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A polar equation $r = f(\theta)$ is a parametric curve with $\theta$ as its parameter, so its slope is $\dfrac{dy/d\theta}{dx/d\theta}$ and never $\dfrac{dr}{d\theta}$ — with the standard circles, cardioids and roses, and the angles at which each reaches the pole.
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 can convert a polar equation into parametric form with $x = r\cos\theta$ and $y = r\sin\theta$, and find its slope as the ratio of the two $\theta$-derivatives rather than as $\dfrac{dr}{d\theta}$. You can recognise a circle, a cardioid and a rose from their equations, evaluate a polar curve at an angle, and find the angles at which a curve reaches the pole and so the range over which it is traced exactly once.
You can convert between polar and Cartesian coordinates from precalculus, and you can differentiate a parametric curve. That is everything this lesson needs: a polar equation turns out to be a parametric curve wearing different clothes, and once it is written as one, no new calculus is required.
Pole: the origin, where $r = 0$.
Polar equation: $r = f(\theta)$, the distance from the pole as a function of the angle.
Conversion: $x = r\cos\theta$, $y = r\sin\theta$.
Cardioid: $r = a(1 + \cos\theta)$, heart-shaped.
Rose: $r = a\cos(n\theta)$, petals that meet at the pole.
A polar equation $r = f(\theta)$ says how far the curve is from the pole at each angle. To do calculus on it, write it as a parametric curve with $\theta$ as the parameter:
$$x(\theta) = f(\theta)\cos\theta, \qquad y(\theta) = f(\theta)\sin\theta.$$
Everything from the last two lessons then applies unchanged.
The slope is the parametric slope.
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$
and it is not $\dfrac{dr}{d\theta}$. That derivative measures how fast the curve is receding from the pole, which is a genuine quantity and not a direction on the page. Both $x$ and $y$ are products, so both derivatives need the product rule — that is the only part of the work that is fiddly.
Reading the shape. Ask what $r$ does as $\theta$ turns:
The pole is where $r = 0$, and the angle there is the direction the curve leaves and enters by. Those angles matter out of all proportion to their difficulty: they are what tells you the range over which a curve is traced exactly once, and the next lesson's areas are wrong without them.
Another way: picture
Picture a radar sweep. The line from the centre turns steadily through $\theta$, and $r = f(\theta)$ says how far along that line the curve sits at each moment. A constant $f$ traces a circle; an $f$ that falls to zero and rises again pulls the trace into the centre and out the other side. The picture on the screen is Cartesian, which is exactly why the slope has to be computed after converting.
Another way: steps
| Equation | Curve | Traced once over |
|---|---|---|
| $r = a$ | circle about the pole | $0$ to $2\pi$ |
| $r = 2a\cos\theta$ | circle through the pole | $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ |
| $r = a(1 + \cos\theta)$ | cardioid | $0$ to $2\pi$ |
| $r = a\cos(n\theta)$ | rose | one petal between consecutive zeros |
The last column is the one people skip and the one that matters. Letting $\theta$ run from $0$ to $2\pi$ out of habit draws the second curve twice, which does no harm to a picture and doubles an area.
Taking $\dfrac{dr}{d\theta}$ as the slope. It is not one. Convert to $x$ and $y$, then take the parametric slope. A curve can be receding from the pole quickly while its tangent on the page is nearly flat.
Assuming every curve is traced once over a full turn. $r = 2a\cos\theta$ is complete after half a turn; a rose petal is complete over a small fraction of one. Deciding the range is part of the problem.
Reading a negative $r$ as impossible. When $f(\theta) < 0$ the point is plotted in the opposite direction, at angle $\theta + \pi$. That is how roses with an even number of petals get the extra ones.
$r = 4\cos\theta$: at $\theta = 0$, $r = 4$, so the point is $(4, 0)$.
Convert each time.
At $\theta = \tfrac{\pi}{3}$, $r = 2$, so $x = 2\cos\tfrac{\pi}{3} = 1$ and $y = 2\sin\tfrac{\pi}{3} = \sqrt3$.
The polar pair is not the point.
At $\theta = \tfrac{\pi}{2}$, $r = 0$: the pole. The curve is a circle of radius $2$ centred at $(2, 0)$, finished in half a turn.
$r = \theta$: convert to $x = \theta\cos\theta$ and $y = \theta\sin\theta$.
A spiral.
Product rule on both: $\dfrac{dx}{d\theta} = \cos\theta - \theta\sin\theta$ and $\dfrac{dy}{d\theta} = \sin\theta + \theta\cos\theta$.
Two products.
The slope is their ratio. Meanwhile $\dfrac{dr}{d\theta} = 1$ everywhere — constant, while the tangent turns through every direction there is.
The pole is where $r = 0$, so solve $2(1 + \cos\theta) = 0$.
Set $r$ to zero.
That needs $\cos\theta = -1$, which happens at $\theta = \pi$ and nowhere else in a full turn.
One solution only.
So the cardioid touches the pole once, at $\theta = \pi$ — the dimple. A rose, whose $r$ oscillates faster, reaches it many times instead.
Match each polar equation to the curve it draws.
| A circle centred at the pole | A circle passing through the pole | A cardioid | A four-petalled rose | |
|---|---|---|---|---|
| $r = 3$ | ||||
| $r = 6\cos\theta$ | ||||
| $r = 2(1 + \cos\theta)$ | ||||
| $r = 3\cos(2\theta)$ |
For the cardioid $r = 2(1 + \cos\theta)$, fill in $r$ and the Cartesian coordinates at each angle.
| $r$ | $x$ | $y$ | |
|---|---|---|---|
| $\theta = 0$ | |||
| $\theta = \tfrac{\pi}{2}$ | |||
| $\theta = \pi$ | |||
| $\theta = \tfrac{3\pi}{2}$ |
Plot the Cartesian points of the polar curve $r = 8\cos\theta$ at $\theta = 0$, $\theta = \tfrac{\pi}{4}$ and $\theta = \tfrac{\pi}{2}$.
Plot your answer on the grid:
For the polar curve $r = 2\theta$, which expression gives the slope $\dfrac{dy}{dx}$ of the curve?
Put the steps of finding $\dfrac{dy}{dx}$ for the polar curve $r = 5\cos\theta$ at a given angle into the order you do them.
Number the steps in order (write the number in the box):
The rose $r = 6\cos(2\theta)$. At how many values of $\theta$ with $0 \le \theta < 2\pi$ is the curve at the pole?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Plot the Cartesian points of the polar curve $r = 4\cos\theta$ at $\theta = 0$, $\theta = \tfrac{\pi}{4}$ and $\theta = \tfrac{\pi}{2}$.
Plot your answer on the grid:
You can read a polar curve and differentiate it. Without looking: what is $\dfrac{dy}{dx}$ for a polar curve, why is it not $\dfrac{dr}{d\theta}$, and where is a curve at the pole?
9. Your turn: where does $r = 2(1 + \cos\theta)$ meet the pole?, step 3