# Parabolas from String Art (Part 10)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

• Prove that string art from two line segments traces a parabola.
• Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
• Prove the reflective property of parabolas.
• Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. In this series of posts, I’d wanted to expand on the article with some pedagogical thoughts about connecting string art to parabolas for algebra students. After all, most mathematical studies of string art curves — formally known as “envelopes” — rely on differential equations or at least limits and calculus.

However, string art is simple enough for a young child to construct, and so this study was inspired by the quest of explaining this phenomenon using only simple mathematical tools.

The article linked above has further thoughts on this problem, including a calculus-free way of deriving the reflective property of parabolas. However, I think the article pretty much has all of my thoughts on this matter, and so I don’t think I need to elaborate upon them here.

This series of posts is dedicated to an inspired and inspiring Algebra I student who wanted to understand string art curves using tools that she could understand… even though she progressed much further into the mathematics curriculum by the time my article was published and this series of posts appeared on my blog.

# Parabolas from String Art (Part 9)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

• Prove that string art from two line segments traces a parabola.
• Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
• Prove the reflective property of parabolas.
• Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

We have shown in the last couple of posts that if the three points that generate the Our explorations of string art led us to consider an arbitrary string $\overline{PQ}$ depicted below. For brevity, this string will be called “string $s$,” matching the (possibly non-integer) $x$-coordinate of its left endpoint $P$. Since $P$ is $s$ units to the right of $A$, the right endpoint $Q$ must correspondingly be $s$ units to the right of $B$. Therefore, the $x$-coordinate of $Q$ is $s + 8$.

Previously, we established that the equation for string $s$ is

$y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$.

We also obtained a bonus result that we obtained using only algebra: string $s$ is tangent to the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$, which is traced by the strings, when $x=2s$. Of course, tangent lines are usually obtained using calculus, and so calculus should be able to confirm this result. The derivative of this function is

$y' = \displaystyle \frac{x}{8} - 1$,

so that the slope of the tangent line when $x=2s$ is $m = \displaystyle \frac{s}{4} - 1 = \frac{s-4}{4}$. We observe that this matches the slope of line segment $\overline{PQ}$ in the above picture:

slope $= \displaystyle \frac{s - (s-8)}{(s+8) - 8} = \frac{2s-8}{8} = \frac{s-4}{4}$.

Therefore, to show that $\overline{PQ}$ is the tangent line, it suffices to show that either $P$ or $Q$ is on the tangent line.

At $x = 2s$, the $y-$coordinate of where the tangent line intersects the curve is

$y = \displaystyle \frac{(2s)^2}{16} - 2s + 8 = \frac{s^2}{4} - 2s + 8$.

Using the point-slope formula for a line, the equation of the tangent line is thus

$y-y_1 = m(x-x_1)$

$y-\displaystyle \left( \frac{s^2}{4} - 2s + 8 \right) = \frac{s-4}{4} (x-2s)$

$y = \displaystyle \frac{s-4}{4} (x-2s) + \frac{s^2}{4} - 2s + 8$.

We now check to see if $P(s,8-s)$ is on the tangent line. Substituting $x =s$, we find

$y = \displaystyle \frac{s-4}{4} (s-2s) + \frac{s^2}{4} - 2s + 8$

$= \displaystyle \frac{s-4}{4} (-s) + \frac{s^2}{4} - 2s + 8$

$= \displaystyle \frac{(s-4)(-s) + s^2}{4} - 2s + 8$

$= \displaystyle \frac{-s^2+4s + s^2}{4} - 2s + 8$

$= \displaystyle \frac{4s}{4} - 2s + 8$

$= s - 2s + 8$

$= -s + 8$

Therefore, the point $(s,8-s)$ is on the tangent line, thus confirming that $P$ is on the tangent line and that $\overline{PQ}$ is the tangent line.

# Parabolas from String Art (Part 8)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

• Prove that string art from two line segments traces a parabola.
• Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
• Prove the reflective property of parabolas.
• Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

We have shown in the last couple of posts that if the three points that generate the Our explorations of string art led us to consider an arbitrary string $\overline{PQ}$ depicted below. For brevity, this string will be called “string $s$,” matching the (possibly non-integer) $x$-coordinate of its left endpoint $P$. Since $P$ is $s$ units to the right of $A$, the right endpoint $Q$ must correspondingly be $s$ units to the right of $B$. Therefore, the $x$-coordinate of $Q$ is $s + 8$.

Previously, we established that the equation for string $s$ is

$y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$.

Finding the curve traced by the strings is a two-step process:

• For a fixed value of $x$, find the value of $s$ that maximizes $y$.
• Find this optimal value of $y$.

Previously, we showed using only algebra that the optimal value of $s$ is $s = \displaystyle \frac{x}{2}$, corresponding to an optimal value of $y$ of $y = \displaystyle \frac{x^2}{16} - x + 8$.

For a student who knows calculus, the optimal value of $s$ can be found by instead solving the equation $\displaystyle \frac{dy}{ds} = 0$ (or, more accurately, $\displaystyle \frac{\partial y}{\partial s} = 0$):

$\displaystyle \frac{dy}{ds} = -\frac{2s}{4} + \frac{x}{4}$

$0 = \displaystyle \frac{-2s+x}{4}$

$0 = -2s + x$

$2s = x$

$s = \displaystyle \frac{x}{2}$,

matching the result that we found by using only algebra.

# Square roots and logarithms without a calculator (Part 12)

I recently came across the following computational trick: to estimate $\sqrt{b}$, use

$\sqrt{b} \approx \displaystyle \frac{b+a}{2\sqrt{a}}$,

where $a$ is the closest perfect square to $b$. For example,

$\sqrt{26} \approx \displaystyle \frac{26+25}{2\sqrt{25}} = 5.1$.

I had not seen this trick before — at least stated in these terms — and I’m definitely not a fan of computational tricks without an explanation. In this case, the approximation is a straightforward consequence of a technique we teach in calculus. If $f(x) = (1+x)^n$, then $f'(x) = n (1+x)^{n-1}$, so that $f'(0) = n$. Since $f(0) = 1$, the equation of the tangent line to $f(x)$ at $x = 0$ is

$L(x) = f(0) + f'(0) \cdot (x-0) = 1 + nx$.

The key observation is that, for $x \approx 0$, the graph of $L(x)$ will be very close indeed to the graph of $f(x)$. In Calculus I, this is sometimes called the linearization of $f$ at $x =a$. In Calculus II, we observe that these are the first two terms in the Taylor series expansion of $f$ about $x = a$.

For the problem at hand, if $n = 1/2$, then

$\sqrt{1+x} \approx 1 + \displaystyle \frac{x}{2}$

if $x$ is close to zero. Therefore, if $a$ is a perfect square close to $b$ so that the relative difference $(b-a)/a$ is small, then

$\sqrt{b} = \sqrt{a + b - a}$

$= \sqrt{a} \sqrt{1 + \displaystyle \frac{b-a}{a}}$

$\approx \sqrt{a} \displaystyle \left(1 + \frac{b-a}{2a} \right)$

$= \sqrt{a} \displaystyle \left( \frac{2a + b-a}{2a} \right)$

$= \sqrt{a} \displaystyle \left( \frac{b+a}{2a} \right)$

$= \displaystyle \frac{b+a}{2\sqrt{a}}$.

One more thought: All of the above might be a bit much to swallow for a talented but young student who has not yet learned calculus. So here’s another heuristic explanation that does not require calculus: if $a \approx b$, then the geometric mean $\sqrt{ab}$ will be approximately equal to the arithmetic mean $(a+b)/2$. That is,

$\sqrt{ab} \approx \displaystyle \frac{a+b}{2}$,

so that

$\sqrt{b} \approx \displaystyle \frac{a+b}{2\sqrt{a}}$.

# Square roots and logarithms without a calculator (Part 10)

This is the fifth in a series of posts about calculating roots without a calculator, with special consideration to how these tales can engage students more deeply with the secondary mathematics curriculum. As most students today have a hard time believing that square roots can be computed without a calculator, hopefully giving them some appreciation for their elders.

Today’s story takes us back to a time before the advent of cheap pocket calculators: 1949.

The following story comes from the chapter “Lucky Numbers” of Surely You’re Joking, Mr. Feynman!, a collection of tales by the late Nobel Prize winning physicist, Richard P. Feynman. Feynman was arguably the greatest American-born physicist — the subject of the excellent biography Genius: The Life and Science of Richard Feynman — and he had a tendency to one-up anyone who tried to one-up him. (He was also a serial philanderer, but that’s another story.) Here’s a story involving how, in the summer of 1949, he calculated $\sqrt[3]{1729.03}$ without a calculator.

The first time I was in Brazil I was eating a noon meal at I don’t know what time — I was always in the restaurants at the wrong time — and I was the only customer in the place. I was eating rice with steak (which I loved), and there were about four waiters standing around.

A Japanese man came into the restaurant. I had seen him before, wandering around; he was trying to sell abacuses. (Note: At the time of this story, before the advent of pocket calculators, the abacus was arguably the world’s most powerful hand-held computational device.) He started to talk to the waiters, and challenged them: He said he could add numbers faster than any of them could do.

The waiters didn’t want to lose face, so they said, “Yeah, yeah. Why don’t you go over and challenge the customer over there?”

The man came over. I protested, “But I don’t speak Portuguese well!”

The waiters laughed. “The numbers are easy,” they said.

They brought me a paper and pencil.

The man asked a waiter to call out some numbers to add. He beat me hollow, because while I was writing the numbers down, he was already adding them as he went along.

I suggested that the waiter write down two identical lists of numbers and hand them to us at the same time. It didn’t make much difference. He still beat me by quite a bit.

However, the man got a little bit excited: he wanted to prove himself some more. “Multiplição!” he said.

Somebody wrote down a problem. He beat me again, but not by much, because I’m pretty good at products.

The man then made a mistake: he proposed we go on to division. What he didn’t realize was, the harder the problem, the better chance I had.

We both did a long division problem. It was a tie.

This bothered the hell out of the Japanese man, because he was apparently well trained on the abacus, and here he was almost beaten by this customer in a restaurant.

Raios cubicos!” he says with a vengeance. Cube roots! He wants to do cube roots by arithmetic. It’s hard to find a more difficult fundamental problem in arithmetic. It must have been his topnotch exercise in abacus-land.

He writes down a number on some paper— any old number— and I still remember it: $1729.03$. He starts working on it, mumbling and grumbling: “Mmmmmmagmmmmbrrr”— he’s working like a demon! He’s poring away, doing this cube root.

Meanwhile I’m just sitting there.

One of the waiters says, “What are you doing?”.

I point to my head. “Thinking!” I say. I write down $12$ on the paper. After a little while I’ve got $12.002$.

The man with the abacus wipes the sweat off his forehead: “Twelve!” he says.

“Oh, no!” I say. “More digits! More digits!” I know that in taking a cube root by arithmetic, each new digit is even more work that the one before. It’s a hard job.

He buries himself again, grunting “Rrrrgrrrrmmmmmm …,” while I add on two more digits. He finally lifts his head to say, “$12.0$!”

The waiter are all excited and happy. They tell the man, “Look! He does it only by thinking, and you need an abacus! He’s got more digits!”

He was completely washed out, and left, humiliated. The waiters congratulated each other.

How did the customer beat the abacus?

The number was $1729.03$. I happened to know that a cubic foot contains $1728$ cubic inches, so the answer is a tiny bit more than $12$. The excess, $1.03$, is only one part in nearly $2000$, and I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. So all I had to do is find the fraction $1/1728$, and multiply by $4$ (divide by $3$ and multiply by $12$). So I was able to pull out a whole lot of digits that way.

A few weeks later, the man came into the cocktail lounge of the hotel I was staying at. He recognized me and came over. “Tell me,” he said, “how were you able to do that cube-root problem so fast?”

I started to explain that it was an approximate method, and had to do with the percentage of error. “Suppose you had given me $28$. Now the cube root of $27$ is $3$ …”

He picks up his abacus: zzzzzzzzzzzzzzz— “Oh yes,” he says.

I realized something: he doesn’t know numbers. With the abacus, you don’t have to memorize a lot of arithmetic combinations; all you have to do is to learn to push the little beads up and down. You don’t have to memorize 9+7=16; you just know that when you add 9, you push a ten’s bead up and pull a one’s bead down. So we’re slower at basic arithmetic, but we know numbers.

Furthermore, the whole idea of an approximate method was beyond him, even though a cubic root often cannot be computed exactly by any method. So I never could teach him how I did cube roots or explain how lucky I was that he happened to choose $1729.03$.

The key part of the story, “for small fractions, the cube root’s excess is one-third of the number’s excess,” deserves some elaboration, especially since this computational trick isn’t often taught in those terms anymore. If $f(x) = (1+x)^n$, then $f'(x) = n (1+x)^{n-1}$, so that $f'(0) = n$. Since $f(0) = 1$, the equation of the tangent line to $f(x)$ at $x = 0$ is

$L(x) = f(0) + f'(0) \cdot (x-0) = 1 + nx$.

The key observation is that, for $x \approx 0$, the graph of $L(x)$ will be very close indeed to the graph of $f(x)$. In Calculus I, this is sometimes called the linearization of $f$ at $x =a$. In Calculus II, we observe that these are the first two terms in the Taylor series expansion of $f$ about $x = a$.

For Feynman’s problem, $n =\frac{1}{3}$, so that $\sqrt[3]{1+x} \approx 1 + \frac{1}{3} x$ if $x \approx 0$. Then $\latex \sqrt[3]{1729.03}$ can be rewritten as

$\sqrt[3]{1729.03} = \sqrt[3]{1728} \sqrt[3]{ \displaystyle \frac{1729.03}{1728} }$

$\sqrt[3]{1729.03} = 12 \sqrt[3]{\displaystyle 1 + \frac{1.03}{1728}}$

$\sqrt[3]{1729.03} \approx 12 \left( 1 + \displaystyle \frac{1}{3} \times \frac{1.03}{1728} \right)$

$\sqrt[3]{1729.03} \approx 12 + 4 \times \displaystyle \frac{1.03}{1728}$

This last equation explains the line “all I had to do is find the fraction $1/1728$, and multiply by $4$.” With enough patience, the first few digits of the correction can be mentally computed since

$\displaystyle \frac{1.03}{500} < \frac{1.03}{432} = 4 \times \frac{1.03}{1728} < \frac{1.03}{400}$

$\displaystyle \frac{1.03 \times 2}{1000} < 4 \times \frac{1.03}{1728} < \frac{1.03 \times 25}{10000}$

$0.00206 < 4 \times \displaystyle \frac{1.03}{1728} < 0.0025075$

So Feynman could determine quickly that the answer was $12.002\hbox{something}$.

By the way,

$\sqrt[3]{1729.03} \approx 12.00238378\dots$

$\hbox{Estimate} \approx 12.00238426\dots$

So the linearization provides an estimate accurate to eight significant digits. Additional digits could be obtained by using the next term in the Taylor series.

I have a similar story to tell. Back in 1996 or 1997, when I first moved to Texas and was making new friends, I quickly discovered that one way to get odd facial expressions out of strangers was by mentioning that I was a math professor. Occasionally, however, someone would test me to see if I really was a math professor. One guy (who is now a good friend; later, we played in the infield together on our church-league softball team) asked me to figure out $\sqrt{97}$ without a calculator — before someone could walk to the next room and return with the calculator. After two seconds of panic, I realized that I was really lucky that he happened to pick a number close to $100$. Using the same logic as above,

$\sqrt{97} = \sqrt{100} \sqrt{1 - 0.03} \approx 10 \left(1 - \displaystyle \frac{0.03}{2}\right) = 9.85$.

Knowing that this came from a linearization and that the tangent line to $y = \sqrt{1+x}$ lies above the curve, I knew that this estimate was too high. But I didn’t have time to work out a correction (besides, I couldn’t remember the full Taylor series off the top of my head), so I answered/guessed $9.849$, hoping that I did the arithmetic correctly. You can imagine the amazement when someone punched into the calculator to get $9.84886\dots$