Print this page and work through it on paper. Revealed answers are never printed, so you can check one question and still print a clean sheet.

1. 1.Computation · 6 min

   For each curve, eliminate the parameter, name the Cartesian curve, and say exactly which part of it is traced and in which direction. The parameter interval is part of the answer.

   1. (a)$x=\sqrt{t},y=1-t,t\ge 0$
   2. (b)$x=3+2\cos t,y=1-2\sin t,0\le t\le 2\pi$
   3. (c)Write a parametrization of the line segment from $(2,-1)$ to $(-1,5)$, and a second parametrization of the same segment traversed in the opposite direction.
   Show the solution to question 1

   (a) Solve for $t$ and substitute. From $x=\sqrt{t}$ we get $t=x^2$, and $x\ge 0$ because a square root is never negative. Then $y=1-t=1-x^2$.

   The Cartesian curve is the parabola $y=1-x^2$, but only its right half, $x\ge 0$, is traced. At $t=0$ the point is $(0,1)$, the vertex, and as $t$ increases $x=\sqrt{t}$ increases, so the curve is traced rightward and downward, away from the vertex, forever. The left half of the parabola is never reached — eliminating the parameter enlarged the curve, and the restriction $x\ge 0$ is what repairs it.

   (b) Use ${\cos }^{2}t+{\sin }^{2}t=1$. Solve for the trigonometric functions: $\cos t=\frac{x-3}{2}$ and $\sin t=\frac{1-y}{2}$, so

   $$\frac{(x-3{)}^2}{4}+\frac{(y-1{)}^2}{4}=1,\text{that is}(x-3{)}^2+(y-1{)}^2=4.$$

   This is the circle of radius $2$ centered at $(3,1)$. At $t=0$ the point is $(5,1)$, the rightmost point; at $t=\pi /2$ it is $(3,-1)$, the lowest point. Rightmost then lowest means the circle is traced clockwise — the minus sign in $y=1-2\sin t$ reverses the standard orientation. Since $\cos t$ and $\sin t$ have period $2\pi$, the interval $0\le t\le 2\pi$ traces the circle exactly once, starting and ending at $(5,1)$.

   (c) Segment from $(2,-1)$ to $(-1,5)$. Using the standard form $x=x_0+(x_1-x_0)t$, $y=y_0+(y_1-y_0)t$:

   $$x=2-3t,y=-1+6t,0\le t\le 1.$$

   Check: $t=0$ gives $(2,-1)$ and $t=1$ gives $(-1,5)$. Eliminating $t=\frac{2-x}{3}$ gives $y=-1+2(2-x)=3-2x$, the line through both points, restricted to $-1\le x\le 2$.

   For the reverse direction swap the roles of the endpoints: $x=-1+3t$, $y=5-6t$, $0\le t\le 1$. (Equivalently, replace $t$ by $1-t$ in the first parametrization.) The set of points is identical; only the orientation has changed, and it is the orientation that a later integral $\int xdy$ would see.

2. 2.Computation · 8 min

   Let $x=2\cos t$, $y=\sin 2t$ for $0\le t\le 2\pi$. (The curve is a figure eight.) Quote the slope formula for parametric curves and check its hypothesis $dx/dt\ne 0$ every time you use it.

   1. (a)Find every point at which the tangent is horizontal, and every point at which it is vertical.
   2. (b)Show that the curve passes through the origin twice, and find the tangent line at each passage.
   3. (c)Compute $\frac{d^{2}y}{d{x}^2}$ at $t=\pi /4$ and say what it tells you about the curve at that point. Check that it is consistent with your answer to (a).
   Show the solution to question 2

   Derivatives first. $\frac{dx}{dt}=-2\sin t$ and $\frac{dy}{dt}=2\cos 2t$, so wherever $\sin t\ne 0$,

   $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2\cos 2t}{-2\sin t}=-\frac{\cos 2t}{\sin t}.$$

   (a) Horizontal tangents need $dy/dt=0$ and $dx/dt\ne 0$. Now $\cos 2t=0$ at $t=\pi /4,3\pi /4,5\pi /4,7\pi /4$, and $\sin t\ne 0$ at all four. The points are

   $$(\sqrt{2},1),(-\sqrt{2},-1),(-\sqrt{2},1),(\sqrt{2},-1),$$

   in that order of $t$ — the four "corners" of the figure eight.

   Vertical tangents need $dx/dt=0$ and $dy/dt\ne 0$: $\sin t=0$ at $t=0$ and $t=\pi$, where $\cos 2t=1\ne 0$. The points are $(2,0)$ and $(-2,0)$, the two ends of the figure.

   (b) The origin. $x=0$ forces $\cos t=0$, so $t=\pi /2$ or $t=3\pi /2$; at both, $y=\sin \pi =\sin 3\pi =0$. So the curve passes through the origin twice, once in each lobe. Both times $\sin t=\pm 1\ne 0$, so the slope formula is legal:

   $$t=\frac{\pi }{2}:\frac{dy}{dx}=-\frac{\cos \pi }{1}=1,t=\frac{3\pi }{2}:\frac{dy}{dx}=-\frac{\cos 3\pi }{-1}=-1.$$

   The two tangent lines at the origin are $y=x$ and $y=-x$; the figure eight crosses itself at right angles.

   (c) Second derivative. Differentiate the slope with respect to $t$ — not $y^{\prime \prime }(t)/x^{\prime \prime }(t)$ — and divide by $dx/dt$ again. With $u=-\cos 2t$ and $v=\sin t$, the quotient rule gives

   $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{2\sin 2t\sin t+\cos 2t\cos t}{{\sin }^{2}t}.$$

   At $t=\pi /4$: $\sin 2t=1$, $\cos 2t=0$, $\sin t=\frac{\sqrt{2}}{2}$, so this equals $\frac{2\cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}}=2\sqrt{2}$. Dividing by $\frac{dx}{dt}=-2\sin \frac{\pi }{4}=-\sqrt{2}$,

   $$\frac{d^{2}y}{d{x}^2}{|}_{t=\pi /4}=\frac{2\sqrt{2}}{-\sqrt{2}}=-2.$$

   Interpretation: the curve is concave down at $(\sqrt{2},1)$. That is consistent with (a), where the tangent at this point was horizontal: a horizontal tangent with negative second derivative is a local maximum of the curve, and $(\sqrt{2},1)$ is indeed the top of the right-hand lobe (since $|y|=|\sin 2t|\le 1$).

3. 3.Computation · 7 min

   Find the length of each curve. State the speed $\sqrt{(dx/dt{)}^2+(dy/dt{)}^2}$ in simplified form before integrating, and say why the curve is traversed only once on the given interval.

   1. (a)$x=\frac{t^3}{3},y=\frac{t^2}{2},0\le t\le 2$
   2. (b)$x=\cos t+t\sin t,y=\sin t-t\cos t,0\le t\le \pi$
   Show the solution to question 3

   (a) Speed, then substitution. $\frac{dx}{dt}=t^2$ and $\frac{dy}{dt}=t$, so

   $$\text{speed}=\sqrt{t^4+t^2}=\sqrt{t^2(t^2+1)}=t\sqrt{t^2+1}(t\ge 0).$$

   Both $x=t^3/3$ and $y=t^2/2$ are increasing on $[0,2]$, so no point is visited twice. Substitute $u=t^2+1$, $du=2tdt$; the limits $t=0,2$ become $u=1,5$:

   $$L={\int }_0^{2}t\sqrt{t^2+1}dt=\frac{1}{2}{\int }_1^5\sqrt{u}du=\frac{1}{2}\cdot \frac{2}{3}[u^{3/2}{]}_1^5=\frac{5\sqrt{5}-1}{3}.$$

   Sanity check: $L\approx 3.39$. The chord from $(0,0)$ to $(\frac{8}{3},2)$ has length $\sqrt{\frac{64}{9}+4}=\frac{10}{3}\approx 3.33$, and the curve is slightly longer than its chord, as it must be.

   (b) The speed is just $t$. By the product rule,

   $$\frac{dx}{dt}=-\sin t+\sin t+t\cos t=t\cos t,\frac{dy}{dt}=\cos t-\cos t+t\sin t=t\sin t.$$

   So the speed is $\sqrt{t^2{\cos }^{2}t+t^2{\sin }^{2}t}=|t|=t$ on $[0,\pi ]$. The speed is positive for $t>0$ and the direction of motion $(\cos t,\sin t)$ turns steadily through half a turn, so the curve never retraces itself (it is the involute of the unit circle — the path of the end of a string unwound from the circle). Hence

   $$L={\int }_0^{\pi }tdt=\frac{{\pi }^2}{2}.$$

   Sanity check: $L\approx 4.93$. The curve starts at $(1,0)$ and ends at $(-1,\pi )$; the chord between them has length $\sqrt{4+{\pi }^2}\approx 3.72$, so a length of about $4.9$ for a curve that bows outward is reasonable.

4. 4.Computation · 7 min

   Convert as asked. Whenever you use $\tan \theta =y/x$, state which quadrant the point lies in and why the angle you chose is the right one.

   1. (a)Find all polar representations $(r,\theta )$ of the point with Cartesian coordinates $(-\sqrt{3},-1)$.
   2. (b)Find the Cartesian coordinates of the point with polar coordinates $(-4,\pi /3)$.
   3. (c)Identify the curve $r=4\sin \theta -2\cos \theta$ by converting it to Cartesian form.
   4. (d)Write the parabola $y=x^2$ in polar form $r=f(\theta )$.
   Show the solution to question 4

   (a) Cartesian to polar, with the quadrant check. $r=\sqrt{(-\sqrt{3}{)}^2+(-1{)}^2}=\sqrt{4}=2$, and $\tan \theta =\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$. The arctangent of $1/\sqrt{3}$ is $\pi /6$, but that angle points into the first quadrant while the point, with $x<0$ and $y<0$, lies in the third. Add $\pi$: $\theta =7\pi /6$. Check: $2\cos \frac{7\pi }{6}=2\cdot (-\frac{\sqrt{3}}{2})=-\sqrt{3}$ and $2\sin \frac{7\pi }{6}=-1$. ✓

   Every polar representation is obtained by adding full turns, or by negating $r$ and adding a half turn:

   $$\left(2,\frac{7\pi }{6}+2\pi n\right)\text{and}\left(-2,\frac{\pi }{6}+2\pi n\right),n\in \mathbb{Z}.$$

   (b) Polar to Cartesian needs no thought. $x=r\cos \theta =-4\cos \frac{\pi }{3}=-2$ and $y=r\sin \theta =-4\sin \frac{\pi }{3}=-2\sqrt{3}$. The point is $(-2,-2\sqrt{3})$: face the direction $60^{\circ }$ and walk four units backwards, into the third quadrant.

   (c) Multiply by $r$ to make the conversions available. $r^2=4r\sin \theta -2r\cos \theta$ becomes $x^2+y^2=4y-2x$. Completing both squares,

   $$(x+1{)}^2+(y-2{)}^2=1+4=5,$$

   the circle of radius $\sqrt{5}$ centered at $(-1,2)$. Multiplying by $r$ adds the pole as a solution, but the pole was already on the curve: $r=0$ when $\tan \theta =\frac{1}{2}$. (Consistent with the circle: its center is at distance $\sqrt{5}$ from the origin, equal to its radius, so it passes through the origin.)

   (d) Substitute $x=r\cos \theta$, $y=r\sin \theta$. $r\sin \theta =r^2{\cos }^2\theta$. Either $r=0$ (the pole, which is on the parabola) or, dividing by $r{\cos }^2\theta$,

   $$r=\frac{\sin \theta }{{\cos }^2\theta }=\tan \theta \sec \theta .$$

   The pole need not be listed separately, since the formula already gives $r=0$ at $\theta =0$. Sanity check: at $\theta =\pi /4$, $r=\sqrt{2}$, which is the point $(1,1)$ — on $y=x^2$. ✓ The equation is uglier in polar form, as expected for a curve that is neither circular nor centered at the origin.

5. 5.Proof · 9 min

   Throughout, $x=f(t)$ and $y=g(t)$ have continuous derivatives on $[\alpha ,\beta ]$.

   1. (a)Suppose the curve is closed: $f(\alpha )=f(\beta )$ and $g(\alpha )=g(\beta )$. Prove, by integration by parts, that ${\int }_{\alpha }^{\beta }x\frac{dy}{dt}dt=-{\int }_{\alpha }^{\beta }y\frac{dx}{dt}dt$, and deduce that the three expressions in the enclosed-area formula always agree.
   2. (b)Find the area enclosed by the right-hand lobe of the figure eight $x=2\cos t$, $y=\sin 2t$. Choose the parameter interval that traces this lobe exactly once and verify that it is traced counterclockwise before you integrate.
   3. (c)Compute $\frac{1}{2}{\int }_0^{2\pi }\left(x\frac{dy}{dt}-y\frac{dx}{dt}\right)dt$ for the whole figure eight, and explain the result. Which hypothesis of the enclosed-area formula fails?
   Show the solution to question 5

   (a) Integrate by parts. Take $u=f(t)$ and $dv=g^{\prime }(t)dt$, so $du=f^{\prime }(t)dt$ and $v=g(t)$:

   $${\int }_{\alpha }^{\beta }f(t)g^{\prime }(t)dt=[f(t)g(t){]}_{\alpha }^{\beta }-{\int }_{\alpha }^{\beta }g(t)f^{\prime }(t)dt.$$

   Because the curve is closed, $f(\beta )=f(\alpha )$ and $g(\beta )=g(\alpha )$, so the boundary term is $f(\beta )g(\beta )-f(\alpha )g(\alpha )=0$. Hence

   $${\int }_{\alpha }^{\beta }x\frac{dy}{dt}dt=-{\int }_{\alpha }^{\beta }y\frac{dx}{dt}dt.■$$

   Call the common value $A$. Then $\frac{1}{2}\int (x{y}^{\prime }-y{x}^{\prime })dt=\frac{1}{2}(A-(-A))=A$ as well: all three expressions agree for every closed curve, whether or not it encloses anything sensibly. (What the proposition on enclosed area adds is that, for a simple counterclockwise curve, this common number is the area.)

   (b) Choose the interval and check the orientation. The right-hand lobe is where $x=2\cos t\ge 0$, that is $-\pi /2\le t\le \pi /2$. At $t=-\pi /2$ the point is $(0,0)$; at $t=-\pi /4$ it is $(\sqrt{2},-1)$; at $t=0$ it is $(2,0)$; at $t=\pi /4$ it is $(\sqrt{2},1)$; at $t=\pi /2$ it is back at $(0,0)$. Origin, then below the axis, then the far right, then above the axis: that is counterclockwise, and the lobe is traced once without crossing itself. The proposition applies.

   With $x^{\prime }=-2\sin t$ and $y^{\prime }=2\cos 2t$,

   $$x{y}^{\prime }-y{x}^{\prime }=4\cos t\cos 2t+2\sin 2t\sin t.$$

   Use the product-to-sum identities $\cos t\cos 2t=\frac{1}{2}(\cos t+\cos 3t)$ and $\sin 2t\sin t=\frac{1}{2}(\cos t-\cos 3t)$:

   $$x{y}^{\prime }-y{x}^{\prime }=2\cos t+2\cos 3t+\cos t-\cos 3t=3\cos t+\cos 3t.$$

   Therefore

   $$A=\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}(3\cos t+\cos 3t)dt=\frac{1}{2}[3\sin t+\frac{\sin 3t}{3}{]}_{-\pi /2}^{\pi /2}=\frac{1}{2}\left[(3-\frac{1}{3})-(-3+\frac{1}{3})\right]=\frac{8}{3}.$$

   Sanity check: eliminating the parameter, $y=2\sin t\cos t=\pm x\sqrt{1-x^2/4}$, so the lobe is symmetric about the $x$-axis and $A=2{\int }_0^{2}x\sqrt{1-x^2/4}dx$. With $u=1-x^2/4$, $du=-\frac{x}{2}dx$, this is $2\cdot 2{\int }_0^1\sqrt{u}du=\frac{8}{3}$. ✓

   (c) The integral over the whole figure is zero. By the same antiderivative,

   $$\frac{1}{2}{\int }_0^{2\pi }(3\cos t+\cos 3t)dt=\frac{1}{2}[3\sin t+\frac{\sin 3t}{3}{]}_0^{2\pi }=0.$$

   The reason is orientation. On $\pi /2\le t\le 3\pi /2$ the pen traces the left lobe: from the origin to $(-\sqrt{2},-1)$ at $t=3\pi /4$, to $(-2,0)$ at $t=\pi$, to $(-\sqrt{2},1)$ at $t=5\pi /4$, back to the origin. Below the axis first, then the far left, then above: in the left half-plane that is clockwise, so this lobe contributes $-\frac{8}{3}$ and the two lobes cancel. The enclosed-area proposition does not apply to the whole curve because the curve crosses itself, and a self-crossing figure eight is exactly a curve whose two halves have opposite orientations. The total area enclosed by the figure eight is $\frac{8}{3}+\frac{8}{3}=\frac{16}{3}$, obtained by treating the lobes separately.

6. 6.Sketch · 8 min

   Let $r=1-2\sin \theta$.

   1. (a)Apply the symmetry tests, find every $\theta$ in $[0,2\pi ]$ at which $r=0$, and find the interval on which $r<0$.
   2. (b)Sketch the curve for $0\le \theta \le 2\pi$. Mark the points at $\theta =0,\pi /2,\pi ,7\pi /6,3\pi /2,11\pi /6$ with their Cartesian coordinates, show the direction in which the curve leaves and enters the pole, and make clear which loop is traced by the negative values of $r$.
   3. (c)The point with Cartesian coordinates $(0,-1)$ lies on the curve, yet substituting $\theta =3\pi /2$ (the direction of that point) gives $r=3$, not $1$. Explain.
   Show the solution to question 6

   (a) Symmetry, zeros, sign. Replacing $\theta$ by $\pi -\theta$ leaves $\sin \theta$, and so the equation, unchanged: the curve is symmetric about the line $\theta =\pi /2$, the $y$-axis. (Replacing $\theta$ by $-\theta$ changes the sign of the sine term, so no symmetry about the $x$-axis is detected — and there is none.)

   $r=0$ when $\sin \theta =\frac{1}{2}$: $\theta =\pi /6$ and $\theta =5\pi /6$. The radius is negative exactly when $\sin \theta >\frac{1}{2}$, that is for $\pi /6<\theta <5\pi /6$, and positive elsewhere on $[0,2\pi ]$. Its largest value is $r=3$ at $\theta =3\pi /2$ and its most negative is $r=-1$ at $\theta =\pi /2$.

   Values at the standard angles: $\theta =0$: $r=1$; $\pi /6$: $0$; $\pi /2$: $-1$; $5\pi /6$: $0$; $\pi$: $1$; $7\pi /6$: $2$; $3\pi /2$: $3$; $11\pi /6$: $2$; $2\pi$: $1$.

   (b) The drawing must show two loops, both hanging below the $x$-axis, meeting at the pole.

   Outer loop. Start at $\theta =5\pi /6$ at the pole and let $\theta$ increase. The radius grows, so the pen leaves the pole along the ray at $150^{\circ }$, rises to a small hump above the $x$-axis and comes back down through $(-1,0)$ at $\theta =\pi$. Then it sweeps through the third quadrant, passing $(-\sqrt{3},-1)$ at $\theta =7\pi /6$ (where $r=2$), reaches the lowest point $(0,-3)$ at $\theta =3\pi /2$, climbs through $(\sqrt{3},-1)$ at $11\pi /6$, crosses the axis at $(1,0)$ at $\theta =2\pi$, makes the mirror-image hump, and returns to the pole along the ray at $30^{\circ }$ as $\theta \to \pi /6$. The humps are genuine: on $5\pi /6<\theta <\pi$ we have $y=r\sin \theta =s-2s^2$ with $s=\sin \theta \in (0,\frac{1}{2})$, which is positive and largest at $s=\frac{1}{4}$, giving $y=\frac{1}{8}$ at $x=\frac{1}{2}\cos \theta =-\frac{\sqrt{15}}{8}\approx -0.48$. So the outer loop pokes up to height $\frac{1}{8}$ on each side of the pole before crossing the axis at $(\pm 1,0)$.

   Inner loop. For $\pi /6<\theta <5\pi /6$ the radius is negative, so every point is plotted on the opposite ray, at directions between $210^{\circ }$ and $330^{\circ }$ — below the axis. Leaving the pole at $\theta =\pi /6$ the pen heads in the direction $210^{\circ }$, curls down to $(0,-1)$ at $\theta =\pi /2$ (the point $(-1,\pi /2)$), and returns to the pole from the direction $330^{\circ }$. This small loop lies inside the outer one, spanning $-1\le y\le 0$.

   At the pole. The two rays $\theta =\pi /6$ and $\theta =5\pi /6$ (and their extensions) are the tangent lines there, by the tangent-at-the-pole theorem: $r=0$ at those angles and $dr/d\theta =-2\cos \theta =\mp \sqrt{3}\ne 0$. So the drawing must show the curve crossing itself at the origin along the lines $y=\pm x/\sqrt{3}$, with each loop leaving and re-entering along these directions.

   Mark on the sketch: $(\pm 1,0)$, $(\pm \sqrt{3},-1)$, $(0,-3)$, $(0,-1)$, the hump tops $(\pm \frac{\sqrt{15}}{8},\frac{1}{8})$, and the two tangent lines at the pole. The whole picture is symmetric about the $y$-axis, as (a) promised.

   (c) Two names for one point. The Cartesian point $(0,-1)$ has polar coordinates $(1,3\pi /2)$, and at $\theta =3\pi /2$ the equation gives $r=3$, not $1$ — so the curve does not pass through $(0,-1)$ at that angle. But the same point is also $(-1,\pi /2)$, and at $\theta =\pi /2$ the equation gives $r=1-2=-1$. ✓ A point is on a polar curve if any of its representations satisfies the equation, and here it is the negative-$r$ representation that does. This is the inner loop's lowest point, reached when the pen faces straight up and walks one unit backwards. Testing a single representation, as one would in Cartesian coordinates, would wrongly exclude it.

7. 7.Proof · 10 min

   Let $r=f(\theta )$ have a continuous derivative, and at a point $P$ of the curve with $r\ne 0$ let $\psi$ be the angle from the radial ray $OP$ (direction $\theta$) to the tangent line, measured counterclockwise. If $\varphi$ is the angle the tangent line makes with the positive $x$-axis, then $\psi =\varphi -\theta$.

   1. (a)Prove that $\tan \psi =\frac{r}{dr/d\theta }$. Start from the slope formula for polar curves and the identity $\tan (\varphi -\theta )=\frac{\tan \varphi -\tan \theta }{1+\tan \varphi \tan \theta }$.
   2. (b)Deduce that the logarithmic spiral $r=a{e}^{b\theta }$ ($a,b>0$) crosses every ray from the pole at the same angle, and find that angle when $b=1$.
   3. (c)For the cardioid $r=1+\cos \theta$, show that $\psi =\frac{\theta }{2}+\frac{\pi }{2}$ for $0<\theta <2\pi$, and use it to find the tangent direction at $\theta =0$ and at $\theta =\pi /3$. Check both against what you know about the shape of the cardioid.
   Show the solution to question 7

   (a) Proof. Write $r^{\prime }$ for $dr/d\theta$. By the slope formula for polar curves, wherever $dx/d\theta \ne 0$,

   $$\tan \varphi =\frac{dy}{dx}=\frac{r^{\prime }\sin \theta +r\cos \theta }{r^{\prime }\cos \theta -r\sin \theta }.$$

   Apply the subtraction formula $\tan (\varphi -\theta )=\frac{\tan \varphi -\tan \theta }{1+\tan \varphi \tan \theta }$ and multiply numerator and denominator by $(r^{\prime }\cos \theta -r\sin \theta )\cos \theta$ to clear both fractions:

   $$\tan \psi =\frac{(r^{\prime }\sin \theta +r\cos \theta )\cos \theta -(r^{\prime }\cos \theta -r\sin \theta )\sin \theta }{(r^{\prime }\cos \theta -r\sin \theta )\cos \theta +(r^{\prime }\sin \theta +r\cos \theta )\sin \theta }.$$

   In the numerator the $r^{\prime }$ terms cancel ($r^{\prime }\sin \theta \cos \theta -r^{\prime }\cos \theta \sin \theta =0$) and what remains is $r({\cos }^2\theta +{\sin }^2\theta )=r$. In the denominator the $r$ terms cancel ($-r\sin \theta \cos \theta +r\cos \theta \sin \theta =0$) and what remains is $r^{\prime }({\cos }^2\theta +{\sin }^2\theta )=r^{\prime }$. Hence

   $$\tan \psi =\frac{r}{dr/d\theta }.■$$

   (If $dx/d\theta =0$ the tangent is vertical and $\tan \varphi$ is undefined; the same identity is obtained by working with $\cot$ instead, so nothing is lost. If $r^{\prime }=0$ the formula says $\tan \psi$ is infinite, that is $\psi =\pi /2$: at the points nearest to and farthest from the pole the tangent is perpendicular to the radius, which is geometrically evident.)

   The formula is the polar counterpart of the slope: $dr/d\theta$ is not a slope, but the ratio $r/(dr/d\theta )$ is the tangent of the angle between curve and radius.

   (b) Logarithmic spiral. For $r=a{e}^{b\theta }$, $dr/d\theta =ab{e}^{b\theta }$, so

   $$\tan \psi =\frac{a{e}^{b\theta }}{ab{e}^{b\theta }}=\frac{1}{b},$$

   which does not depend on $\theta$. So the spiral meets every ray from the pole at the same angle $\psi =\arctan (1/b)$ — this is the defining property of the logarithmic spiral, and the reason it appears in nautilus shells and in the flight path of a moth circling a lamp. For $b=1$, $\tan \psi =1$ and $\psi =\pi /4$: the spiral $r=e^{\theta }$ crosses every ray at $45^{\circ }$.

   Consistency check: for $r=e^{\theta }$ the arc-length integrand is $\sqrt{r^2+r^{\prime 2}}=\sqrt{2}{e}^{\theta }=\sqrt{2}r$, and $\sqrt{2}$ is exactly $1/\cos (\pi /4)$, the factor by which a path at $45^{\circ }$ to the radial direction is longer than its radial component.

   (c) Cardioid. With $r=1+\cos \theta$ and $dr/d\theta =-\sin \theta$, use the half-angle forms $1+\cos \theta =2{\cos }^2(\theta /2)$ and $\sin \theta =2\sin (\theta /2)\cos (\theta /2)$:

   $$\tan \psi =\frac{2{\cos }^2(\theta /2)}{-2\sin (\theta /2)\cos (\theta /2)}=-\cot \frac{\theta }{2}=\tan (\frac{\theta }{2}+\frac{\pi }{2}),$$

   since $\tan (u+\pi /2)=-\cot u$. Hence $\psi =\frac{\theta }{2}+\frac{\pi }{2}$ (modulo $\pi$, which is all a line's direction determines) for $0<\theta <2\pi$.

   Check at $\theta =0$. $\psi =\pi /2$: the tangent is perpendicular to the radial line, which at $\theta =0$ is the $x$-axis. So the tangent at $(2,0)$ is vertical. ✓

   Check at $\theta =\pi /3$. $\psi =\pi /6+\pi /2=2\pi /3$, so the tangent direction is $\varphi =\theta +\psi =\pi /3+2\pi /3=\pi$: the tangent is horizontal at the point $(3/2,\pi /3)$, which is the highest point of the cardioid. ✓ Both agree with the direct computation of $dy/dx$ from the slope formula, but the angle $\psi$ delivers them with almost no algebra.

8. 8.Critique · 8 min

   A student is asked for the area of the region inside the outer loop but outside the inner loop of the limaçon $r=\sqrt{2}+2\cos \theta$, and submits the following.

   Step 1. The curve is traced once as $\theta$ runs over $[0,2\pi ]$, and $r^2=2+4\sqrt{2}\cos \theta +4{\cos }^2\theta =4+4\sqrt{2}\cos \theta +2\cos 2\theta$. So the area enclosed by the outer loop is

   $$A_{\text{outer}}=\frac{1}{2}{\int }_0^{2\pi }{r}^{2}d\theta =\frac{1}{2}[4\theta +4\sqrt{2}\sin \theta +\sin 2\theta {]}_0^{2\pi }=4\pi .$$

   Step 2. The inner loop is traced where $r<0$, that is where $\cos \theta <-\frac{1}{\sqrt{2}}$, so for $3\pi /4<\theta <5\pi /4$. Its area is

   $$A_{\text{inner}}=\frac{1}{2}[4\theta +4\sqrt{2}\sin \theta +\sin 2\theta {]}_{3\pi /4}^{5\pi /4}=\frac{1}{2}[(5\pi -3)-(3\pi +3)]=\pi -3.$$

   Step 3. The required area is $A_{\text{outer}}-A_{\text{inner}}=4\pi -(\pi -3)=3\pi +3$.

   Every integral is evaluated correctly.

   1. (a)Identify the line that is wrong and explain, in terms of what the polar area formula actually measures, why it is wrong.
   2. (b)Give the correct area, and verify it by computing the area of the outer loop directly over the angles that trace it.
   Show the solution to question 8

   (a) The wrong line is the first sentence of Step 1: "the area enclosed by the outer loop is $\frac{1}{2}{\int }_0^{2\pi }{r}^{2}d\theta$".

   The area formula measures the region swept from the pole outward as $\theta$ runs from $\alpha$ to $\beta$, one thin sector $\frac{1}{2}{r}^{2}d\theta$ at a time. On $[0,2\pi ]$ every point of the curve is visited once, so the curve is traced once — but that is not the hypothesis that matters. On $3\pi /4<\theta <5\pi /4$ the radius is negative and the sectors reach from the pole to the inner loop; those sectors lie inside the inner loop, and the inner loop itself sits inside the outer loop, so this part of the plane is being counted a second time. What $\frac{1}{2}{\int }_0^{2\pi }{r}^{2}d\theta$ actually computes is $A_{\text{outer}}+A_{\text{inner}}$: the outer loop's area with the inner loop's area added on top, not the outer loop's area alone.

   Step 2 is correct as it stands: the formula does not mind $r<0$, since $r$ is squared, and the sectors for $3\pi /4\le \theta \le 5\pi /4$ fill the inner loop exactly once. Step 3 subtracts the right thing from the wrong thing.

   (b) Repair. From Step 1, correctly read,

   $$A_{\text{outer}}+A_{\text{inner}}=4\pi ⟹A_{\text{outer}}=4\pi -(\pi -3)=3\pi +3.$$

   The student's final number is the area of the outer loop, inner loop included — a true statement attached to the wrong question. The required area is

   $$A_{\text{outer}}-A_{\text{inner}}=(3\pi +3)-(\pi -3)=2\pi +6.$$

   Direct verification: the outer loop is traced where $r\ge 0$, that is for $-3\pi /4\le \theta \le 3\pi /4$. With the same antiderivative $F(\theta )=4\theta +4\sqrt{2}\sin \theta +\sin 2\theta$,

   $$A_{\text{outer}}=\frac{1}{2}[F(3\pi /4)-F(-3\pi /4)]=\frac{1}{2}[(3\pi +3)-(-3\pi -3)]=3\pi +3,$$

   using $F(3\pi /4)=3\pi +4\sqrt{2}\cdot \frac{\sqrt{2}}{2}+\sin \frac{3\pi }{2}=3\pi +4-1$ and the fact that $F$ is odd. This agrees with the repaired Step 1, so $2\pi +6\approx 12.28$ stands.

   Sanity check: the outer loop reaches $x=\sqrt{2}+2\approx 3.41$ on the right and $y=\pm \sqrt{2}$ on the $y$-axis, and $r=0$ only at the pole where the loops meet, so it is roughly an oval of length $3.4$ and height a little over $2.8$; an area around $12$ is right, and the inner loop is tiny ($\pi -3\approx 0.14$), so removing it changes little.

   The general rule: integrate $\frac{1}{2}{r}^2$ only over the angles that sweep the region you want, and read a negative $r$ as a warning that the sectors have jumped to the other side of the pole.

9. 9.Computation · 9 min

   Consider the lemniscate $r^2=8\cos 2\theta$ and the circle $r=2$.

   1. (a)Find every point where the two curves intersect. Carry out all three checks: same coordinates, opposite representations, and the pole.
   2. (b)Describe the region that lies inside the lemniscate and outside the circle: where it is, how many pieces it has, and which curve bounds it on which side.
   3. (c)Find the area of that region.
   4. (d)Without setting up a new integral, find the area of the region inside both curves.
   Show the solution to question 9

   (a) Intersections, all three checks. Solving $r^2=8\cos 2\theta$ together with $r^2=4$ gives $\cos 2\theta =\frac{1}{2}$, so $2\theta =\pm \pi /3+2\pi n$ and

   $$\theta =\pm \frac{\pi }{6},\theta =\pi \pm \frac{\pi }{6}.$$

   Second check, opposite representations: a point $(2,\theta )$ of the circle is also $(-2,\theta +\pi )$, and $r=-2$ satisfies $r^2=8\cos 2\theta$ under exactly the same condition $\cos 2\theta =\frac{1}{2}$ (the lemniscate's equation already contains both signs of $r$, and $\cos 2(\theta +\pi )=\cos 2\theta$). No new points. Third check, the pole: the circle never has $r=0$, so the pole is not an intersection. There are exactly four intersection points, with Cartesian coordinates

   $$(\sqrt{3},1),(\sqrt{3},-1),(-\sqrt{3},1),(-\sqrt{3},-1).$$

   (b) The picture. The lemniscate is a figure eight along the $x$-axis: it exists only where $\cos 2\theta \ge 0$, that is for $|\theta |\le \pi /4$ (right loop) and $|\theta -\pi |\le \pi /4$ (left loop), and each loop reaches out to $r=2\sqrt{2}\approx 2.83$ at the tip. The circle of radius $2$ is centered at the pole. Since $2<2\sqrt{2}$, each loop of the lemniscate pokes out through the circle: inside the circle near the pole, outside it near the tip. The crossing angles $\pm \pi /6$ say where: for $|\theta |<\pi /6$ we have $8\cos 2\theta >4$, so the lemniscate is outside the circle; for $\pi /6<|\theta |<\pi /4$ it is inside. The required region is the two "caps" of the loops that stick out past the circle, one on each side, symmetric about both axes.

   (c) Area of the caps. Over $-\pi /6\le \theta \le \pi /6$ the outer curve is the lemniscate and the inner curve is the circle, so by the area-between formula the right-hand cap has area

   $$\frac{1}{2}{\int }_{-\pi /6}^{\pi /6}(8\cos 2\theta -4)d\theta =\frac{1}{2}[4\sin 2\theta -4\theta {]}_{-\pi /6}^{\pi /6}=\frac{1}{2}\left[(2\sqrt{3}-\frac{2\pi }{3})-(-2\sqrt{3}+\frac{2\pi }{3})\right]=2\sqrt{3}-\frac{2\pi }{3}.$$

   By the symmetry of both curves under $\theta \mapsto \theta +\pi$, the left cap is congruent, and

   $$A=2\left(2\sqrt{3}-\frac{2\pi }{3}\right)=4\sqrt{3}-\frac{4\pi }{3}\approx \mathrm{2.74.}$$

   Sanity check: a cap is roughly a triangle with base the chord from $(\sqrt{3},-1)$ to $(\sqrt{3},1)$ (length $2$) and apex at $(2\sqrt{2},0)$ (height $\approx 0.83$), area $\approx 0.83$; the cap bulges outside that triangle, so about $1.37$ per cap is plausible.

   (d) Inside both, by subtraction. The area enclosed by the whole lemniscate is, by the polar area formula applied to each loop once,

   $$2\cdot \frac{1}{2}{\int }_{-\pi /4}^{\pi /4}8\cos 2\theta d\theta =[4\sin 2\theta {]}_{-\pi /4}^{\pi /4}=8.$$

   The part of the lemniscate that is inside the circle is everything except the two caps:

   $$A_{\text{both}}=8-(4\sqrt{3}-\frac{4\pi }{3})=8-4\sqrt{3}+\frac{4\pi }{3}\approx \mathrm{5.26.}$$

   Sanity check: the disc has area $4\pi \approx 12.6$ and the lemniscate $8$; their overlap must be less than $8$ and must contain the lemniscate's neighbourhood of the pole, so $5.26$ is sensible. Doing this directly would have needed $\frac{1}{2}\int$ of $8\cos 2\theta$ over $\pi /6\le |\theta |\le \pi /4$ plus $\frac{1}{2}\int 4$ over $|\theta |\le \pi /6$; subtraction is quicker and uses only integrals already done.

10. 10.Computation · 10 min

    Check the hypotheses of the polar arc length theorem (continuous derivative, curve traced once) before each integral, and simplify $r^2+(dr/d\theta {)}^2$ completely before integrating.

    1. (a)Find the length of the spiral $r={\theta }^2$ for $0\le \theta \le \sqrt{5}$.
    2. (b)Find the area of the surface obtained by rotating one loop of the lemniscate $r^2=\cos 2\theta$ about the polar axis. One of the theorem's hypotheses fails at an endpoint; say which, and why the answer is nevertheless correct.
    3. (c)Explain what goes wrong, and what number you would get, if you integrated in (b) over the whole loop $-\pi /4\le \theta \le \pi /4$ instead of the half you used.
    Show the solution to question 10

    (a) Spiral. $dr/d\theta =2\theta$ is continuous, and $r={\theta }^2$ is strictly increasing on $[0,\sqrt{5}]$, so the curve moves steadily outward and no point is visited twice. Then

    $$r^2+(\frac{dr}{d\theta }{)}^2={\theta }^4+4{\theta }^2={\theta }^2({\theta }^2+4),\sqrt{\cdots }=\theta \sqrt{{\theta }^2+4}(\theta \ge 0).$$

    Substitute $u={\theta }^2+4$, $du=2\theta d\theta$; the limits $\theta =0,\sqrt{5}$ become $u=4,9$:

    $$L={\int }_0^{\sqrt{5}}\theta \sqrt{{\theta }^2+4}d\theta =\frac{1}{2}{\int }_4^9\sqrt{u}du=\frac{1}{2}\cdot \frac{2}{3}[u^{3/2}{]}_4^9=\frac{1}{3}(27-8)=\frac{19}{3}.$$

    Sanity check: the curve runs from the pole to the point at distance $5$ in the direction $\theta =\sqrt{5}\approx 128^{\circ }$, so it is longer than $5$ but, since it sweeps through only about a third of a turn while growing, not hugely so; $19/3\approx 6.33$ fits.

    (b) Surface swept by one loop of the lemniscate. The right-hand loop is $-\pi /4\le \theta \le \pi /4$. Rotating the whole loop about the $x$-axis produces the same surface as rotating its upper half, $0\le \theta \le \pi /4$, where $y=r\sin \theta \ge 0$ as the formula requires; so integrate over the upper half only.

    On $0\le \theta <\pi /4$, $r=\sqrt{\cos 2\theta }$ and $\frac{dr}{d\theta }=\frac{-\sin 2\theta }{\sqrt{\cos 2\theta }}$, so

    $$r^2+(\frac{dr}{d\theta }{)}^2=\cos 2\theta +\frac{{\sin }^{2}2\theta }{\cos 2\theta }=\frac{{\cos }^{2}2\theta +{\sin }^{2}2\theta }{\cos 2\theta }=\frac{1}{\cos 2\theta },$$

    and the integrand of the surface formula is

    $$2\pi r\sin \theta \sqrt{r^2+r^{\prime 2}}=2\pi \sqrt{\cos 2\theta }\sin \theta \cdot \frac{1}{\sqrt{\cos 2\theta }}=2\pi \sin \theta .$$

    A word about hypotheses. As $\theta \to \pi /4^{-}$ the derivative $dr/d\theta$ is unbounded, so the theorem's "continuous derivative on $[\alpha ,\beta ]$" fails at the endpoint and the integral is, strictly, improper: $S={\lim }_{b\to \pi /4^{-}}{\int }_0^{b}2\pi \sin \theta d\theta$. But the integrand has simplified to something continuous on the closed interval, so the limit exists and equals the ordinary integral:

    $$S=2\pi {\int }_0^{\pi /4}\sin \theta d\theta =2\pi [-\cos \theta {]}_0^{\pi /4}=2\pi (1-\frac{\sqrt{2}}{2})=\pi (2-\sqrt{2})\approx \mathrm{1.84.}$$

    Sanity check: the loop runs from the pole to $(1,0)$, and its greatest height is where $y^2=\cos 2\theta {\sin }^2\theta =(1-2s^2)s^2$ ($s=\sin \theta$) is largest, at $s^2=\frac{1}{4}$: $y_{\max }=\sqrt{1/8}\approx 0.35$. That widest point is at $\theta =\pi /6$, where $x=\sqrt{1/2}\cos (\pi /6)\approx 0.61$. The double cone with apexes at the pole and at $(1,0)$ and that radius at $x\approx 0.61$ has slant heights about $0.71$ and $0.53$, hence lateral area about $\pi \cdot 0.35\cdot (0.71+0.53)\approx 1.37$; the lemniscate's rounded profile bulges outside this cone, so $1.84$ is the right size.

    (c) Why not $-\pi /4\le \theta \le \pi /4$? Two reasons, one geometric and one formal. Geometrically, the lower half of the loop ($-\pi /4\le \theta \le 0$) sweeps out exactly the same surface as the upper half, so integrating over both halves would count every point of the surface twice — if the integrand were positive. Formally it is not: the hypothesis $r\sin \theta \ge 0$ fails for $\theta <0$, and the simplified integrand $2\pi \sin \theta$ is negative there. Integrating over the full loop gives

    $$2\pi {\int }_{-\pi /4}^{\pi /4}\sin \theta d\theta =0,$$

    a surface of area zero: the lower half's contribution cancels the upper half's instead of doubling it. The formula $\int 2\pi yds$ is a signed quantity when $y$ changes sign, and the hypothesis $y\ge 0$ is what makes it an area.

11. 11.Proof · 10 min

    Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be an ellipse with $a>b>0$, foci $F_1=(-c,0)$ and $F_2=(c,0)$ where $c^2=a^2-b^2$, and eccentricity $e=c/a$, which you may assume satisfies $0<e<1$. Let $P=(x_0,y_0)$ be any point of the ellipse.

    1. (a)Prove that $|P{F}_2|=a-e{x}_0$. Take care to justify the sign when you remove the square root.
    2. (b)Let $l$ be the line $x=a/e$. Prove that $\frac{|P{F}_2|}{|Pl|}=e$, so that the ellipse is the set of points whose distances to the focus $F_2$ and to the directrix $l$ are in the ratio $e$ — the focus–directrix description, with the same eccentricity.
    3. (c)Using (a) and its mirror image for $F_1$, recover the defining property of the ellipse and identify the constant.
    Show the solution to question 11

    (a) The focal distance. Since $P$ lies on the ellipse, $y_0^2=b^2(1-\frac{x_0^2}{a^2})$, and $b^2=a^2-c^2$. Then

    $$|P{F}_2{|}^2=(x_0-c{)}^2+y_0^2=x_0^2-2c{x}_0+c^2+(a^2-c^2)(1-\frac{x_0^2}{a^2}).$$

    Expand the last product: $a^2-c^2-x_0^2+\frac{c^2{x}_0^2}{a^2}$. The $x_0^2$ and $c^2$ terms cancel against the first three, leaving

    $$|P{F}_2{|}^2=a^2-2c{x}_0+\frac{c^2{x}_0^2}{a^2}=(a-\frac{c}{a}{x}_0{)}^2=(a-e{x}_0{)}^2.$$

    To take the square root we need the sign of $a-e{x}_0$. Every point of the ellipse has $|x_0|\le a$ (the equation forces $x_0^2/a^2\le 1$), so $|e{x}_0|\le ea<a$ because $e<1$. Hence $a-e{x}_0>0$ and

    $$|P{F}_2|=a-e{x}_0.■$$

    The distance to a focus is a linear function of $x_0$: as $P$ moves from the far vertex $(-a,0)$, at distance $a+c$, to the near vertex $(a,0)$, at distance $a-c$, the focal distance decreases uniformly with $x_0$.

    (b) Focus and directrix. Let $l$ be the vertical line $x=a/e$. Since $e<1$, $a/e>a\ge x_0$, so $l$ lies to the right of the whole ellipse and the distance from $P$ to $l$ is

    $$|Pl|=\frac{a}{e}-x_0.$$

    Multiply by $e$: $e|Pl|=a-e{x}_0=|P{F}_2|$ by (a). So every point of the ellipse satisfies $\frac{|P{F}_2|}{|Pl|}=e$, with $F_2$ as focus and $x=a/e$ as directrix. $■$

    This is the focus–directrix description with the same eccentricity $e=c/a$, established here directly from the Cartesian equation rather than by converting the polar form; the two definitions of $e$ agree because they are describing the same number. (The mirror-image statement pairs $F_1$ with the directrix $x=-a/e$.)

    The hypothesis $e>0$ is needed: for a circle, $c=0$, the "directrix" $x=a/e$ recedes to infinity and the ratio description degenerates. That is why the focus–directrix definition covers ellipses, parabolas and hyperbolas but not circles, which it reaches only as a limit.

    (c) The defining property, recovered. Replacing $c$ by $-c$ in (a) — equivalently, reflecting the picture in the $y$-axis, which sends $x_0$ to $-x_0$ and $F_2$ to $F_1$ — gives $|P{F}_1|=a+e{x}_0$, positive for the same reason. Adding,

    $$|P{F}_1|+|P{F}_2|=(a+e{x}_0)+(a-e{x}_0)=2a,$$

    independent of $P$. This is the sum-of-distances definition of the ellipse, and it identifies the constant as $2a$, the length of the major axis — as the vertex $(a,0)$ shows directly: its distances to the foci are $a-c$ and $a+c$. So (a) is a sharpening of the definition: not only is the sum constant, but each summand is known separately, and the two-square-root algebra that derived the standard equation from the definition can be run backwards in one line.

12. 12.Computation · 10 min

    In every part the focus is at the pole. Give all coordinates exactly.

    1. (a)Identify the conic $r=\frac{8}{2-3\sin \theta }$. Find its eccentricity, its directrix, both vertices in Cartesian coordinates, its center, and $a$, $b$, $c$. Say which values of $\theta$ trace which branch.
    2. (b)An ellipse has one focus at the pole; its nearest point to that focus is at distance $2$ in the direction $\theta =0$, and its farthest point is at distance $6$ in the direction $\theta =\pi$. Find its eccentricity and a polar equation for it.
    3. (c)Show that for any conic $r=\frac{ed}{1+e\cos \theta }$ the chord through the focus perpendicular to the axis has length $2ed$, and evaluate it for the curves in (a) and (b).
    Show the solution to question 12

    (a) Divide by the constant first. $r=\frac{8}{2-3\sin \theta }=\frac{4}{1-\frac{3}{2}\sin \theta }$, so $e=\frac{3}{2}>1$: a hyperbola. From $ed=4$, $d=\frac{8}{3}$; the minus-sine form puts the directrix on the negative side, at $y=-\frac{8}{3}$. (Reading $e=3$ straight off the original denominator is the standard slip; it would still say hyperbola, but every number after it would be wrong.)

    The vertices lie on the axis perpendicular to the directrix, $\theta =\pi /2$ and $3\pi /2$:

    $$\theta =\frac{\pi }{2}:r=\frac{8}{2-3}=-8⟹(0,-8),\theta =\frac{3\pi }{2}:r=\frac{8}{2+3}=\frac{8}{5}⟹(0,-\frac{8}{5}).$$

    The first has negative $r$, so it is plotted $8$ units in the opposite direction, straight down. Both vertices are on the negative $y$-axis, $8-\frac{8}{5}=\frac{32}{5}$ apart, so $a=\frac{16}{5}$ and the center is their midpoint, $(0,-\frac{24}{5})$. The focus at the origin is $c=\frac{24}{5}$ from the center, and $c/a=\frac{24}{16}=\frac{3}{2}=e$. ✓ Then $b^2=c^2-a^2=\frac{576-256}{25}=\frac{64}{5}$, and the Cartesian equation, with the transverse axis vertical, is

    $$\frac{(y+\frac{24}{5}{)}^2}{256/25}-\frac{x^2}{64/5}=1.$$

    Which branch is which: $r>0$ where $\sin \theta <\frac{2}{3}$, and those angles trace the branch through $(0,-\frac{8}{5})$, the one that wraps around the focus at the origin; the angles with $\sin \theta >\frac{2}{3}$ give $r<0$ and trace the far branch through $(0,-8)$. Check that the directrix is where it should be: a hyperbola's directrix lies between center and vertex at distance $a/e=\frac{16}{5}\cdot \frac{2}{3}=\frac{32}{15}$ from the center, and $-\frac{24}{5}+\frac{32}{15}=-\frac{8}{3}$. ✓

    (b) Two vertices determine $e$. Nearest and farthest points on the polar axis means vertices at $\theta =0$ and $\theta =\pi$, so the cosine form $r=\frac{ed}{1+e\cos \theta }$ applies (plus sign, because the near vertex is at $\theta =0$, on the directrix side). The two conditions are

    $$\frac{ed}{1+e}=2,\frac{ed}{1-e}=6.$$

    Dividing, $\frac{1+e}{1-e}=3$, so $1+e=3-3e$ and $e=\frac{1}{2}$: an ellipse, as it had to be for a closed orbit. Then $ed=2(1+\frac{1}{2})=3$, so $d=6$ and

    $$r=\frac{3}{1+\frac{1}{2}\cos \theta }=\frac{6}{2+\cos \theta }.$$

    Check: $\theta =0$ gives $r=2$ and $\theta =\pi$ gives $r=6$. ✓ Reading off the geometry: the major axis has length $8$, so $a=4$; the center is at $(-2,0)$, the focus at the origin is $c=2$ from it, and $c/a=\frac{1}{2}=e$ ✓; $b^2=a^2-c^2=12$. The same equation comes from $r=\frac{a(1-e^2)}{1+e\cos \theta }=\frac{4\cdot \frac{3}{4}}{1+\frac{1}{2}\cos \theta }$, and the nearest and farthest distances are $a(1-e)=2$ and $a(1+e)=6$ — a planet with those distances from its sun has $e=\frac{6-2}{6+2}=\frac{1}{2}$.

    (c) The chord through the focus. In the cosine form the axis is the $x$-axis, and the chord through the focus perpendicular to it lies along $\theta =\pi /2$ and $\theta =3\pi /2$. At both angles $\cos \theta =0$, so $r=ed$ on each side of the focus, and the chord has length $2ed$, whatever $e$ is — the same for a parabola as for an ellipse or hyperbola. (The sine forms behave identically with $\theta =0$ and $\pi$.)

    For the ellipse in (b): $2ed=6$. For the hyperbola in (a): the chord lies along $\theta =0$ and $\theta =\pi$, where $r=\frac{8}{2}=4$, so its length is $2ed=8$; both endpoints have positive $r$, so both lie on the near branch, as a chord through a focus of a hyperbola must.

    Cross-check with the Cartesian data: for an ellipse $ed=a(1-e^2)=\frac{b^2}{a}$, so the chord is $\frac{2b^2}{a}=\frac{2\cdot 12}{4}=6$. ✓ For the hyperbola, $ed=a(e^2-1)=\frac{b^2}{a}=\frac{64/5}{16/5}=4$, giving $8$. ✓