Back to the on-screen lesson ·
Radius and angle instead of across and up; conversion is substitution, and the cost is that a point has infinitely many names.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to convert between polar and Cartesian descriptions in both directions, multiplying through by $r$ to manufacture the three combinations that translate directly, recognise the standard polar curves and tell a circle about the pole from a circle through it, sketch $r$ against $\theta$ on ordinary axes to find where a curve is traced and where $r$ is negative, choose the angle range that traces a curve exactly once, and find intersections by solving, checking the pole, and sketching.
The last lesson described a curve by where a point is at each time. Polar coordinates describe it by how far out it reaches at each angle: $r = f(\theta)$.
That is a parametric curve with $\theta$ as the parameter — $x = f(\theta)\cos\theta$, $y = f(\theta)\sin\theta$ — so nothing about the calculus is new, and the next lesson will say so explicitly. What is new is the coordinate system itself, and in particular that a point no longer has one name.
The radius $r$ is the signed distance out along the ray at angle $\theta$, and the angle is measured anticlockwise from the positive $x$-axis. The origin is called the pole, and its angle is undetermined.
A negative $r$ is allowed and means: face in the direction $\theta$ and step backwards. So $(-2, 0)$ is the point $(-2, 0)$ in Cartesian terms, the same as $(2, \pi)$.
A curve is traced once over an angle range if no point is visited twice. Finding that range is part of every polar problem and cannot be done by algebra alone.
Conversion. $$x = r\cos\theta, \quad y = r\sin\theta, \quad r^2 = x^2 + y^2, \quad \tan\theta = \frac yx.$$ Going from polar to Cartesian is pure substitution. Going the other way often needs a trick: multiply through by $r$ so that $r^2$, $r\cos\theta$ or $r\sin\theta$ appears, since those are the three combinations that translate directly.
The naming is not unique. The point with polar coordinates $(r, \theta)$ is also $(r, \theta + 2k\pi)$ for every integer $k$, and $(-r, \theta + \pi)$. The pole is $(0, \theta)$ for every $\theta$. Cartesian coordinates name each point exactly once; polar coordinates never do.
Standard curves.
Finding the range that traces once. Sketch $r$ against $\theta$ as an ordinary graph first, and read off where $r$ is zero, where it is negative, and where the pattern repeats. For a rose, one petal is traced between consecutive zeros of $r$.
Intersections. Solving $f(\theta) = g(\theta)$ finds only the meetings that happen at the same angle. Check the pole separately, and sketch — two curves can meet at a point each reaches under a different name.
Another way: picture
A radar sweep. The beam turns steadily through the angle $\theta$ and reaches out a distance $r$ that depends on where it is pointing; the tip of the beam traces the curve. A negative $r$ is the beam reaching out behind itself. The picture explains why a full turn is usually enough and why the pole is special — it is where the beam has no direction at all.
Another way: steps
$r = \cos(n\theta)$ has $n$ petals when $n$ is odd and $2n$ when $n$ is even, which looks backwards until you follow the negative values of $r$.
Take $n = 3$. As $\theta$ runs from $-\pi/6$ to $\pi/6$, $r$ goes from $0$ up to $1$ and back to $0$: one petal, pointing along $\theta = 0$. Continue, and $r$ goes negative between $\pi/6$ and $\pi/2$ — the beam reaches backwards, drawing a petal on the opposite side, which is the petal already drawn... no, which is a petal at angle $\pi$ from the middle of that range. By the time $\theta$ has run over $[0, \pi]$, three petals exist; running on to $2\pi$ retraces all three exactly. So three petals, traced twice over a full turn, and the range that traces once is $[0, \pi]$.
Take $n = 2$. Between consecutive zeros of $\cos 2\theta$ a petal is drawn, and there are four such intervals in a full turn; the negative-$r$ petals land in the gaps rather than on top of existing ones. So four petals, traced once over $[0, 2\pi]$.
The moral is that the range cannot be guessed and cannot be got from the formula by algebra. Graph $r$ against $\theta$ on ordinary axes — zeros, maxima, sign changes — and read the answer off that. It matters because the next lesson's area formula integrates over exactly that range, and a range that traces twice doubles the answer.
Using $\theta = \arctan(y/x)$ without checking the quadrant. The arctangent returns values in $(-\pi/2, \pi/2)$ and cannot distinguish $(1,1)$ from $(-1,-1)$.
Assuming $r \ge 0$. Negative $r$ is permitted and is how half the petals of an even-order rose get drawn.
Taking the wrong angle range. A cardioid needs a full turn; a three-petal rose needs half of one. Integrating over the wrong range gives a correct answer to a different question.
Solving for intersections and stopping. The pole must be checked separately, and a sketch catches the meetings at different angles.
Converting a product without multiplying by $r$ first. $r = 2a\cos\theta$ says nothing until both sides are multiplied by $r$.
Confusing the polar graph with the graph of $r$ against $\theta$. They are different pictures and both are useful; the second is a tool for drawing the first.
Treating $\theta = \alpha$ as a ray. It is a whole line, because $r$ may be negative.
Cartesian coordinates are a naming system in which every point has exactly one name and every name denotes exactly one point. That property is so basic it is invisible, and polar coordinates do not have it.
$(2, \pi/3)$, $(2, \pi/3 + 2\pi)$ and $(-2, \pi/3 + \pi)$ are three names for one point, and there are infinitely many more. The pole is worse: it is $(0, \theta)$ for every angle whatever, so it has a continuum of names and its angle is simply undefined.
Most of the time this is harmless, and it is what makes negative $r$ usable rather than an error. Where it bites is any question that compares two descriptions. Solving $f(\theta) = g(\theta)$ asks when do these two curves have the same name at the same angle, and two curves can perfectly well pass through the same point without ever doing that. The equation is not wrong; it is answering a narrower question than the one asked.
So polar intersection problems have a three-part method — solve, check the pole, sketch — and the sketch is not a check on the algebra but a genuine part of the argument. That is unusual in this course, and it is a direct consequence of choosing coordinates that name points more than once.
Convert $x^2 + y^2 = 4y$ to polar. The left side is $r^2$ immediately.
$r^2$ is one of the three direct translations.
The right side is $4r\sin\theta$, so $r^2 = 4r\sin\theta$, giving $r = 4\sin\theta$ — after dividing by $r$, which discards the solution $r = 0$.
Divide carefully, and note what is lost.
Does anything go missing? $r = 0$ is the pole, and $r = 4\sin\theta$ reaches it at $\theta = 0$, so nothing is lost after all. That check is worth making every time a division by $r$ happens.
$r = 1$ and $r = 2\cos\theta$. Solving: $2\cos\theta = 1$, so $\theta = \pm\pi/3$, giving two intersection points.
The algebraic answer.
Now sketch. The first is the unit circle about the pole; the second is a circle of radius $1$ centred at $(1,0)$. They cross at exactly two points, and those are the two found.
Here the algebra was complete.
But try $r = \cos\theta$ and $r = 1 - \cos\theta$: solving gives $\cos\theta = \tfrac12$, two points. The sketch shows a third — the pole, reached by the first at $\theta = \pi/2$ and by the second at $\theta = 0$. Different angles, same place, and no equation would ever have found it.
Graph $r$ against $\theta$ on ordinary axes: $\cos 4\theta$ has period $\pi/2$ and eight zeros in a full turn.
The auxiliary graph, before the polar one.
Between consecutive zeros $r$ has one sign and one hump, drawing one petal. Eight intervals, eight petals — the even case, where the negative-$r$ petals fall in the gaps.
Count the intervals between zeros.
So $n = 4$ gives $2n = 8$ petals, traced exactly once over $[0, 2\pi]$. Compare $r = \cos 3\theta$: three petals, traced once over $[0, \pi]$ and twice over a full turn.
What curve is $r = 4$?
Match each polar equation to the curve it describes.
| A circle of radius $5$ about the origin | A line through the origin | A circle of radius $5$ through the origin, centred at $(5, 0)$ | The vertical line $x = 5$ | |
|---|---|---|---|---|
| $r = 5$ | ||||
| $\theta = \pi/4$ | ||||
| $r = 10\cos\theta$ | ||||
| $r\cos\theta = 5$ |
Find $r$ for the point $(8, 6)$.
Answer:
For the cardioid $r = 3(1 + \cos\theta)$, fill in $r$ at three angles.
| $r$ | |
|---|---|
| $\theta = 0$ | |
| $\theta = \pi/2$ | |
| $\theta = \pi$ |
Plot the three points given in polar form: $(2, 0)$, $\left(2, \dfrac{\pi}{2}\right)$ and $\left(2, \pi\right)$, each written as $(r, \theta)$.
Plot your answer on the grid:
The curves $r = 8$ and $r = 8\,(1 - \cos\theta)$ both pass through some point where solving $8 = 8(1 - \cos\theta)$ finds them. Why can solving the equations miss an intersection?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
What curve is $r\cos\theta = 3$?
You can read and draw a polar curve and choose the range that traces it once. Say in your own words why solving two polar equations can miss an intersection. Next: area and length in polar coordinates, where the range you just found decides the answer.
10. Your turn: how many petals has $r = \cos(4\theta)$, and over what range is it traced once?, step 3