Introduction

Hi, It's Kevin again. In the last video, we introduced the mathematical concepts of a pencil and an envelope. I recall that as a child I was fascinated with pencils and envelopes, and amazed that simply drawing a set of lines or circles could result in beautiful sweeping curves. The cardioid was one of the first such curves that I ran across. That fascination persisted, because I learnt a number of ways that a simple pencil would yield a cardioid. This video will present several of them and even prove a couple. I hope that at the end of it, you'll understand why the cardioid is such a favorite of mine, so stick around!

Review: pencils and envelopes

Last time - there's a link to the video somewhere nearby - we started looking at mathematical pencils and envelopes. We defined a pencil as a family of curves defined using an implicit formula by varying a single parameter u, and the envelope of a pencil as the set of points for which, for some value of u, both the function and its partial derivative with respect to u are zero. We looked at the envelope of a pencil of lines. Now let's stretch a little bit farther and look at the envelope of a pencil of circles.

Envelope of a pencil of circles

Start by drawing a unit circle, and choose a point on its circumference To make an element of the pencil, choose another point on the circumference of the circle Draw a circle centered on this new point and passing through the first point. Going around the circle gives us a pencil of circles, varying only by the positions of the chosen centers along the circumference. Let's see if you can spot what the envelope of these circles might look like. It might be a little easier if I didn't erase the circles every time. It looks awfully like a cardioid, doesn't it? Let's prove it. As usual, we'll center our coordinate system on the cusp of the cardioid. I'll erase the pencil and envelope for now, and reason about just one circle. Since we said that we started with a unit circle, its center will be -1 unit on the x axis. We can find a point on the circle by specifying a central angle. I'm going to call this angle 2 theta, and in fact I'm going to bisect it now, and we'll see why in just a moment. The center of the circle in the pencil will be at coordinates -1+cos(2 theta) and sin(2 theta) And the radius of the circle in the pencil is 2 sin theta We know the center and radius of the circle, so we can write out the formula for the pencil, with parameter theta. Let's get everything in terms of 2 theta. A double angle substitution does the job. I'd like to work with polynomials rather than trig functions. We can do this by making the Weierstrass substitution u = tan theta. I discussed Weierstrass substitution in an earlier video, and there should be a link somewhere nearby. Once we grind through all the algebra, we are left with a rational function. We can multiply both sides of the equation by the denominator, and now the pencil is defined as a polynomial, with no pesky trig functions. It's still the same set of circles! Now that I have this formula, I'm going to make room on the chalkboard by erasing how I got it. For a point to be on the envelope, both the defining function and its partial derivative with respect to the parameter must be zero. We differentiate the function. solve the derivative identity for u substitute this value into the equation for f, multiply both sides by the denominator and we get a polynomial formula for the envelope. But does this formula describe a cardioid?

Implicit Cartesian formula of a cardioid

This is where I was going to insert a link to a previous video deriving the formula of a cardioid in Cartesian coordinates, but it appears that I never troubled to indroduce that formula. Oh well, it doesn't matter. It's simple enough to derive. I'll do that now. Remember that the formula for a cardioid in polar coordinates is r = 2a ( 1 - cos theta ) And remember the basic trigonometry of polar coordinates. If you substitute first r and then x into the polar formula, you get an implicit formula in Cartesian coordinates. I'll run through this one on "algebra autopilot" The equation of the cardioid is a fourth-degree polynomial with six terms.

Envelope of pencil of circles is a cardioid

Now we can go back to see whether the envelope of the circles is anything like this. Bringing in the formula that we just derived, we see that our guess was correct. The envelope is a cardioid, and the defining circle is the incircle of the cardioid.

String art in the circle

Now that we understand pencils and envelopes, let's come back to string art. We'll arrange the n pins in a circle this time. We'll number them from zero to n-1. Stretch a string from pin 1 to pin 2 from pin 2 to pin 4 from pin 3 to pin 6 and so on, connecting pin i to pin 2i When 2i reaches n, continue using arithmetic modulo n. That envelope looks pretty familiar, doesn't it?

String art and coffee cup are the same figure

Not only do you immediately recognize it as a cardioid, but you may have seen it, in exactly the same proportions, in your coffee cup. Let's take a mathematical look at how this comes about. Place a light source on the rim of a circular reflector. Trace a light ray in an arbitrary direction from that source. Let theta be the angle it makes with the axis. Draw the radius from the center of the reflector to the point where the light ray strikes The triangle that it makes is isosceles, so the angle of incidence of the light ray is equal to theta. High school physics tells us that the angle of incidence is equal to the angle of reflection. Plot the reflected ray. Draw another radius to meet the point where the second ray meets the reflector, and you get a pair of congruent triangles, so the central angles are equal. Call the central angle phi. The reflected ray runs from an angle of phi to an angle of 2 phi, which is just the same as a string starting from phi and running to 2 phi, so the geometry of the strings and the reflected rays is exactly the same. Tracing all the rays demonstrates this.

String art forms a cardioid

So let's return to the string art now, and prove that we're looking at a cardioid. Because the circle that holds the strings looks like the excircle of the cardioid, whose diameter is 3/2 the diameter of the cardioid, I'm going to guess that's the case. I'll draw axes with the origin on what appears to be the cusp and give the circle a radius of 3 units, centered on the coordinates (1, 0) Let's look at the equation for one string, which is a chord of the circle running from an angle of 2 theta to an angle of 4 theta. (I've used 2 theta as the angle because I'll be bisecting it in the next step) Bisect the central angle that subtends the string, which forms a perpendicular bisector of the string. Since the radius of the circle is 3, the altitude is 3 cos theta. Some straightforward trigonometry gives us a parametric representation of the string. This needs to be made into an implicit representation We've done this a couple of times, and the procedure is similar each time. Solve for t in the equation for y. Substitute into the equation for x, and rearrange to make it simpler. Engage algebra autopilot! We now have a fairly tidy implicit representation. Now let's calculate the envelope! Well, let's tidy the blackboard first. As before, the envelope is the locus of points where both the function and its partial derivative with respect to the parameter are zero. We differentiate the function, and then start our algebraic manipulations again to extract x and y. The conventional way to do that is to solve one of these equations for x or y, and then substitute the result into the other, but when we do that, we get a mess of quotients of trig functions. But I notice that there's only one appearance of x in the two equations. I can make a linear combination of the two that will get x all by itself. Let's see what happens when we do that and simplify. We have more work to do to get rid of those triple angles, but we'll push this equation to one side for a moment, clean up our scratch work, and start on y. We can use the same trick with y that we did with x. And we have successfully isolated both x and y. It's not obvious that these equations define a cardioid, so we're going to have to try to simplify them further. I had to spend more time than I'd have liked, playing around with possible transformations. What I'll show here is one line of reasoning that worked. If you can make it more elegant, I'd love to hear about it! Let's tidy up and keep just the equations of the envelope. We'll start by using the angle sum formulas to reduce to single and double angles. That gives us the expansions on the right. Wow, what a mess! But if we collect the terms involving sin(2 theta) and cos(2 theta), we'll start to see opportunities to simplify. Regrouping leaves us with the double-angle sine formula, plus a couple of terms with the common factor 3 cos squared theta + sin squared theta. Can we make that common factor nicer? Let me step over to the other chalkboard and quickly run through that calculation. We begin by splitting both terms in such a way that the two yellow terms have the same coefficient, and the two blue terms also have the same coefficient. Now we regroup. And a couple of basic trig identities finish the simplification. This rewriting can be generalized. Can you come up with a general formula for A cos squared theta + B sin squared theta? Leave your answer in the comments! Now we can come back to our equations and make the substitution that we just derived. We can get the rest of the way on autopilot. After that tedious proof, we conclude that the envelope of the strings is indeed a cardioid.

Envelope of normals to the cardioid (evolute)

Before we go, let's look at just one more envelope. Start with a cardioid this time. Consider the pencil of lines that meet the cardioid at right angles. A mathematician would say that they are 'normal' to the cardioid, and define them as being perpendicular to the tangent line at the point of tangency. The envelope of these lines is a cardioid exactly 1/3 the size, in a mirror image to the original cardioid, with its vertex against the original cardioid's cusp. There's a closely related fact that can be shown from the cardioid arc length formula, which we derived in a previous episode. If we attach a string that's half the circumference of a cardioid to the cusp, wind it around the cardiod, and then unwind it starting from the vertex, keeping it taut as we go, the string will trace out a cardioid three times as large.

Cardioidal coordinate system

As a final look at pencils, let's look at two pencils of cardioids. The first are all the cardioids with the cusp on the origin and the vertex somewhere on the positive x axis. They're shown here in yellow. The second are all the cardioids with the cusp on the origin and the vertex on the negative x axis. They're shown in blue. These sets define coordinate systems for the two half planes above and below the x axis. Every pair of yellow and blue cardioids meets at right angles, at mirror images across the axis. Every point on one of the half planes belongs to exactly one pair of the cardioids. And starting from any point on the half plane, following a path that is perpendicular to each yellow cardioid in turn traces out a blue cardioid, and vice versa. I'm told that this coordinate system is useful occasionally for calculations in electromagnetic field theory, but I confess that I'm hazy on the details here.

Summary

If a viewer can enlighten me, I'd be happy to learn! So, in summary, we've seen in this video a number of ways in which cardioids emerge from pencils of lines, circles, or other cardioids. The way that these structures formed curvilinear figures fascinated me as a child. And later I learnt that acoustics, geometric optics, navigation, field theory, and other scientific analyses make use of them.

Coming attractions: the Mandelbrot cardioid

Before we wrap up the series, there's still one more cardioid that I want to examine. That one is the cardioid that is at the heart of the Mandelbrot set, the famous mathematical structure that represents in a deep way the boundary between order and chaos, between the predictable and the random. We'll just barely be scratching the surface of its beauties, but we'll be exploring one facet that I don't think I've seen covered in a YouTube video, so stay tuned for that! Until then, stay safe, stay healthy, and keep calculating!