2}$ is best remembered as a picture: the endpoint $(0, b)$ of the minor axis is at distance $a$ from each focus (its two distances are equal and add to $2 a$ ), so $a$ , $b$ and $c$ form a right triangle with hypotenuse $a$ . The eccentricity $e = c / a$ lies in $[0, 1)$ and measures how far the ellipse is from a circle: $e = 0$ is a circle, and as $e \to 1$ the ellipse flattens toward its major axis.

Example 8.71 (Foci and vertices of an ellipse). Find the foci and vertices of $9 x^{2} + 16 y^{2} = 144$ , and then find the equation of the ellipse with foci $(0, \pm 2)$ and vertices $(0, \pm 3)$ .

Solution.

Divide by $144$ : $x^{2} /16 + y^{2} /9 = 1$ . The larger denominator is under $x^{2}$ , so the major axis is horizontal with $a = 4$ , $b = 3$ , and $c^{2} = 16 - 9 = 7$ . The vertices are $(\pm 4, 0)$ and the foci $(\pm 7, 0)$ ; the eccentricity is $7 /4 \approx 0.66$ .
For the second ellipse the foci lie on the $y$ -axis, so it is of the form $x^{2} / b^{2} + y^{2} / a^{2} = 1$ with $a = 3$ and $c = 2$ , hence $b^{2} = 9 - 4 = 5$ :

\frac{x ^{2}}{5} + \frac{y ^{2}}{9} = 1.

Check: the point $(0, 3)$ is at distances $1$ and $5$ from the foci, summing to $6 = 2 a$ .□

Definition 8.72 (Hyperbola). A hyperbola is the set of points whose distances from two fixed points $F_{1}$ and $F_{2}$ (the foci) have a constant difference in absolute value. The center, vertices and axis are defined as for the ellipse.

Theorem 8.73 (Standard equation of a hyperbola). The hyperbola with foci $(\pm c, 0)$ and constant difference of distances $2 a$ , where $0 < a < c$ , has equation

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, b^{2} = c^{2} - a^{2} .

Its vertices are $(\pm a, 0)$ and its asymptotes are the lines $y = \pm \frac{b}{a} x$ . If the foci are $(0, \pm c)$ the equation is $y^{2} / a^{2} - x^{2} / b^{2} = 1$ , with vertices $(0, \pm a)$ and asymptotes $y = \pm \frac{a}{b} x$ .

Proof. The derivation of the equation is the ellipse computation with the sum replaced by a difference: the same two squarings lead to $(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2})$ , and now the triangle inequality gives $c > a$ , so $b^{2} = c^{2} - a^{2} > 0$ .

For the asymptotes, solve for $y$ on the right-hand branch: $y = \pm \frac{b}{a} x^{2} - a^{2}$ for $x \geq a$ . The vertical gap between the upper half and the line $y = (b / a) x$ is

\frac{b}{a} (x - x^{2} - a^{2}) = \frac{b}{a} \cdot \frac{a ^{2}}{x + x ^{2} - a ^{2}} ⟶ 0 as x \to \infty,

after multiplying by the conjugate. The other three half-branches are handled by symmetry. The equation also shows there are no points with $∣ x ∣ < a$ , which is why the curve has two separate branches.∎

The asymptotes are the diagonals of the rectangle with corners $(\pm a, \pm b)$ , and drawing that rectangle first is the quickest way to sketch a hyperbola accurately. Note the difference from the ellipse: here $c$ is the largest of the three numbers, $c^{2} = a^{2} + b^{2}$ , and the eccentricity $e = c / a$ exceeds $1$ .

Example 8.74 (Foci and asymptotes of a hyperbola). Find the foci, vertices and asymptotes of $9 x^{2} - 16 y^{2} = 144$ , and the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $(0, \pm 1)$ .

Solution.

Divide by $144$ : $x^{2} /16 - y^{2} /9 = 1$ , so $a = 4$ , $b = 3$ and $c^{2} = 16 + 9 = 25$ . The vertices are $(\pm 4, 0)$ , the foci $(\pm 5, 0)$ and the asymptotes $y = \pm \frac{3}{4} x$ .
For the second hyperbola the foci are on the $y$ -axis: $a = 1$ , $c = 3$ , $b^{2} = 9 - 1 = 8$ , so

y^{2} - \frac{x ^{2}}{8} = 1,

with asymptotes $y = \pm \frac{1}{2 2} x = \pm \frac{2}{4} x$ .□

Intuition. A satellite dish is a parabola rotated about its axis, with the receiver at the focus: every incoming ray parallel to the axis reflects to the focus, which is a consequence of the equidistance definition. A dish described by $x^{2} = 4 p y$ with $p = 0.3$ m needs its receiver $30$ cm above the vertex. The same geometry, run backwards, is a car headlight; and the whispering gallery of an elliptical room sends a whisper at one focus to a listener at the other.

A conic need not be centered at the origin. Replacing $x$ by $x - h$ and $y$ by $y - k$ shifts every point by $(h, k)$ , so

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1, \frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1, (x - h)^{2} = 4 p (y - k)

are an ellipse, a hyperbola and a parabola with center (or vertex) $(h, k)$ ; foci, vertices and asymptotes shift by the same $(h, k)$ . Any equation of the form $A x^{2} + C y^{2} + D x + E y + F = 0$ can be brought to one of these shapes, or shown to be degenerate, by completing the square.

Method 8.75 (Identifying a conic by completing the square). Given $A x^{2} + C y^{2} + D x + E y + F = 0$ with no $x y$ term:

If exactly one of $A$ , $C$ is zero, the curve is a parabola: complete the square in the squared variable and solve for the other.
If $A$ and $C$ have the same sign, the curve is an ellipse (a circle if $A = C$ ): complete the square in both variables and divide by the constant. A zero or negative constant means a single point or no points.
If $A$ and $C$ have opposite signs, the curve is a hyperbola: complete the square in both and divide. A zero constant means a pair of intersecting lines.

Example 8.76 (A shifted hyperbola). Sketch the conic $9 x^{2} - 4 y^{2} - 72 x + 8 y + 176 = 0$ and find its foci and asymptotes.

Solution.

The coefficients $9$ and $- 4$ have opposite signs: a hyperbola. Group and complete the square:

9 (x^{2} - 8 x) - 4 (y^{2} - 2 y) = - 176 ⟹ 9 (x - 4)^{2} - 144 - 4 (y - 1)^{2} + 4 = - 176,

so $9 (x - 4)^{2} - 4 (y - 1)^{2} = - 36$ , and dividing by $- 36$ ,

\frac{( y - 1 ) ^{2}}{9} - \frac{( x - 4 ) ^{2}}{4} = 1.

The positive term is in $y$ , so the branches open up and down from the center $(4, 1)$ , with $a = 3$ , $b = 2$ and $c = 13$ . The vertices are $(4, 4)$ and $(4, - 2)$ , the foci $(4, 1 \pm 13)$ , and the asymptotes $y - 1 = \pm \frac{3}{2} (x - 4)$ .

□

Example 8.77 (A shifted ellipse). Identify $x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0$ and find its foci.

Solution. Same signs, unequal coefficients: an ellipse, if anything. Completing the square,

(x - 3)^{2} - 9 + 4 (y + 1)^{2} - 4 + 9 = 0 ⟹ (x - 3)^{2} + 4 (y + 1)^{2} = 4 ⟹ \frac{( x - 3 ) ^{2}}{4} + (y + 1)^{2} = 1.

The center is $(3, - 1)$ , the major axis is horizontal with $a = 2$ and $b = 1$ , so $c = 3$ and the foci are $(3 \pm 3, - 1)$ . Sanity check: the point $(5, - 1)$ , a vertex, is at distances $2 - 3$ and $2 + 3$ from the foci, which sum to $4 = 2 a$ .□

Example 8.78 (A shifted parabola). Find the vertex, focus and directrix of $y^{2} - 6 y - 8 x + 1 = 0$ .

Solution. Only $y$ is squared, so this is a parabola with a horizontal axis. Complete the square in $y$ :

(y - 3)^{2} - 9 - 8 x + 1 = 0 ⟹ (y - 3)^{2} = 8 (x + 1) .

Compare with $(y - k)^{2} = 4 p (x - h)$ : the vertex is $(- 1, 3)$ and $4 p = 8$ , so $p = 2$ . The parabola opens to the right; the focus is $2$ units right of the vertex at $(1, 3)$ , and the directrix is $2$ units left, the line $x = - 3$ .□

Exercises for §8.12 →

8.13Conic sections in polar coordinates

The three conics were defined by three different rules. There is a single rule that produces all of them, with one number deciding which, and in polar coordinates with the focus at the pole it gives a single equation for all three. That is the form in which orbits are written.

Definition 8.79 (Conic by focus, directrix and eccentricity). Let $F$ be a point, $l$ a line not through $F$ , and $e > 0$ . The set of points $P$ with

\frac{∣ PF ∣}{∣ Pl ∣} = e,

where $∣ Pl ∣$ is the distance from $P$ to $l$ , is a conic section with focus $F$ , directrix $l$ and eccentricity $e$ .

Theorem 8.80 (Polar equation of a conic). With the focus at the pole and the directrix the vertical line $x = d$ (where $d > 0$ ), the conic with eccentricity $e$ has polar equation

r = \frac{e d}{1 + e cos θ} .

It is an ellipse if $e < 1$ , a parabola if $e = 1$ and a hyperbola if $e > 1$ ; in every case the pole is a focus, and for the ellipse and hyperbola the eccentricity $e$ agrees with the Cartesian eccentricity $c / a$ .

Proof. Let $P = (r, θ)$ with $r > 0$ . Then $∣ PF ∣ = r$ and, since $P$ has $x$ -coordinate $r cos θ$ , its distance to the line $x = d$ is $∣ Pl ∣ = d - r cos θ$ for points on the focus side of the directrix. The condition $∣ PF ∣ = e ∣ Pl ∣$ becomes

r = e (d - r cos θ) ⟹ r (1 + e cos θ) = e d,

which is the equation claimed. (Points on the far side of the directrix, which arise only for the hyperbola, are covered by the negative- $r$ convention.)

To identify the curve, convert to Cartesian coordinates. From $r = e (d - x)$ , squaring gives $x^{2} + y^{2} = e^{2} (d - x)^{2}$ , that is,

(1 - e^{2}) x^{2} + 2 d e^{2} x + y^{2} = e^{2} d^{2} .

If $e = 1$ the $x^{2}$ term drops out and $y^{2} = d^{2} - 2 d x = - 2 d (x - \frac{d}{2})$ : a parabola with vertex $(d /2, 0)$ opening to the left, with $4 p = - 2 d$ , so its focus is $p = - d /2$ to the left of the vertex — at the origin, as it should be.

If $e < 1$ , divide by $1 - e^{2}$ and complete the square in $x$ :

(x + \frac{e ^{2} d}{1 - e ^{2}})^{2} + \frac{y ^{2}}{1 - e ^{2}} = \frac{e ^{2} d ^{2}}{( 1 - e ^{2} ) ^{2}} .

This is an ellipse with center $(h, 0)$ , $h = - e^{2} d / (1 - e^{2})$ , and

a^{2} = \frac{e ^{2} d ^{2}}{( 1 - e ^{2} ) ^{2}}, b^{2} = \frac{e ^{2} d ^{2}}{1 - e ^{2}}, c^{2} = a^{2} - b^{2} = \frac{e ^{2} d ^{2}}{( 1 - e ^{2} ) ^{2}} (1 - (1 - e^{2})) = \frac{e ^{4} d ^{2}}{( 1 - e ^{2} ) ^{2}} .

So $c = e^{2} d / (1 - e^{2}) = ∣ h ∣$ : one focus sits at $h + c = 0$ , the pole. And $c / a = e$ , so the two notions of eccentricity agree. If $e > 1$ the same algebra, with $e^{2} - 1$ in place of $1 - e^{2}$ and a minus sign before the $y^{2}$ term, gives a hyperbola with $c = e^{2} d / (e^{2} - 1)$ , again with a focus at the pole and $c / a = e$ .∎

Turning the directrix produces three more forms, and the pattern is easy to read: a cosine means a vertical directrix, a sine a horizontal one; a plus sign puts the directrix on the positive side of the focus, a minus sign on the negative side.

Summary. With the focus at the pole and eccentricity $e$ :

r = \frac{e d}{1 + e cos θ} (directrix x = d), r = \frac{e d}{1 - e cos θ} (x = - d),

r = \frac{e d}{1 + e sin θ} (y = d), r = \frac{e d}{1 - e sin θ} (y = - d) .

To identify a given equation, divide numerator and denominator by the constant term of the denominator so that the denominator begins with $1$ ; the coefficient of the trigonometric function is then $e$ , and the numerator is $e d$ . The vertices lie on the axis through the focus perpendicular to the directrix, at $θ = 0$ and $θ = π$ for the cosine forms, at $θ = π /2$ and $3 π /2$ for the sine forms.

For an ellipse the polar form is often written without $d$ . Since $a = e d / (1 - e^{2})$ from the proof, $e d = a (1 - e^{2})$ and

r = \frac{a ( 1 - e ^{2} )}{1 + e cos θ} .

The nearest and farthest points from the focus are then $r = a (1 - e)$ at $θ = 0$ and $r = a (1 + e)$ at $θ = π$ .

Intuition. Think of $e$ as a dial. At $e = 0$ the curve is a circle around the focus. Turn the dial up and the circle stretches into a longer and longer ellipse, the focus staying near one end. At exactly $e = 1$ the far end of the ellipse reaches infinity and the curve opens into a parabola. Turn further and the curve opens wider still, into one branch of a hyperbola, the far end now pointing along the asymptotes. A comet on a hyperbolic orbit ( $e > 1$ ) visits the sun once and never returns; on an elliptical one ( $e < 1$ ) it comes back.

Example 8.81 (Writing a polar equation). Find a polar equation for the parabola with focus at the origin and directrix $y = - 6$ .

Solution. A parabola has $e = 1$ , and a horizontal directrix below the focus at distance $d = 6$ calls for the form with $- sin θ$ :

r = \frac{6}{1 - sin θ} .

Check: at $θ = 3 π /2$ (straight down) $r = 3$ , the vertex, midway between focus and directrix; as $θ \to π /2$ the denominator tends to $0$ and $r \to \infty$ , the parabola opening upward.□

Example 8.82 (Identifying an ellipse). Identify the conic $r = \frac{10}{3 - 2 cos θ}$ and find its eccentricity, directrix, vertices, center and semi-axes.

Solution.

Divide numerator and denominator by $3$ :

r = \frac{10/3}{1 - \frac{2}{3} cos θ} .

So $e = 2/3 < 1$ : an ellipse. From $e d = 10/3$ , $d = 5$ , and the minus-cosine form puts the directrix at $x = - 5$ . 2. The vertices are at $θ = 0$ and $θ = π$ : $r = 10/ (3 - 2) = 10$ and $r = 10/ (3 + 2) = 2$ , that is, the Cartesian points $(10, 0)$ and $(- 2, 0)$ . 3. The major axis has length $12$ , so $a = 6$ and the center is the midpoint $(4, 0)$ . The focus at the origin is then $c = 4$ from the center, and $c / a = 2/3 = e$ , confirming the eccentricity. Finally $b^{2} = a^{2} - c^{2} = 36 - 16 = 20$ , so $b = 25$ .□

Example 8.83 (Identifying a hyperbola). Identify the conic $r = \frac{12}{2 + 4 sin θ}$ and find its vertices.

Solution.

Divide by $2$ : $r = 6/ (1 + 2 sin θ)$ , so $e = 2 > 1$ : a hyperbola, with $e d = 6$ , $d = 3$ , and directrix $y = 3$ (plus-sine form).
The vertices are at $θ = π /2$ and $3 π /2$ . At $θ = π /2$ , $r = 6/3 = 2$ : the point $(0, 2)$ . At $θ = 3 π /2$ , $r = 6/ (1 - 2) = - 6$ : negative, so the point is $6$ units in the opposite direction, at $(0, 6)$ .
The two vertices $(0, 2)$ and $(0, 6)$ are $4$ apart, so $a = 2$ and the center is $(0, 4)$ ; the focus at the origin is $c = 4$ from the center, and $c / a = 2 = e$ . The branch through $(0, 2)$ wraps around the focus at the origin; the other branch, through $(0, 6)$ , is traced by the angles where $1 + 2 sin θ < 0$ , that is, where $r$ is negative.

□

Remark (Kepler's laws). Newton showed that a body moving under an inverse-square force from a fixed center travels on a conic with the center at one focus, and this is why the polar form matters. Kepler's three laws, found empirically from Tycho Brahe's observations sixty years earlier, read directly off it. First law: a planet's orbit is an ellipse with the sun at one focus, so $r = a (1 - e^{2}) / (1 + e cos θ)$ with $θ$ measured from perihelion. Second law: the line from the sun to the planet sweeps equal areas in equal times, which in the language of this chapter says that $\frac{1}{2} r^{2} d θ / d t$ is constant — the polar area formula, differentiated. Third law: the square of the period is proportional to $a^{3}$ .

Example 8.84 (Perihelion and aphelion). Halley's comet has an orbit of eccentricity $e \approx 0.967$ with semi-major axis $a \approx 17.8$ astronomical units. Find its closest and farthest distances from the sun.

Solution. Perihelion is $a (1 - e) = 17.8 \times 0.033 \approx 0.59$ AU, inside the orbit of Venus; aphelion is $a (1 + e) = 17.8 \times 1.967 \approx 35.0$ AU, beyond Neptune. The ratio of the two, $(1 + e) / (1 - e) \approx 60$ , is what an eccentricity close to $1$ looks like: a long thin ellipse with the sun close to one end.□

Exercises for §8.13 →

Common Mistakes

Treating the parameter interval as the curve. $x = sin 2 t$ , $y = cos 2 t$ on $[0, 2 π]$ traces the circle twice; arc length and enclosed area computed over that interval are doubled. Find the interval that traces the curve once before integrating.
Computing the second derivative as $y^{''} (t) / x^{''} (t)$ . It is $\frac{d}{d t} (\frac{d y}{d x})$ divided by $\frac{d x}{d t}$ , and the wrong version gets the concavity backwards.
Declaring "no tangent" when $d x / d t = d y / d t = 0$ . Take the limit of $d y / d x$ ; the cycloid has vertical tangents at its cusps and $x = t^{2}$ , $y = t^{3}$ has a horizontal one.
Using $θ = tan^{- 1} (y / x)$ in the second or third quadrant without adding $π$ .
Forgetting the negative- $r$ convention when plotting. The point $(- 1, π)$ of $r = 1 + 2 cos θ$ is on the positive $x$ -axis, and the inner loop of the limaçon lives there, not on the left.
Miscounting rose petals. $r = cos n θ$ has $n$ petals for odd $n$ and $2 n$ for even $n$ , because for odd $n$ the negative- $r$ petals retrace the positive ones.
Writing the polar area integrand as $r$ or as $\frac{1}{2} r$ . It is $\frac{1}{2} r^{2}$ , and it measures area from the pole, not from the $x$ -axis.
Finding intersections of polar curves by solving $f (θ) = g (θ)$ alone. Check the pole separately and solve $f (θ) = - g (θ + π)$ too; the rose $r = cos 2 θ$ meets the circle $r = 1/2$ at eight points, not four.
Choosing the outer curve by looking at the equations. Test an angle, and remember that the roles can swap between intersection points.
Dropping the absolute value in $4 cos^{2} (θ /2) = 2∣ cos (θ /2) ∣$ . Integrating $2 cos (θ /2)$ over $[0, 2 π]$ gives a cardioid of length zero.
Mixing up $b^{2} = a^{2} - c^{2}$ (ellipse) with $b^{2} = c^{2} - a^{2}$ (hyperbola). For the ellipse $a$ is the largest number; for the hyperbola $c$ is.
Reading the eccentricity off $r = 10/ (3 - 2 cos θ)$ as $2$ . Divide by the constant in the denominator first: $e = 2/3$ .
Assuming a shifted conic's foci are on the coordinate axes. Complete the square, find the center, and shift the foci by the same amount.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed. This chapter also has a printable problem set: proofs, sketches and find-the-error work to do on paper, with full solutions on a separate page.

Practice this chapter Printable problem set