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Semi-circles and right angle dilemmas . . .

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Daily Mail reports on a class assignment for seven year olds that happened to be set for the daughter of a Mathematics Lecturer at Oxford.

Maths lecturer is left baffled by his seven-year-old daughter’s geometry homework and turns to Twitter for help – so can YOU work out if it’s true or false?

Dr Kit Yates shares his seven-year-old daughter’s maths homework to Twitter

The question asked students whether a semi-circle had ‘two right angles’ or not

The maths lecturer, from Oxford, admitted that he was stumped by the problem 

People were left baffled by the question and came up with conflicting answers 

By Kate Dennett For Mailonline

Published: 17:40 GMT, 25 February 2021 | Updated: 17:40 GMT, 25 February 2021

Here is the question and the “expected” seven year old School Math answer:

Why would a PhD level Maths Lecturer disagree? Indeed, why would he suggest — and this is where it gets interesting for us here at UD — “there was a strong case for claiming that the answer was ‘true’ as well as ‘false’“?

Sounds familiar?

Obviously, a Math Lecturer — I switch to the American style abbreviation — will be familiar with the world of the infinitesimal. So, he would be aware that at the point of intersection of diameter and circumference, there is a required right angle to the tangent and that at that point curve and tangent coincide . . . the tangent and the curve share the same slope at that point, but not at any identifiable real value on the circumference beyond — thus, distinct from — that point.

Arguably, for an infinitesimal spread about the point where the tangent touches the curve and the diameter cuts it, we have a straight line segment c_0 +/- dc in the circumference but at right angles to the diameter. That is, the hyperreals are peeking in and giving us a language to constructively talk about structures and quantities.

Let’s cross check, through Wikipedia:

In geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain or polygonal circuit. The solid plane region, the bounding circuit, or the two together, may be called a polygon.

The segments of a polygonal circuit are called its edges or sides, and the points where two edges meet are the polygon’s vertices (singular: vertex) or corners. The interior of a solid polygon is sometimes called its body. An n-gon is a polygon with n sides; for example, a triangle is a 3-gon.

Notice, the requirement, “finite number of straight line segments”? Other common sources will omit finiteness, and it is commonly recognised that a circle is a beyond finite case of a regular n-gon. So, we have reasonable grounds for referring to infinitesimal line segments and removing the finitude. For example, that is one way to approach the question, what is the area of a circle.

I should note that an infinitesimal value can be seen as a number say h near to 0 in the number line, closer to 0 than any 1/n, where n is a finite counting number. (This opens up also the concept of transfinite hyperreals. That is h = 1/H, a number such that H > n for any finite counting number.)

Shifting to Coordinate Geometry, let’s put our circle at the origin, O, and give it radius r:

With these ideas in hand, we can see that by using infinitesimals, a circle is a limiting case of a regular polygon, but the length of sides is too short to be captured by any distinct point on the circumference c_1 at (x1,y1) distinctly removed from C_0, at (x0,y0) such that x1 – x0 is a real value, and the same for y1 – y0. (That is, we here mark a distinction between the infinitesimally altered x0 + dx or the similarly altered y0 + dy and any neighbouring x1 and y1, constrained to be strictly real.)

This allows us to seemingly have our cake and eat it.

That is, no distinctly removed point c1 with strictly real coordinates different from c0 will lie on the tangent to the circle at c0 at the point where the diameter cuts the circle. But at the same time we can speak reasonably of points along a short line segment at c0, that is straight but in the circumference, as the points in question c0 + dc, are infinitesimally removed from c0. Where, dc^2 = dx^2 + dy^2.

That is, further, we are looking at an infinitesimal right angle triangle with the hypoteneuse dc being on the tangent line for the circle at c0.

If we can swallow this camel, then we have an answer, yes there is a right angle at the intersections of diameter with circumference, and yes the circle is bounded by a closed curve where any distinct c_1 at (x1,y1) that is such that x1 and y1 are strictly real, will fall away from the tangent at c_0.

Is such reasonable?

Well, let us see how we arrived at the differential coefficient dy/dx, as an expression of the slope of y = f(x) at some point on a curved line, through the slope of the tangent-limit to the slopes of the secants:

Here, we see how the slopes of secants, in the limit, become the slope of the tangent at M, as h tends to 0, i.e. becomes an infinitesimal.

Now, let us ponder the curvature of a continuous, differentiable line:

Here, curvature K = 1/R, R the radius of curvature. Observe that the tangent line is perpendicular to the curve and its osculating circle, which near P is coincident with the curve. All of this begs for interpretation on infinitesimals. BTW, for a straight line, the radius of curvature is infinite, yielding a curvature of 0.

With these in mind, we can see that no, this is not a superposition of two contradictory states but a case that is surprisingly relevant, regarding the value of the infinitesimal perspective.

Now, where does this all lead? Interesting places . . . END

Comments
VL, I had in mind the point of inflexion -- momentary straightness -- case, which is necessarily present when sense of curvature changes. What I learned way back specifically does not include that a tangent does not cross a curve; which a source implicitly used above seemed to say. KFkairosfocus
March 7, 2021
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Yes, KF, I discussed the point of inflection case. Also, yes the fact that the tangent line has a slope given by the derivative is the best mathematical way to describe a tangent line. This covers the endpoint case and the point of inflection case. You write, "An issue arises on a contrast with what I learned, which leaves off the does not cross: tangent at an inflexion point." I'm not sure what that means. Are you saying there is some other case where a tangent line crosses the curve (a "standard" continuous differentiable curve) other than at a point of inflexion? If so, can you describe the case?Viola Lee
March 7, 2021
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VL, that "tangent touches a curve at a single point (and does not cross)" is a case of a studied seemingly simple but deeply loaded remark that I think embeds a mistake on a "pathological" case. Any other real coordinates point on the curve is already falling away from the tangent. But, does such have to be on the same side? An issue arises on a contrast with what I learned, which leaves off the does not cross: tangent at an inflexion point. I suspect, touches at one point and shares the same slope there, may cover this case. Of course, on apolitical matters Wiki can be helpful:
In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve.[1] More precisely, a straight line is said to be a tangent of a curve y = f(x) at a point x = c if the line passes through the point (c, f(c)) on the curve and has slope f'(c), where f' is the derivative of f. A similar definition applies to space curves and curves in n-dimensional Euclidean space. As it passes through the point where the tangent line and the curve meet, called the point of tangency, the tangent line is "going in the same direction" as the curve, and is thus the best straight-line approximation to the curve at that point. Similarly, the tangent plane to a surface at a given point is the plane that "just touches" the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometry and has been extensively generalized; see Tangent space. The word "tangent" comes from the Latin tangere, "to touch".
The idea of infinitely close points of course is a peeking in of infinitesimals. (Onward, transfinite hyperreals: 1/h --> H, > any finite n in N.) KFkairosfocus
March 7, 2021
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KF writes, " I think teachers need to be more aware of the big issues next door to what we talk about question. It is hard to be simple, clear and building of core capabilities without occasionally going in over our heads. " This is important. That is why the remark in the picture that "There is a difference between the correct answer and the answer at this learning stage” is poor pedagogy.Viola Lee
March 7, 2021
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VL & SA2, yes, that's why I thought of the tiny ant metaphor. At the corner, it pivots through an angle, in the limit a right angle. We can argue, two branched function on a range specified by the corners shared in common. There are workarounds for the one result rule. Also, the value of limits and tamed infinitesimals becomes evident. I think teachers need to be more aware of the big issues next door to what we talk about question. It is hard to be simple, clear and building of core capabilities without occasionally going in over our heads. Things may not be as simple as we have been led to believe, e.g. I recall counting by tens and getting out of depth with my second dad then a next door neighbour: . . . nine-ty . . . nine-ty nine, ten-ty. I followed the rule not knowing the scale transition, this was the first time I counted to 100, and it was of an evening after school. I was corrected, one hundred. I immediately saw, counting by hundreds as extension of counting by tens. I did not go on to ten hundred that day, but lurking there is an infinity as well as the convention-riddled way we group numbers. I think in this case we are looking at, the ant metaphor might be helpful. Somebody tried to give a counter example to an angle and got in over the head where the great whites swim. KFkairosfocus
March 7, 2021
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Q, yes, all of these are feasible. The similar triangles was used to deduce things about distance to moon and sun way back. Shadow of earth on moon in an eclipse shows 3:1 ratio of diameters of shadow and object, mix in Eratosthenes' work and we have our first sol system scale yardsticks. A pity work on the Sun was marred by issues on refraction etc. My bet is, lowering on a string and similar triangles with shadows are the best methods. KFkairosfocus
March 7, 2021
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KairosFocus @70, Yes, exactly. You got two of the student's options. Other solutions he suggested included - Lowering the barometer on a string and then measuring the length of the barometer plus the string. - Swinging the string to determine the period of the pendulum (the difference in G being negligible). - Measuring the lengths of the shadow of the barometer and the building and applying the ratio to the height of the barometer. - Setting up a line of sight between the top of the barometer and the top of the building and measuring the ratios of the resulting similar triangles. And a bunch more like these! You're right that barometric pressure difference is probably the poorest method of trying to measure the height of the building, a fact that eluded the teacher. Can anyone here find any other methods that the student might have listed? -QQuerius
March 6, 2021
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I understand your questions, Steve, and will try to explain. I wish I could draw pictures, though. Here are several relevant points 1. We are not thinking of the semicircle and the diameter as a single curve, but rather as two separate curves. The diameter has angle of 0° in respect to the horizontal. The angle of a tangent line on the semicircle is different everyplace. At the bottom of the semicircle the tangent line is also horizontal. Image a point P starting at the bottom of the semicircle and moving to the right along the circle as it approaches the intersection with the diameter: it gets steeper all the time. At the exact moment P reaches the diameter the tangent line would be perfectly vertical, and thus perpendicular to the diameter. If we were thinking of the shape as a single function (which we are not), then the "point" where the diameter and semicircle meet would not have a single tangent line, because the value you get approaching that point along the diameter would be different than the value you get approaching along the semicircle. 2. Also, you offer this definition of a tangent line "“a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.” I see you got this from Encyclopedia.com. There are two ways this is incomplete. One is at the endpoint of a curve, which we are discussing here in respect to the semicircle. This is more obvious in respect to a smaller part of a circle. Imagine a point P at the 45° point on a circle, starting with 0° at the far right. It has a tangent at that point perpendicular to the radius at a 45° angle also. Now throw away all the circle except the part from 0° to 45°, so we just have an eighth of a circle. The tangent at 45° is still exactly what it was even though the circle doesn't continue on past that point. The second situation is what is called a point of inflection. Image the letter S. The letter curves one way at the top, and then back the other way at the bottom. Right where the curvature changes (the point of inflection) there is a tangent line that is"outside" and below the top part of the S but "outside" and above the bottom part. This is a tangent line that does cross the curve at that point. To really get a tangent line you have to use the process KF showed in the OP about turning a secant line into a tangent line using the process of a limit.Viola Lee
March 6, 2021
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Steve Alten2: The tangent is undefined at corners or places where more than one straight line can go through only that point on the 'curve'.JVL
March 6, 2021
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Viola Lee ” But the definition of the angle between two curves (and the line is a curve, mathematically speaking) is that it is the angle between the tangent lines.” I apologize for not quite getting this. I understand that for any point along an arc there can only be one tangent. But how would this apply to the shape displayed at the point where the arc changes to a straight line (the vertices)? Especially given the definition of a tangent. “ a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.” At the vertice, a tangent could be 90 degrees to the straight line, but it could also be greater than 90. What I am saying is that you can pivot the tangent around this point an additional 90 degrees without crossing any plane.Steve Alten2
March 6, 2021
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Q, I suspect, unless you have a skyscraper, the barometer will not be an effective index of difference in height. Most obvious, time how long it takes to hit ground or use it as weight on a string. KFkairosfocus
March 6, 2021
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VL, yes. I would use a glass standing in for a bucket and a lab tap, to give rates and accumulations of change as flows and stocks. Then, slopes and areas under curves, then fundamental theorem on mutually inverse operations. I would do a bell shaped pulse and show the sigmoid. That points onward to a lot of other things. KF PS: How that definition came to be settled is the root of the matter. Hint, Calculus in key part arose out of need to analyse curved motion of planets etc.kairosfocus
March 6, 2021
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Educators are sometimes guilty of discouraging students from thinking outside the box. There was a famous test question in the 1950s (IIRC) that a creative student answered in an out-of-the-box way, refusing to regurgitate the obvious. The question was 1. How could you measure the height of a building using a barometer? The student suggested that there is a large number of options available and provided a list. Can you see the student's point? Oh, and the teacher didn't know whether to give the student credit and so on advice, gave the student the *same* question--a second chance to conform to "science." The student then provided more options not provided previously, none of which were the desired, obvious answer. -QQuerius
March 6, 2021
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JVL, ah, i rounded from a rounding teach me to underestimate the old girl, Penny, reporting a 4-place value. KFkairosfocus
March 6, 2021
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Steve, you write, "The only angles possible would be those between the straight line and tangents of the arc at the point they meets the straight line. But that is not the question that was asked." But the definition of the angle between two curves (and the line is a curve, mathematically speaking) is that it is the angle between the tangent lines. Of course the kids wouldn't know that, but it is true that the semicircle and the diameter meet at a right angle.Viola Lee
March 6, 2021
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WJM, here is a thought experiment about this issue that I used in my very first introduction to the derivative in my "Intro to Calc" class. Imagine a falling ball. We know that gravity constantly accelerates a falling object, so at every moment the velocity of the ball is greater than it was the moment before. Suppose we want to measure the velocity of the ball at time t(0): call this desired number v(0). Also suppose we have extremely accurate measuring devices. We could measure the height at t(0) and the height at some other time a short while later, t(1). Then we could compute the average velocity between t(0) and t(1) by dividing the change in the height by the change in the time. Call this number v(1). Since the ball is falling faster and faster we know that v(1) is somewhat faster than v(0), so we have a least an estimate for v(0) that we know is a little too big. How could we get a better estimate? Obviously, we could measure the height at a different time t(2) that is closer to t(0). (That is, the change in time t(2) - t(0) is smaller.) This would give us a new, and better estimate, of the velocity v(0) because the moments in time were closer together. How could we do better? Pick moments in time closer and closer to t(0). However, we always have to have second point: we can't let the second point also be t(0), because then we have no change in height and no change in distance, and that doesn't produce a velocity. But we can let our second point get closer and closer to t(0), indefinitely. and what we find is that our computed average velocity gets closer and closer to a particular velocity. At some point we realize that the computed velocity approaches a limit, so we say that the limit of the computed velocity, as the time interval gets indefinitely small, is in fact the instantaneous velocity v(0) we were looking for. So even though we can't compute the instantaneous velocity directly because at a moment in time there is no change in height or change in time, we can still determine the instantaneous velocity by taking the limit formed by considering smaller and smaller intervals of time. In calculus language, the instantaneous rate of change is the limit, as change in time approaches 0, of the quotient of the change in height divided by change in time. (This is the formula in KPF's OP.) This works, and in this and countless other places both leads to a consistent mathematical system and agrees with the real world when properly applied. Obviously the ball is falling. It is not frozen in time just because we can't compute zero divided by zero. This method is the heart of calculus, without which our modern world of science and technology would not exist.Viola Lee
March 6, 2021
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Let’s get back to the question that was asked of the student. I agree with WJM that there are no angles in the shape displayed. The only angles possible would be those between the straight line and tangents of the arc at the point they meets the straight line. But that is not the question that was asked.Steve Alten2
March 6, 2021
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Viola Lee “ Steve, I assume you know you are talking about an entirely different problem?” Sorry, just my lame attempt at humour, playing on Kairosfocus’ frequently used cliches that you can’t square a circle.Steve Alten2
March 6, 2021
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Kairosfocus: About 1.773 Not the right rounding for sqrt(pi) which is approximately 1.772453850905516 . . .JVL
March 6, 2021
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SA2, yes you can use Calculus to determine the area of a circle. A = pi * r^2. Which then gives the side of a square of equivalent area, root-pi times r. About 1.773 times r. KFkairosfocus
March 6, 2021
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Steve, I assume you know you are talking about an entirely different problem?Viola Lee
March 6, 2021
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Some responses: First, let's be clear what we are talking about: the semicircle is one curve and the diameter is another (assuming that is what they are.) We are not talking about a semicircle in isolation nor are we talking about the picture as one single function. So, it is true that if we consider just a semicircle, there are no right angles, or angles of any sort, on the semicircle. And if we consider the shape as a single function, with the place where the diameter and the semicircle meet as P, the function is undifferentiable at P because the right and left hand derivatives are different. But if we ask clearly what angle does the semicircle make with the diameter, the answer is 90°. The reason is, and this is official calculus, the angle between two curves is the angle between their tangent lines at the point of intersection. The central issue that calculus solves is how to find the instantaneous rate of change (or as KF says, instantaneous velocity) at a moment when the rate of change is continuously changing. This is done via the concept of limit. WJM expresses the dilemma of pre-calculus understanding in 52 and JVL responds correctly at 56, "Calculus is the mathematics of change. One of the greatest intellectual achievements of the human race." We would have to have better software, and lots of time, to give WJM a beginning lesson, although KF summarized the approach in the OP in the section "From Slopes of a Secant to Slopes of a Curve."Viola Lee
March 6, 2021
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See, you can square a circle. :)Steve Alten2
March 6, 2021
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WJM, VL was right to point out that an object constrained to move in a circle, at each point of the arc is moving along the tangent. This connects to the issue, instantaneous velocity: at a given point, but in motion and under acceleration. In the case, release and our bung flies off along the tangent, then of course is influenced by the ballistics forces. This is how the old fashioned sling works. KFkairosfocus
March 6, 2021
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William J Murray: To put it simply: a circle has no straight line segments, so when it intersects any other line, curved or straight, there is no angle produced unless one is talking about something else. This is the point of calculus: what is a function doing at a single point (in time?) even if it's changing. Calculus is the mathematics of change. One of the greatest intellectual achievements of the human race.JVL
March 6, 2021
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JVL, no, there are two different slopes, approaching the corner from two different paths and pivoting at the corner. The pivoting is real, the angle of pivot is real and the angle at the corner has a legitimate meaning. All of this points to the infinitesimals and limits, thence R*. KFkairosfocus
March 6, 2021
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To put it simply: a circle has no straight line segments, so when it intersects any other line, curved or straight, there is no angle produced unless one is talking about something else.William J Murray
March 6, 2021
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William J Murray: the second being “right next to it” on the circle but off the dissecting line. There is no point right next to another one. Time to move on folks.JVL
March 6, 2021
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Maybe calculus covers this, but I never took calculus. To me, it seem obvious that the "angle" cannot be anything other than that created by an abstract straight line using two plot points on the circle, one being where it intersects with the dissecting line and the second being "right next to it" on the circle but off the dissecting line. The one "off" the dissecting line cannot be perpendicular to the line (nature of a circle), so there cannot be a "right angle" there. But further, you're not even talking about the line of the circle any more, you're talking about a straight line plotted from two points on the circle. Not only is there not a right angle there, there isn't any angle whatsoever. There is no "angle" produced by the circle dissected by the straight line. This goes back to logic and identities. You can't measure that angle because no measurable angle is produced; you have to be measuring the angle of something else. As it turns out, what you're measuring is the angle generated by two intersecting straight lines.William J Murray
March 6, 2021
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Kairosfocus: the shift in slope at the corners gives an angle as the ant metaphor will showk The slope approaching the corner from the two 'sides' gives two different answers. Therefore, the slope at the 'corner' is undefined. Basic calculus stuff. Time to move on.JVL
March 6, 2021
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