Corrected exercises: polar coordinates, 201-SN3-RE (201-NYB)

$r=\sqrt{1+3}=2$. The point is in the SECOND quadrant ($x<0$, $y>0$). The calculator gives $\arctan\left(\frac{\sqrt{3}}{-1}\right)=\arctan\left(-\sqrt{3}\right)=-\frac{\pi}{3}$, which designates a point in the fourth quadrant: that is wrong. You must add $\pi$: $\theta=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}$. Check: $2\cos\frac{2\pi}{3}=-1$ and $2\sin\frac{2\pi}{3}=\sqrt{3}$.

For $\left(3, -3\right)$: $r=3\sqrt{2}$, fourth quadrant, $\arctan(-1)=-\frac{\pi}{4}$ is correct, and in $[0, 2\pi)$ we write $\theta=\frac{7\pi}{4}$. For $\left(-2, -2\sqrt{3}\right)$: $r=\sqrt{4+12}=4$, THIRD quadrant, $\arctan\left(\frac{-2\sqrt{3}}{-2}\right)=\arctan\sqrt{3}=\frac{\pi}{3}$, a first-quadrant value: once again you must add $\pi$, hence $\theta=\frac{4\pi}{3}$. Rule: add $\pi$ whenever $x<0$, whatever the sign of $y$.

$x=4\cos\frac{5\pi}{6}=4\cdot\left(-\frac{\sqrt{3}}{2}\right)=-2\sqrt{3}$ and $y=4\sin\frac{5\pi}{6}=4\cdot\frac{1}{2}=2$. The Cartesian point is $\left(-2\sqrt{3},2\right)$. This direction of conversion causes no difficulty at all: it is always the other one that sets the trap.

$\left(2,\frac{\pi}{3}\right)$, $\left(2,\frac{\pi}{3}+2\pi\right)=\left(2,\frac{7\pi}{3}\right)$, $\left(2,\frac{\pi}{3}-2\pi\right)=\left(2,-\frac{5\pi}{3}\right)$, and with $r<0$: $\left(-2,\frac{\pi}{3}+\pi\right)=\left(-2,\frac{4\pi}{3}\right)$. The pole is an extreme case: $\left(0,\theta\right)$ designates the origin for EVERY $\theta$, so it has uncountably many representations. That is precisely why it escapes systems of equations, as we will see in exercise 8.

In Cartesian coordinates, $d^{2}=\left(r_{1}\cos\theta_{1}-r_{2}\cos\theta_{2}\right)^{2}+\left(r_{1}\sin\theta_{1}-r_{2}\sin\theta_{2}\right)^{2}$. Expanding: $r_{1}^{2}\left(\cos^{2}\theta_{1}+\sin^{2}\theta_{1}\right)+r_{2}^{2}\left(\cos^{2}\theta_{2}+\sin^{2}\theta_{2}\right)-2r_{1}r_{2}\left(\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos\left(\theta_{1}-\theta_{2}\right)$. This is the law of cosines in the triangle formed by the pole and the two points, the angle at the pole being $\theta_{1}-\theta_{2}$. Application: $d^{2}=9+25-2\cdot 15\cos\frac{\pi}{3}=34-15=19$, so $d=\sqrt{19}\approx 4.36$.

$r=c$ is the circle centred at the pole with radius $\left|c\right|$. $\theta=\alpha$ is the HALF-line (ray) from the pole making the angle $\alpha$ if we require $r\geq 0$; if $r<0$ is allowed, we get the whole line, because $\left(-r,\alpha\right)$ describes the opposite ray.

Multiply by $r$: $r^{2}=4r\cos\theta$, that is $x^{2}+y^{2}=4x$, so $\left(x-2\right)^{2}+y^{2}=4$. This is the circle with centre $\left(2, 0\right)$ and radius 2, passing through the pole. It is traced exactly once for $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, the interval on which $r\geq 0$. On $[0, 2\pi]$ it would be traced TWICE, because for $\theta\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ the radius becomes negative and redraws the same circle. This is the mistake that doubles every area of a polar circle.

$r^{2}=6r\sin\theta$ gives $x^{2}+y^{2}=6y$, that is $x^{2}+\left(y-3\right)^{2}=9$: the circle with centre $\left(0, 3\right)$ and radius 3, tangent to the polar axis at the pole. It is traced once for $\theta\in[0,\pi]$. Remember: $r=2a\cos\theta$ is centred on the polar axis, $r=2a\sin\theta$ on the line $\theta=\frac{\pi}{2}$.

$r\cos\theta=x$, so $r\cos\theta=3$ is the vertical line $x=3$, with polar equation $r=\dfrac{3}{\cos\theta}=3\sec\theta$. For $x=d$: $r=d\sec\theta$. For a line whose foot of the perpendicular from the pole is at distance $d$ in the direction $\alpha$, projecting gives $r\cos\left(\theta-\alpha\right)=d$, that is $r=\dfrac{d}{\cos\left(\theta-\alpha\right)}$.

From $r\left(1+\cos\theta\right)=2$ we get $r=2-r\cos\theta=2-x$, then $\sqrt{x^{2}+y^{2}}=2-x$. Squaring: $x^{2}+y^{2}=4-4x+x^{2}$, so $y^{2}=4-4x=4\left(1-x\right)$. This is a PARABOLA with vertex $\left(1, 0\right)$, opening towards negative $x$, with its focus at the pole. More generally, $r=\dfrac{ed}{1+e\cos\theta}$ is a conic with focus at the pole and eccentricity $e$: an ellipse if $e<1$, a parabola if $e=1$, a hyperbola if $e>1$. This is the form in which Kepler writes orbits, and the very reason polar coordinates exist in celestial mechanics.

Symmetry about the POLAR AXIS: the curve is unchanged if $\left(r,\theta\right)$ is replaced with $\left(r,-\theta\right)$, or with $\left(-r,\pi-\theta\right)$. Symmetry about the line $\theta=\frac{\pi}{2}$: if it is replaced with $\left(r,\pi-\theta\right)$ or with $\left(-r,-\theta\right)$. Symmetry about the POLE: if it is replaced with $\left(-r,\theta\right)$ or with $\left(r,\theta+\pi\right)$.

$r=3\cos 2\theta$. Polar axis: $3\cos\left(-2\theta\right)=3\cos 2\theta$, the test succeeds. Line $\theta=\frac{\pi}{2}$: $3\cos\left(2\pi-2\theta\right)=3\cos 2\theta$, it succeeds. Pole: $3\cos\left(2\theta+2\pi\right)=3\cos 2\theta$, it succeeds. The curve has all three symmetries. Since $k=2$ is EVEN, the rose has $2k=4$ petals, of length 3, pointing along the axes, and it is traced exactly once on $[0, 2\pi]$.

$r=3\cos 3\theta$: the polar axis test succeeds, the other two fail in this form. Since $k=3$ is ODD, there are $k=3$ petals, and the whole curve is already traced on $[0,\pi]$; on $[0, 2\pi]$ it would be traced TWICE. The reason is the sign: $\cos k\left(\theta+\pi\right)=\cos\left(k\theta+k\pi\right)$ equals $\cos k\theta$ if $k$ is even, and $-\cos k\theta$ if $k$ is odd. When $k$ is odd, the next half-turn gives back the same points through the representation $\left(-r,\theta+\pi\right)$: the petals land on top of each other instead of fitting in between.

$r=2+\sin\theta$. Polar axis: $2+\sin\left(-\theta\right)=2-\sin\theta\neq r$, and the second form gives $-2-\sin\left(\pi-\theta\right)$, which fails too: no symmetry about the polar axis, as confirmed by the fact that $r\left(\frac{\pi}{2}\right)=3$ while $r\left(-\frac{\pi}{2}\right)=1$. Line $\theta=\frac{\pi}{2}$: $2+\sin\left(\pi-\theta\right)=2+\sin\theta=r$, the test SUCCEEDS. It is a convex limaçon, symmetric about the vertical axis only.

For $r=\sin 2\theta$, the test “$-r$ in place of $r$” gives $-\sin 2\theta$, which differs from $\sin 2\theta$: it FAILS. Yet the second form of the same test, “$\theta+\pi$ in place of $\theta$”, gives $\sin\left(2\theta+2\pi\right)=\sin 2\theta$: it SUCCEEDS. So the curve is indeed symmetric about the pole, as its graph with four petals on the diagonals shows. Conclusion: the tests are SUFFICIENT conditions. A test that succeeds proves the symmetry; a test that fails proves nothing at all, and you must then try the other form, or go back to the graph. This is a direct consequence of the fact that a point has several polar representations.

If $a<b$: a limaçon with an INNER LOOP, because $r$ changes sign, for example $r=1+2\cos\theta$. If $a=b$: a CARDIOID, $r$ vanishes without changing sign, for example $r=3+3\cos\theta$. If $b<a<2b$: a DIMPLED limaçon, $r$ stays positive but the curve has an inward dent, for example $r=3+2\cos\theta$. If $a\geq 2b$: a CONVEX limaçon, for example $r=4+2\cos\theta$.

$1+2\cos\theta=0$ gives $\cos\theta=-\frac{1}{2}$, that is $\theta=\frac{2\pi}{3}$ and $\theta=\frac{4\pi}{3}$. Between these two values, $\cos\theta<-\frac{1}{2}$ so $r<0$: the points are plotted in the direction OPPOSITE to $\theta$, which produces the small inner loop. The inner loop therefore corresponds exactly to $\theta\in\left[\frac{2\pi}{3},\frac{4\pi}{3}\right]$, and the outer loop to $\theta\in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. Remember: the bounds of a loop are ALWAYS consecutive zeros of $r$.

$r=3+3\cos\theta$ is largest when $\cos\theta=1$, at $\theta=0$: the point $\left(6, 0\right)$ is at distance 6 from the pole. $r=0$ when $\cos\theta=-1$, at $\theta=\pi$: the curve reaches the pole. Since $r\geq 0$ everywhere and vanishes without changing sign, the curve arrives there and leaves on the same side: it is a CUSP, the characteristic point of the cardioid.

For $r=4+2\cos\theta$, $\frac{a}{b}=2$: $r$ ranges from 2 to 6, never zero, so there is no loop; and the convexity threshold $\frac{a}{b}\geq 2$ is reached, so there is no dimple either, the curve is a convex oval. For $r=3+2\cos\theta$, $\frac{a}{b}=\frac{3}{2}$ lies strictly between 1 and 2: $r$ ranges from 1 to 5 without vanishing, so there is no loop, but convexity is lost and the curve has a dimple on the side $\theta=\pi$, where the minimum radius is 1.

RULE: if $r\left(\theta_{0}\right)=0$ and $r'\left(\theta_{0}\right)\neq 0$, the tangent line to the curve at the pole is the line $\theta=\theta_{0}$. Justification: the slope is $\dfrac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$, which at $\theta_{0}$ reduces to $\dfrac{r'\left(\theta_{0}\right)\sin\theta_{0}}{r'\left(\theta_{0}\right)\cos\theta_{0}}=\tan\theta_{0}$, since $r\left(\theta_{0}\right)=0$ and $r'\left(\theta_{0}\right)$ cancels. Application: $r'=-2\sin\theta$ vanishes neither at $\frac{2\pi}{3}$ nor at $\frac{4\pi}{3}$. The two tangents at the pole are therefore the lines $\theta=\frac{2\pi}{3}$ and $\theta=\frac{4\pi}{3}$: they are in fact the SAME line, with slope $\tan\frac{2\pi}{3}=-\sqrt{3}$, along which the inner loop enters and leaves.

Parametrize by $\theta$: $x=r\left(\theta\right)\cos\theta$ and $y=r\left(\theta\right)\sin\theta$. Differentiating as products, $\dfrac{dx}{d\theta}=r'\cos\theta-r\sin\theta$ and $\dfrac{dy}{d\theta}=r'\sin\theta+r\cos\theta$. The parametric quotient of derivatives then gives directly $\dfrac{dy}{dx}=\dfrac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$.

$r=1+\cos\theta$, $r'=-\sin\theta$. Then $\dfrac{dy}{d\theta}=-\sin^{2}\theta+\left(1+\cos\theta\right)\cos\theta=-\left(1-\cos^{2}\theta\right)+\cos\theta+\cos^{2}\theta=2\cos^{2}\theta+\cos\theta-1=\left(2\cos\theta-1\right)\left(\cos\theta+1\right)$. The zeros are $\cos\theta=\frac{1}{2}$, that is $\theta=\pm\frac{\pi}{3}$, and $\cos\theta=-1$, that is $\theta=\pi$. At $\theta=\frac{\pi}{3}$: $r=\frac{3}{2}$ and the point is $\left(\frac{3}{4},\frac{3\sqrt{3}}{4}\right)$; at $\theta=-\frac{\pi}{3}$, its mirror image $\left(\frac{3}{4},-\frac{3\sqrt{3}}{4}\right)$. At $\theta=\pi$, $\frac{dx}{d\theta}$ vanishes too: it is the cusp, not a horizontal tangent.

$\dfrac{dx}{d\theta}=-\sin\theta\cos\theta-\left(1+\cos\theta\right)\sin\theta=-\sin\theta\left(2\cos\theta+1\right)$. The zeros are $\sin\theta=0$, that is $\theta=0$ or $\theta=\pi$, and $\cos\theta=-\frac{1}{2}$, that is $\theta=\pm\frac{2\pi}{3}$. At $\theta=0$: $r=2$, the point $\left(2, 0\right)$. At $\theta=\frac{2\pi}{3}$: $r=\frac{1}{2}$, the point $\left(-\frac{1}{4},\frac{\sqrt{3}}{4}\right)$, and its mirror image at $\theta=-\frac{2\pi}{3}$. The case $\theta=\pi$ is the cusp again.

Let $\phi$ be the angle of the tangent line with the polar axis; then $\psi=\phi-\theta$. Setting $\tan\phi=\dfrac{dy}{dx}$ and using the tangent of a difference, $\tan\psi=\dfrac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$. Replacing $\tan\phi$ with the expression from a and multiplying top and bottom by $\cos\theta$, all the terms simplify and what remains is $\tan\psi=\dfrac{r}{r'}$. For the cardioid: $\dfrac{r}{r'}=\dfrac{1+\cos\theta}{-\sin\theta}=\dfrac{2\cos^{2}\frac{\theta}{2}}{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=-\cot\dfrac{\theta}{2}=\tan\left(\dfrac{\theta}{2}+\dfrac{\pi}{2}\right)$, so $\psi=\dfrac{\theta}{2}+\dfrac{\pi}{2}$. At $\theta=0$, $\psi=\frac{\pi}{2}$: the tangent is perpendicular to the radius at the farthest point, which is visually obvious.

For $r=e^{a\theta}$, we have $r'=ae^{a\theta}=ar$, so $\tan\psi=\dfrac{r}{ar}=\dfrac{1}{a}$: the angle $\psi$ is CONSTANT, equal to $\arctan\frac{1}{a}$, at every point. The spiral therefore crosses every ray from the pole at the same angle, hence the name equiangular spiral. This is the property that makes it appear in nautilus shells and in approach paths flown on a constant heading, and the one Jacob Bernoulli had engraved on his tombstone with the motto “eadem mutata resurgo”.

Cut $\left[\alpha,\beta\right]$ into small angles $d\theta$. Over such an angle, the region is almost a circular sector of radius $r\left(\theta\right)$, whose area is the fraction $\frac{d\theta}{2\pi}$ of the full disk $\pi r^{2}$, that is $\frac{1}{2}r^{2}d\theta$. Adding up, $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. The quantity $\int r\,d\theta$ has no meaning here: $r\,d\theta$ is the length of an ARC, not an area, and its dimension is a length, not a length squared. The dimensional check is what should alert you if you forget the square.

The circle is traced once for $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. $A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}16\cos^{2}\theta\,d\theta=8\int_{-\pi/2}^{\pi/2}\cos^{2}\theta\,d\theta=8\cdot\frac{\pi}{2}=4\pi$. This is indeed $\pi R^{2}$ with $R=2$. On $[0, 2\pi]$ we would find $8\pi$: twice as much, because the circle is traced twice, once with $r>0$ and once with $r<0$. The area formula counts the passages, exactly as with parametric curves.

$A=\frac{1}{2}\int_{0}^{2\pi}\left(1+\cos\theta\right)^{2}d\theta=\frac{1}{2}\int_{0}^{2\pi}\left(1+2\cos\theta+\cos^{2}\theta\right)d\theta$. The three pieces are $2\pi$, $0$ and $\pi$, so $A=\frac{1}{2}\left(3\pi\right)=\frac{3\pi}{2}$. For a cardioid $r=a\left(1+\cos\theta\right)$, the area is $\frac{3\pi a^{2}}{2}$, one and a half times the area of the disk of radius $a$.

One petal of $r=3\cos 2\theta$ lies between two consecutive zeros of $r$: $\cos 2\theta=0$ gives $\theta=\pm\frac{\pi}{4}$. $A_{1}=\frac{1}{2}\int_{-\pi/4}^{\pi/4}9\cos^{2}2\theta\,d\theta=\frac{9}{2}\cdot\frac{\pi}{4}=\frac{9\pi}{8}\approx 3.53$. Since $k=2$ is even, there are 4 petals: total area $\frac{9\pi}{2}\approx 14.14$.

For $r=3\cos 3\theta$, $\cos 3\theta=0$ gives $\theta=\pm\frac{\pi}{6}$ for the central petal. $A_{1}=\frac{1}{2}\int_{-\pi/6}^{\pi/6}9\cos^{2}3\theta\,d\theta=\frac{9}{2}\cdot\frac{\pi}{6}=\frac{3\pi}{4}\approx 2.36$. Since $k=3$ is odd, there are 3 petals: total area $\frac{9\pi}{4}\approx 7.07$. The two totals differ by a factor of 2, although the two curves have the same maximum radius 3: the larger $k$ gets, the thinner the petals, and the total is not conserved. There is no shortcut of the kind “total area $=$ number of petals times something fixed”: each time, you must find the zeros of $r$ and count the petals according to the parity of $k$.

$r=0$ for $\cos\theta=-\frac{1}{2}$, that is $\theta=\frac{2\pi}{3}$ and $\theta=\frac{4\pi}{3}$. These are the moments when the curve PASSES THROUGH THE POLE: between two consecutive passages through the pole, the curve traces one complete closed loop, and that is exactly what $\frac{1}{2}\int r^{2}d\theta$ needs to measure a well-defined region. Taking other bounds would measure a truncated sector.

The inner loop is traced for $\theta\in\left[\frac{2\pi}{3},\frac{4\pi}{3}\right]$. Expanding, $\left(1+2\cos\theta\right)^{2}=1+4\cos\theta+4\cos^{2}\theta=3+4\cos\theta+2\cos 2\theta$, an antiderivative of which is $3\theta+4\sin\theta+\sin 2\theta$. At $\frac{4\pi}{3}$: $4\pi-2\sqrt{3}+\frac{\sqrt{3}}{2}=4\pi-\frac{3\sqrt{3}}{2}$. At $\frac{2\pi}{3}$: $2\pi+2\sqrt{3}-\frac{\sqrt{3}}{2}=2\pi+\frac{3\sqrt{3}}{2}$. The difference is $2\pi-3\sqrt{3}$, so $A_{\text{in}}=\frac{1}{2}\left(2\pi-3\sqrt{3}\right)=\pi-\frac{3\sqrt{3}}{2}\approx 0.544$. The result is positive, as it should be, even though $r$ is negative there: it is $r^{2}$ that is integrated.