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Polar Curves

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.5 · lesson 49 of 5410 min · free

← The Polar Coordinate SystemNext lesson: Areas and Arc Length in Polar Coordinates →

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In this lesson

Roses, cardioids, limacons - one equation each.

  • Recognise the standard polar families on sight
  • Use symmetry tests to halve the plotting work
  • Classify a limacon by the ratio a/b
  • Find tangent slopes by converting to parametric form

Transcript

Everything said in the video, section by section (1,484 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 9.5

Polar Curves

Curves with circular or rotational symmetry are cumbersome in Cartesian form. Flower petals. Spiral shells. Antenna radiation patterns. Try writing a four-petaled rose as y equals f of x. You can't — it fails the vertical line test everywhere. But in polar coordinates, each of those shapes reduces to a single short equation. This section catalogues the standard families, and shows how to analyze them.

The Gallery

Shapes a Cartesian Equation Would Struggle With

Here are four of the standard families. First, a circle. r equals three cosine theta. Note that it passes through the pole — polar circles through the origin have this simple form. Second, a cardioid. r equals two plus two cosine theta. Heart-shaped, with a cusp at the pole. Third, a limacon with an inner loop. r equals one plus two cosine theta. You can see the small loop nested inside. And fourth, a rose with three petals. r equals two cosine three theta. Now — the petal-counting rule deserves a moment of thought, because it's genuinely counterintuitive. For r equals a cosine n theta, or a sine n theta, the curve has n petals when n is odd. And two n petals when n is even. So r equals two cosine three theta gives three petals. But r equals two sine two theta gives four. Why does an even coefficient double the count? Because when n is odd, the second half of the interval from zero to two pi retraces petals already drawn during the first half. So the count stays at n. When n is even, no such retracing occurs. And all two n petals are distinct.

Symmetry

Three Tests That Halve the Work

Detecting symmetry before plotting cuts the work substantially. If a curve is symmetric about the polar axis, for instance, you only need to compute angles from zero to pi. The remainder is obtained by reflection. There are three tests. One. Polar axis symmetry — that's the x-axis. If replacing theta by minus theta produces an equivalent equation. Two. Symmetry about the line theta equals pi over two — the y-axis. If replacing theta by pi minus theta produces an equivalent equation. Three. Symmetry about the pole. If replacing r by minus r — or equivalently theta by theta plus pi — produces an equivalent equation. Now, one important caveat. These tests are sufficient, but not necessary. A curve may possess a symmetry that the corresponding test fails to detect. Because a single curve can have more than one valid polar equation. So failing a test proves nothing. Passing one proves the symmetry. Let's test the rose, r equals two cosine three theta, for all three. Polar axis. Replace theta with minus theta. Cosine is an even function, so cosine of minus three theta equals cosine three theta. The equation is unchanged. So the curve is symmetric about the polar axis. That halves the plotting work. Line theta equals pi over two. Replace theta with pi minus theta. Expanding the argument gives three pi minus three theta. Apply the cosine difference identity, using cosine three pi equals minus one and sine three pi equals zero. You get minus cosine three theta. So the equation becomes r equals minus two cosine three theta. Not equivalent. This test fails. And as the theorem cautions — that is not proof of asymmetry. Pole. Replace r with minus r. You get the same non-equivalent equation. Also fails. So: symmetry about the polar axis is confirmed. Which is all we needed.

Limacons

One Form, Four Shapes - Decided by a Ratio

Now the limacon family. And here's the elegant part. All four limacon shapes come from the same equation form. r equals a plus b cosine theta. Or with sine. Only the ratio a over b distinguishes them. So computing that ratio before plotting reveals the shape immediately. If a over b is less than one: a limacon with an inner loop. And it passes through the pole at two angles. If a over b equals exactly one: a cardioid, with a cusp at the pole. It passes through the pole at one angle. If a over b is between one and two: a dimpled limacon. It does not reach the pole. And if a over b is two or more: a convex limacon, with no dimple at all. The cardioid row is the boundary — between curves that loop, and curves that don't. Let's identify one. r equals one plus two cosine theta. Match the form: a equals one, b equals two. So the ratio is one half. Less than one. That's a limacon with an inner loop. Find the pole crossings by setting r equal to zero. One plus two cosine theta equals zero, so cosine theta is minus one half. On zero to two pi, that gives theta equals two pi over three, and four pi over three. Now sample some key values to confirm the shape. At theta equals zero, r is three. At theta equals pi over two, r is one. At theta equals pi, r is one plus two times minus one. Minus one. That negative value at theta equals pi is the signature of the inner loop. Because a negative r means the point is plotted in the direction opposite theta. So this point lands a distance one along the positive x-axis. Rather than the negative one.

Tangent Lines

No New Theory Required

Finding the slope of a polar curve requires no new theory at all. The strategy is to convert the polar equation into a parametric one — and then apply the method of section nine point two, with theta serving as the parameter. Recall the conversion equations. x equals r cosine theta, and y equals r sine theta. Replace r by f of theta, and both become parametric equations in the single variable theta. x of theta equals f of theta cosine theta. y of theta equals f of theta sine theta. Section nine point two established that d y d x equals d y d theta, over d x d theta. Differentiating each of those with the product rule, and substituting, produces the polar slope formula. d y d x equals d r d theta sine theta, plus r cosine theta — all over d r d theta cosine theta, minus r sine theta. Valid wherever the denominator is nonzero. And the tangent conditions carry over exactly. Horizontal tangent where d y d theta is zero and d x d theta isn't. Vertical where the reverse holds. And if both vanish at the same theta, the formula is indeterminate and the point requires separate analysis. Now — a practical note. You need not memorize that formula. Writing x equals f of theta cosine theta, and y equals f of theta sine theta, and differentiating directly, reproduces it in two lines. And that route is considerably less error-prone than recalling the arrangement of four terms and their signs.

Worked Example 3

Slope at a Specified Angle

Let's find the slope of the tangent line to the circle r equals three cosine theta, at theta equals pi over six. Step one. Write the parametric equations. Multiply r by cosine theta, and by sine theta. x equals three cosine squared theta. And y equals three cosine theta sine theta. That second one simplifies with the double-angle identity, to three halves sine two theta. Step two. Differentiate each with respect to theta. d x d theta is minus three sine two theta. And d y d theta is three cosine two theta. Step three. Evaluate at theta equals pi over six. So two theta is pi over three. Sine of pi over three is root three over two. Cosine of pi over three is one half. So d x d theta is minus three root three over two. And d y d theta is three halves. Step four. Divide. d y d x equals three halves, over minus three root three over two. The three halves cancel, leaving minus one over root three. Which rationalizes to minus root three over three. Two lines of setup. Then ordinary section nine point two machinery. No polar-specific formula had to be recalled at all.

Key Takeaways

What to Carry Forward

Five takeaways. Polar equations describe rotationally symmetric shapes compactly. Roses, cardioids, limacons, lemniscates and spirals each take one short equation. Roses have n petals for odd n, and two n petals for even n. Because odd n retraces its petals over the second half of the interval. Three symmetry tests can halve the plotting work. But they're sufficient, not necessary — failing a test proves nothing. A limacon's shape is decided entirely by the ratio a over b. Inner loop below one. Cardioid at one. Dimpled between one and two. Convex at two or above. And tangent slopes need no new theory. Set x equals f of theta cosine theta, y equals f of theta sine theta, and use the parametric method of section nine point two. Next up is section nine point six — Areas and Arc Length in Polar Coordinates. Where the area element is a sector, rather than a rectangle.

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FAQ

Questions students ask us

About Polar Curves

How many petals does a polar rose curve have?
For r = a\cos n\theta or r = a\sin n\theta with integer n \ge 2, the rose has n petals when n is odd and 2n petals when n is even. So r = 2\cos 3\theta has three petals, while r = 2\sin 2\theta has four. When n is odd, the second half of 0 \le \theta \le 2\pi retraces petals already drawn. When n is even, no retracing occurs.
How do you tell which type of limacon a polar equation gives?
Compute the ratio a/b for r = a + b\cos\theta or r = a + b\sin\theta with a, b > 0. If a/b < 1, it has an inner loop. If a/b = 1, it is a cardioid. If 1 < a/b < 2, it is dimpled. If a/b \ge 2, it is convex. For example, r = 1 + 2\cos\theta has ratio 1/2, so it has an inner loop and passes through the pole at \theta = 2\pi/3 and 4\pi/3.
How do you find the slope of a tangent line to a polar curve?
Rewrite the curve parametrically as x = f(\theta)\cos\theta and y = f(\theta)\sin\theta, then use dy/dx = (dy/d\theta)/(dx/d\theta). For r = 3\cos\theta, x = 3\cos^2\theta and y = \tfrac{3}{2}\sin 2\theta, so dx/d\theta = -3\sin 2\theta and dy/d\theta = 3\cos 2\theta. At \theta = \pi/6, the slope is \tfrac{3}{2} \div \left(-\tfrac{3\sqrt{3}}{2}\right) = -\tfrac{\sqrt{3}}{3}.
If a polar curve fails a symmetry test, does that mean it isn't symmetric?
No. Failing a symmetry test proves nothing, while passing one does prove the symmetry. The tests are sufficient but not necessary, because one curve can have several polar equations. For example, r = \sin 2\theta fails the \theta \to -\theta test. Yet replacing (r, \theta) with (-r, \pi - \theta) returns the original equation, which shows the curve is symmetric about the polar axis after all.

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