Topic 3.15: Rates of Change in Polar Functions

One more song

On one side of a record, every song lives in a single groove. Just one. It’s a spiral. It starts at the outside edge and winds all the way in toward the center.

And it’s long. If you could unwind one side of an album and lay that groove out straight, it would stretch about half a kilometer. This is where a groove gets made. A cutting lathe carves the music into a soft lacquer disc as it turns, and tonight this one has a problem. This side of the album runs twenty-four minutes, but with the turns spaced the way the lathe is set right now, the groove runs out of room at twenty minutes, and the last song doesn’t fit.

By the end of this lesson, you’ll be able to work out exactly how tightly to wind this spiral so the whole side fits. Then you’ll set the lathe yourself, cut the record, and drop the needle on that last song. To get there, we’ll read this groove as a polar function. So let’s follow the cutter for a few turns and watch what its radius does.

Reading the groove

Here’s the record from its own point of view. In the room the disc spins under the cutter, but riding along with the disc, the cutter is the one going around, counterclockwise, and slowly closing in.

So let’s read it like a polar graph. The input theta is how far the disc has turned since the cut began, from this white line, and the output r is the cutter’s distance from the center, in millimeters. At the start theta is zero and r is a hundred forty-six. After one full turn, theta is two pi, and r is down to a hundred forty-five point eight eight.

Up close you can see why. Each new turn lands twelve hundredths of a millimeter inside the last, and theta keeps counting, so after two turns it’s four pi, and r is a hundred forty-five point seven six. So r is positive and getting smaller, and a positive radius that’s decreasing means the point is getting closer to the center.

In a table, every time theta goes up by two pi, one more turn, r goes down by the same point one two millimeters. Plot r against theta and the points sit on a straight line heading down, but wrap that same function around the center and the line becomes a spiral.

How steep is that line? The average rate of change of r with respect to theta is the change in r over the change in theta, so for one turn it’s negative point one two over two pi.

That’s about negative point zero one nine millimeters per radian. Every radian the disc turns, the cutter’s radius shrinks by about nineteen thousandths of a millimeter. The negative sign says r is decreasing, and the units say what it’s a rate of, millimeters of radius per radian of turning.

Now suppose the engineer tightens the spacing to zero point zero eight millimeters per turn. What’s the average rate of change of r with respect to theta now?

The music is the wiggle

Now let’s go much closer, because a real groove isn’t a smooth spiral at all. Under a microscope its walls wiggle in and out, and those wiggles are the music. Follow one wiggle with this probe. Here r is decreasing, so the groove is closing in on the center, until r bottoms out and starts to climb again. Right at that low point, the groove is closer to the center than anywhere around it, so it’s a relative minimum of r.

Now you keep going up the other side. Drag the probe along this wiggle. Find the spot that’s farthest from the center. Pull back, and all those tiny ups and downs vanish into the slow drift inward, which is all an average rate over a whole turn would ever see.

Your shadow has a negative radius

On a record, r is always positive, so it’s just the distance from the center. But the output of a polar function can be negative too, and there’s a negative radius you’ve seen on every sunny day of your life. This is the giant sundial at Quitsato, on the equator in Ecuador, and you might remember its orange column from our lesson on cotangent. Today is the June solstice, so the sun swings across the northern sky instead of passing overhead.

Let’s lay a polar grid on the platform, with the pole at the column and theta equals zero pointing east. For our input, theta is the direction of the sun. Now look at the shadow. It points away from the sun, so to reach its tip you face the sun and go backwards, which means its radius is negative, minus the length of the shadow.

At a quarter to nine the shadow is thirteen point one meters long, so r is negative thirteen point one, and its distance from the pole, this ring, is just the length, thirteen point one. Now let the morning run. The sun swings around toward the north, so theta increases, and r goes up, from negative thirteen to negative seven, to about negative five. But watch the ring, because it’s shrinking the whole time.

Now you take the sun. Keep the morning going. Then drag it to the moment when r is as large as it gets today.

Think about what that means. The largest value of r gives the point closest to the pole, because r is negative, and the closer it gets to zero, the shorter the shadow. After noon it runs the other way. r decreases, back down toward negative thirteen, and the shadow grows, so negative and decreasing means the point is moving farther away.

Here’s the morning as a table. As theta goes from pi over six to pi over two, r climbs from negative thirteen point one to negative four point three, while the distance falls. On a graph of r against theta, the shadow is this hump below the axis, and its distance is the mirror image above, so where r peaks, the distance dips.

So now we have all four cases. When r is positive, like on the record, increasing means farther and decreasing means closer. When r is negative it flips, so increasing means closer and decreasing means farther. And one idea sits behind all four. The distance from the pole is the absolute value of r, so all that matters is whether r is heading toward zero or away from it.

Four ways to move

Let’s try all four on one curve, r equals two plus four cosine theta, which starts at six, drops to zero at two pi over three, bottoms out at negative two at pi, and climbs back the same way. From zero to two pi over three, r is positive and decreasing, so the point closes in, all the way to the pole. Then r turns negative and keeps decreasing, heading away from zero, so the point moves away from the pole and starts drawing this little inner loop.

From pi to four pi over three, r climbs from negative two back up to zero, so it’s negative and increasing, and the point comes back in to the pole. And on the last stretch r is positive and increasing, so the point swings back out to six. On the exam you would justify each stretch in a sentence like this one: r is negative and decreasing, so the distance from the origin is increasing.

Where r switches between increasing and decreasing, the point is relatively closest or farthest. At pi, r hits its minimum, negative two, yet the point sits at the tip of the inner loop, two units from the pole and farther out than its neighbors. So the smallest r isn’t the closest point. The closest points are where r is zero and the curve passes through the pole, at two pi over three and four pi over three.

Here’s one for you, r equals one plus two sine theta, graphed both ways. On which interval is r increasing while the point gets closer to the pole? The exam also likes to ask where the points are. And a negative r puts the point across the pole. It lands on the opposite side from the direction theta.

This is exactly how the exam asks it. As theta increases from zero to pi over two, which statement is true about the points on the graph of r equals two cosine theta minus three?

Rate per radian, and estimates

Back to r equals two plus four cosine theta, to measure how fast r changes. Between pi over two and pi, r goes from two down to negative two. So the average rate of change is negative two minus two, over pi minus pi over two, which is negative four divided by pi over two, or negative eight over pi, about negative two point five five.

That means on average the signed radius drops about two and a half units per radian, the slope of this secant line on the graph of r against theta. But at both ends the point is two units from the pole, and in between it went right through the pole. The average can’t see that, because it only compares the two ends.

An average rate also lets you estimate values in between. Start at the radius you know, and add the rate times how far you’ve turned. At two pi over three, we’ve turned pi over six past pi over two, so that’s two plus negative eight over pi times pi over six, which is two minus four thirds, or two thirds.

But the true value is zero, because r doesn’t change at a steady rate here. On our record the cutter’s carriage moves in at a steady rate, so for the spiral’s path the method is exact.

Real lathes vary the spacing with the music, though, so from a real lathe’s log it only gives an estimate. Use these two readings to estimate r at theta equals five hundred pi.

What to remember

When r is positive, increasing means farther from the pole and decreasing means closer. When r is negative it’s the other way around, because the distance is the absolute value of r. Where r switches between increasing and decreasing, the point is relatively closest or farthest, and the smallest r isn’t always the closest point. And the average rate of change of r is the change in r over the change in theta, per radian. That lets you estimate r at an angle in between.

Cut the record

Back in the cutting room, the music has to fit between a radius of a hundred forty-six millimeters and the inner limit at sixty-six, so over the whole side r drops eighty millimeters. The record turns thirty-three and a third times a minute, so twenty-four minutes is eight hundred turns, and eight hundred turns is sixteen hundred pi radians.

So the rate we need is negative eighty over sixteen hundred pi, about negative point zero one five nine millimeters per radian, which works out to exactly a tenth of a millimeter per turn. The lathe is all yours now, so turn the feed knob until the side ends exactly at the inner limit.

Before we cut, let’s predict where song four begins. It starts twelve minutes in, which is four hundred turns, so theta equals eight hundred pi. So r should be a hundred forty-six plus negative point zero one five nine times eight hundred pi, which takes it forty millimeters in, to a hundred six millimeters. Let’s cut it. There’s the gap before song four, landing right on a hundred six, and the groove reaches the inner limit at exactly twenty-four minutes.

Every song fits, and with the turns this close together, the engineer cuts the side a little quieter, so the wiggles never crash into the next turn. Now let’s drop the needle on that last song. You just worked out how tightly to wind a spiral so a whole album fits. And it all came down to how fast the radius changes for every radian the record turns.