Parametric Equations and Polar Coordinates

(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.

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 $|y|=|\sin 2t|\le 1$).

(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.

(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.

(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.

(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.

(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.

(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.

(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.

(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.

(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.

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