1 Arc Length and Surface Area Calculus Techniques Meet History David W. Stephens The Bryn Mawr School Baltimore, MD NCTM – Baltimore 2004 15 October 2004.

1

Arc Length and Surface AreaCalculus Techniques Meet History

David W. StephensThe Bryn Mawr School

Baltimore, MD NCTM – Baltimore 2004

15 October 2004

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Contact Information

Email:[email protected]

The post office mailing address is:David W. Stephens109 W. Melrose AvenueBaltimore, MD 21210410-323-8800

The PowerPoint slides will be available on my school website: http://207.239.98.140/UpperSchool/math/stephensd/StephensFirstPage.htm

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Why is Arclength a Fascinating Topic?

This is a late topic in BC Calculus.The seniors are getting near the end of their

high school years … and the AP exam is on the doorstep.

Calculus is a great capstone course in high school, because it brings together all of the mathematics that the students have previously learned.

4

How is Arclength a Fascinating Topic?

Calculus students already know about arclength on a circle from their geometry class

They understand radians (although perhaps they still struggle with the importance of radians) … and radians are crucial for calculus.

It is valuable to tie in new methods to ones they already know. Calculus topics often lend themselves to doing this.

5

Calculus Strategies: Integration

The definite integral is an accumulation of products … that is the sum of products of two quantities, so definite integrals can be thought of as measurements of areas.

4

2

-5 5

f x = 0.125-x2

6


In any application of integration (such as areas under a curve, volumes, arclength, work, distances, or total costs), there is a three step strategy: Cut the <area, volume, arclength, work, etc> into small

pieces. Code the quantity to be measured on a representative

small piece, because we understand the geometry of the small parts.

Recombine the parts (with sums / definite integrals).

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Step 1 (Cut the desired result into small pieces.)

2

-2

-5 5

f x = 0.125-x2

4

2

-5 5

f x = 0.125-x2

8


Step 2 (Code the quantity to be measured on a representative small piece) It looks like this:

dA = y dx

The width (x) is cut into infinitesimally small parts, and the height (y) depends on the function under which the area is to be measured.

4

2

-5 5

f x = 0.125-x2

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Step 3 (Recombine the parts with sums / definite integrals ) dA = y dx

Adding up all of these simpler parts becomes

b

a

ydxA 5

0

)225( dxxA

4

2

-5 5

f x = 0.125-x2

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A Whirlwind Histo-Mathematical TourHow Do We Calculate Length?

300 BC Euclidean GeometryEuclid (325 – 265 BC, probably at Alexandria, Egypt)

The subject of plane geometry was known as far back as 2000 BC – 2500 BC. Perhaps the Chinese and

other Asian cultures knew this information independently at about this same time as well.

Distance is measured with a straightedge.History of mathematics available at:

http://www-gap.dcs.st-and.ac.uk/~history/BiogIndex.html

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1629 to 1640’s Cartesian coordinates

Rene Descartes (France 1596 – 1650)

Points were located with numbers, “marrying” geometry and algebra. Fermat knew these

results in about 1629 as well.2)21(2)21( yyxxlength

Length is now calculated, rather than measured.

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1660-1670 Integral CalculusIsaac Newton (1643 – 1727, England) and Gottfried Leibniz (1646 – 1716, Germany)

Ideas of cutting a length into small pieces and measuring the small pieces with plane geometry methods and then recombining the pieces was a

new strategy.

(Details to be shown later.)

2)(2)( dydxds

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1920 – 1945 Measurement of Coastlines

Lewis F. Richardson (England 1881 - 1953)

Richardson investigated to find out that the reported length of coastlines in Europe (and he is known

especially for a discussion of the coastline of England) varied by as much as 20%.

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1975 Fractals

Benoit Mandelbrot (Poland 1924 - )(His family was Lithuanian Jewish. He now resides in the USA.)

Methods were developed to look at the similarity of small pieces of a line or surface to the whole line or surface.

Measurements (and the accumulation of parts of the measurements) seemed to depend on the scale of the

measurement tool.

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300 AD Theorems of PappusPappus ( 290-350 AD, Alexandria, Egypt)

Pappus stated two useful theorems, long before the methods of calculus were in existence, which help to calculate volume and surface area. In uncanny ways, these ancient theorems

are verified by the much newer methods of the integral calculus and the fractals.

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1980’s Gaussian QuadratureTexas Instruments Calculator Algorithm

The method for performing the numerical integration fnInt is a fast, usually accurate, but

complicated and fascinating algorithm.

(This is a method used for any integration, not just for calculating length, but it has a connection to the other methods.)

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Arclength Meets History

Here is how a class might proceed, building up the ideas for calculus in a historical-mathematical way.

This discussion will proceed as if all of you are not actually familiar with the calculus topic of arclength.

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What is an “arc”?

arc – Middle English word derived from Latin arcus meaning bow, as in bow-and-arrow, and, later, arch or curve. In his 1551 Pathway to Knowledge, Recorde used arche, arche lyne (also spelled archline), and bowe lyne (also spelled bowline) for the arc of a circle. Billingsley uses the word arke in his 1570 translation of Euclid’s Elements.

This is from Historical Modules for the teaching and Learning of Secondary Mathematics (December 2002, Mathematical Association of America). This definition comes from “Lengths, Areas and Volumes” (page 193).

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What do little pieces of most functions look like?

Most functions have a curve to them, so the question of the length of an arc amounts to calculating the length of a piece of a function.

Calculus students have been well trained to say that little pieces of most functions look like ….

line segments, because functions are usually locally linear.

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Setting up Arclength

So calculating the length of a curve comes down to methods to measure the length of a line segment.Cut the <area, volume, arclength, work, etc> into

small pieces.Code the quantity to be measured on a

representative small piece, because we understand the geometry of the small parts.

Recombine the parts (with sums / definite integrals).

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Setting up Arclength :Now we follow the history …

Pythagoras (569 – 475 BC, Samos, Ionia) No coordinates available

22 bac a

b

c

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Setting up Arclength:Use Pythagoras in a calculus class

We want to know the length of y = x2 on the interval [0 , 4].

We do this in four pieces to begin.

15

10

5

-20 -10 10 20

f x = x2

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Setting up Arclength: Use Pythagoras in a calculus class

1

1

1

3

1

5

1

7

15

10

5

-20 -10 10 20

f x = x2

Here are the four triangles whose hypotenuses are straight, under the assumption that curves are locally linear.

2 10 26 50s

So s = about 16.747

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Setting up Arclength: Use Pythagoras in a calculus class

Notice that we have (1) cut the curve into small pieces even though

Pythagoras would not have understood the idea of a function with coordinates,

(2) used the geometry of Pythagoras to calculate the lengths of the four pieces, and

(3) recombined with addition. No calculus was used, but the ideas of calculus were employed.

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Setting up Arclength:Add Descartes to the question

For each of the triangles, coordinates are used to locate the points on the function, and the distance formula is that of Pythagoras with adaptations for the coordinates.

2)21(2)21( yyxxlength

15

10

5

-20 -10 10 20

f x = x2

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Setting up Arclength: Add Descartes to the question

The coordinates of the points are

(4, 16) ,

(3 , 9) ,

(2 , 4),

( 1, 1),

and (0, 0)

15

10

5

-20 -10 10 20

f x = x2

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Setting up Arclength: Add Descartes to the question

Using the distance formula on each of the triangles gives the same results as before.

1

1

1

3

1

5

1

7

2 10 26 50s

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Setting up Arclength: A Detour to the 20th Century

Calculus students accept the idea of local linearity fairly easily, even though it is a novel idea at first.

To challenge their acceptance of this idea (and recall it is late in the senior year at the end of a long and challenging AP course), let’s move to Richardson and Mandelbrot … and the coastline of England (and other places).

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Length of a Coastline

From Chaos by James Gleick (Penguin Books 1987, page 95)

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Some of the stories told about the measurement of coastlines include the importance of knowing the length of the coastlines of England and Norway during World War II, so that the navies knew how long a coastline they needed to defend.

Later it became a fascinating mathematical topic.

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We can actually do some measurements now to see how this paradox of Lewis Richardson goes.We will simulate this with the maps of

Jaggedland and SmootherlandWe can measure with different “smallest” units

available.

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Use a 3 inch straightedge. Start at some point on the map.Swing the 3 inch straightedge until it first hits

another point on the map.Move the end of the 3 inch straightedge until it

is at the last endpointCount how many 3 inch measurements you can

make, continuing until you are back at the starting point.

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Use a 1 inch straightedge.Do the same process as above.

Use a ½ inch straightedge.Do the same process as above.

Use the scale on the map to convert the total number of inches to miles.

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# of # of # of3 inch 1 inch 1/2 inchparts Miles parts Miles parts Miles

FloridaEnglandNorwayChesapeake BayMississippi River

Observations:

A Student Worksheet

Use actual maps of Florida, Norway, England, the Chesapeake Bay, and the Mississippi River in classes.

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Actual mileages …whatever “actual” means (since we are now skeptical about whether there is a real answer ????)

Florida ….1,350 miles England … 5,581miles (6261 including islands)

(11,072 miles for Great Britain, 19491 including islands)

Norway … Chesapeake Bay … 11,864 miles of shoreline Mississippi River … 2,350 to 2,552 miles

(depending on who you ask)

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What seems to be the results and connections?As the measuring tool gets shorter, the total

length gets longer, but not always!What measurement tool does a geological survey

use? Why?Actual length seems to be the result of practical

methods, but they are not definite answers.

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Small pieces on the maps are measured as the Greeks would have done it (!!), and the Pythagorean theorem could have been used to calculate from the vertical and horizontal.

Old meets new.Mathematics is still evolving and new

methods and ideas are still being added.It is OKAY to combine new and old ideas!

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Coastline Paradox Determining the length of a country's coastline is not as simple as it first

appears, as first considered by L. F. Richardson (1881-1953). In fact, the answer depends on the length of the ruler you use for the measurements. A shorter ruler measures more of the sinuosity of bays and inlets than a larger one, so the estimated length continues to increase as the ruler length decreases.

In fact, a coastline is an example of a fractal, and plotting the length of the ruler versus the measured length of the coastline on a log-log plot gives a straight line, the slope of which is the fractal dimension of the coastline (and will be a number between 1 and 2).

from http://mathworld.wolfram.htm

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How Long is the Coast of Great Britain?

Figure 1: The coastline of Great Britain In 1967, Benoit Mandelbrot published [7] ``How Long is the Coastline of Great Britain'' in Nature. In it, he posed the simple question of how one measures the length of a coastline. As with any curve, the obvious answer for the mathematician is to approximate the curve with a polygonal path, each side of which is of length є . (See Figure 2.)

Then by evaluating the length of these polygonal paths as є0 , we expect to see the length estimate approach a limit. Unfortunately, it appears that for coastlines, as є0 , the approximated length L(є) infinity as well.

Figure 2: Approximating the coastline of Great Britain

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In a later book, [10, pp 28-33,] Mandelbrot discusses the extensive experimental work on this problem which was done by Lewis Fry Richardson. Richardson discovered that for any given coastline, there were constants F and D such that to approximate the coastline with a polygonal path, one requires roughly Fe-D intervals of length . Thus, the length estimate can be given as L(є) = Fe 1-D

The reason has to do with the inherent ``roughness'' of a coastline. In general, a coastline is not the type of curve we are usually used to seeing in mathematics. Although it is a continuous curve, it is not smooth at any point. In fact, at any resolution, more inlets and peninsulas are visible that were not visible before. (See Figure 3.) Thus as we look at finer and finer resolutions, we reveal more and more lengths to be approximated, and our total estimate of length appears to increase without bound.

http://www.math.vt.edu/people/hoggard/FracGeomReport/node2.html

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Contrast this idea with the foundations of calculus which assert that a limit is attained when we cut the length into smaller and smaller pieces.

We make the assumption …and conclusion… that there is a finite length and that our methods of the integral calculus will help calculate that length.

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What is the length of the coastline of Britain? Benoit Mandelbrot proposed this question to demonstrate the complexity of measurement and scale. There are a number of almanacs that provide this information. However, if one examines the measuring techniques used to determine the length of Britain's coastline, it becomes obvious that this measurement is only an estimate based on the accuracy of the measuring device. Smaller units mean greater accuracy. But we can continue that line of thinking indefinitely, just as we do with fractions. There are always smaller fractions, an infinite number. Therefore, the coastline of Britain is an infinite length, however, it is confined within a finite space. We can begin to understand then that perimeter can have an infinite length confined within a finite area. http://home.inreach.com/kfarrell/measure.html

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How Long Is Australia's Coastline? (an explanation) At first blush the question seems eminently reasonable, but it is as open-ended

as the classical "how long is a piece of string?" The answer to both is the same: it all depends.

Dr Robert Galloway of the CSIRO Division of Land Use Research in Canberra was recently confronted with the question when compiling an inventory of Australia's coastal lands. Looking up the published figures he found the following answers:

The great disparity has to do with the precision with which the measurement is made. The larger and more detailed the map, and the more finely the measurement is made, the longer will be the coastline. Ultimately one could walk around the coast itself with a measuring stick, but the answer still depends on whether you use seven-league boots or a metre rule.

(It's a philosophical point whether the coastline tends to any limit as precision improves. Some say it does, others not.)

To settle on a reliable, repeatable figure, Dr. Galloway got together 162 maps covering the Australian coast and enlisted the help of Ms Margo Bahr of the Division.

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A few points of methodology had to be agreed on before the exercise could begin:

How far up estuaries should the coastline be taken? It was decided that all inlets would be arbitrarily (but consistently) cut off whenever their mapped width was less than 1 km. Within Sydney Harbour, for example, Kirribilli Point was joined to Garden Island. Straits less than 1 km wide were ignored, treating the island as though it were part of the mainland.

Islands less than 12 ha. were ignored. Measuring the coastline of the 2600 islands larger than that would be tedious in the extreme. Instead, a 16% sample was taken and a graph of coast length against area drawn.

This plot gave a good correlation, allowing island coastlines to be derived simply from their area. However, the ten largest islands (including Tasmania) were, for accuracy, measured directly. (Macquarie Island and Lord Howe Island were not included.)

Mangroves were regarded as part of the land, with the coastline following their seaward fringe; channels between mangroves were treated as estuaries. All coral reefs were excluded.

http://www.maths.mq.edu.au/numeracy/tutorial/cts2.htm

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Finally came the question, which tools to use: a pair of dividers, a map measuring wheel, or a length of string or fine wire? On a test run, dividers gave consistent results only if the same starting point was used; the wheel was rapid but inaccurate. Fine wire (not string) laid on the drawn coastline proved surprisingly consistent and accurate (as good as dividers set to a 0.7 km interval) and so was chosen for the task. Down to work!

When the 162nd map was put aside, the total length of the mainland coast plus Tasmania worked out to be 30 270 km. Adding on the length of the coast of all the islands greater than 12 ha, about 16 800 km, gave a grand total of 47 070 km.

As a matter of interest and undeterred by their prior efforts, the two workers examined the effect of different divider lengths on the measured coastline. As expected, the apparent length of the coast of the mainland diminished steadily as the divider length was increased; shrinking to 10 830 km at a 1000-km intercept.

A simple formula was derived that linked coast length to the measuring intercept. Using this formula to extrapolate a divider length of just 1 mm, gave a length of about 132 000 km for the mainland of Australian rather more than three times the circumference of the earth!

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How Long Is Australia's Coastline?The correct answer is .... it depends! It depends on which source you read, apparently.

Source LengthYear Book of Australia (1978) 36 735 kmAustralian Encyclopedia 19 658 kmAustralian Handbook 19 320 km

So who is correct? The answer is: all of them! Each source used a ruler with different sized increments on it. If you measure the coastline with a ruler that is just 1 mm long, you would get a length of 132 000 km!

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4.2 Calculating coastline and population [in New Zealand for Maori tribes] 4.2.1 Coastline calculations Under the proposed allocation method, inshore fishstocks and 60% of deepwater

fishstocks would be allocated according to the length of an Iwi’s coastline. Exactly how would coastline lengths be worked out?

It is proposed that a 1:50,000 scale map of New Zealand would be used. Iwi would have to reach agreement with neighbouring Iwi as to their respective coastline lengths. The exact coastline length for a quota management area would then be calculated as follows:

rivers would be cut off at the coast and the distance across the river mouth included in the coastline measurement;

the coastline length of harbours and bays whose natural entrance points are greater than 10 km apart would be included in the coastline measurement;

the juridical bay formula (see below) would be applied to harbours and bays whose natural entrance points are less than 10 km apart in order to determine whether those harbours and bays would be included in the coastline measurement; and

with the exception of the islands in the Chatham Islands group, coastline measurements would not include the coastline of islands claimed by Iwi to be part of their traditional takiwa.

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The juridical bay formula The juridical bay formula is applied to bays where the natural entrance

points are less than 10 km apart in order to determine whether the distance across the entrance of the bay or the actual coastline of the bay should be added to the coastline measurement. The formula works as follows (see Figure 3):

a straight line is drawn between the natural entrance points of the bay; a semicircle is drawn on the straight line (using the straight line as the

diameter of the circle) and the surface area of the semicircle is calculated; the surface area of the bay enclosed by the straight line is also calculated

using map information software; if the surface area of the semicircle is smaller than the water surface area of

the bay, then the distance between the natural entrance points is included in the coastline measurement, (see figure 3a, the shoreline of the bay is not included) ; or

if the surface area of the semicircle is bigger than the water surface area of the bay, then the shoreline of the bay is measured and included in the coastline measurement.( see figure 3b)

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A drawing of the New Zealand juridical bay formula

http://www.tokm.co.nz/allocation/1997/implementing.htm

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The Florida Shoreline and its Measurement

The FLDEP is responsible for monitoring and managing approximately 680 miles of Florida’s coastline.

This includes the state’s entire coastline except for Monroe County (Florida Keys) and Federal sites. These management efforts are the result of two contributing factors. In addition, nature constantly changes the shoreline through normal coastal processes and occasional storm events. Other shoreline changes result from man’s engineering activities associated with ports and harbors and shoreline stabilization. The FLDEP must identify and quantify these changes and manage the coastline to preserve Florida’s most important natural resource – its beaches.

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Traditional Method to Measure the Florida Shoreline

Measurements are taken at fifty feet intervals and at points of slope change along these cross sections. The nearshore survey extends from the waterline to either the 30 ft contour line or 2,400 ft from the shoreline, whichever is closer. Survey monuments on the baseline are maintained so that subsequent surveys may be taken in the same locations providing a common reference point to allow comparisons. Because of the time required to collect these data, only three to four counties’ shorelines are surveyed each year using this technique. Aerial photography is taken in conjunction with the surveys to provide visual record and supplement the coastline management process.

http://64.233.161.104/search?q=cache:77CGD4GeIfgJ:www.thsoa.org/pdf/hy99/9_3.pdf+coastline+measurement+Florida&hl=en

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Advancements in survey technology have provided new tools for gathering data to manage the coastline. Airborne lidar is one such advancements. Lidar is an acronym for LIght Detection And Ranging. Lidar works similar to radar, but a laser is used instead of radio waves for distance measurements. Each laser pulse is transmitted from the airborne platform to the surface below. Some of the light energy is reflected from the water surface and detected by onboard optical sensors. The remaining energy continues through the water column, is reflected from the bottom and is detected by the onboard sensors. The time difference between the two energy returns indicates the water depth.

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The traditional methods would require about 2 years time to complete a measurement of the Florida coastline, where the new methods require about 1 month.

though the use of airborne lidar the areas were surveyed in approximately one month producing data on an 8m by 8m grid spacing as opposed to the historic 1,000 ft cross sections. The speed and cost effectiveness of SHOALS will enhance FLDEP’s ability to “resurvey more beach areas on a more frequent basis”

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The sophistication of methods, as well as the extraordinary effort that is expended to measure coastlines, indicates to students that (1) the process is important, (2) the assumptions about how to accomplish

the measurement are heeded, and (3) mathematical methods and practical

considerations are both part of the process.

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Arclength using calculus

Now we get to the calculus methods to measure arclength.

Step 1: Cut the length of the curve into small pieces. (“Small” is undefined) We get little triangles, and we

still believe that the curves are usually locally linear, so that the hypotenuse is very close to linear.

dx

dy

ds

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Step 2: Code the quantity to be measured on a representative small piece, because we understand the geometry of the small parts

2)(2)( dydxds

dx

dy

ds

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Step 3: Recombine the parts (with sums / definite integrals)

2)(2)( dydxds

b

a

dydxs 2)(2)(

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This is not quite in the form that we prefer, since we need a dx or dy outside the square root. So …

b

a

dydxs 2)(2)(

b

a

dxdx

dydx

dx

dy

dx

dxb

a

s 2)(12)(*2)(2)(

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Similar derivations from the original

can be done by dividing by either (dy)2 or (dt)2 to get the companion arclength formulas:

b

a

dydxs 2)(2)(

b

a

dydy

dxs 12)(

b

a

dtdt

dy

dt

dxs 2)(2)(

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Examples of arclength

Let’s look at an example:f(x) = x2 on [0 , 4 ]

S =

=

42

0

1 (2 )x dx

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Here’s another example:For f(x) = sin -1 (x), suppose that

students have not learned how to integrate the arctrig functions.

So y = sin -1(x) becomes x = sin y

S = = 2

2

0

(cos( )) 1y dy

62


Here’s an example using parametric functions:

x =cos(t)

y = sin(3t)

S =

=

22 2

0

( sin( )) (3cos(3 ))t t dt

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Surface area: Back to the past

There is an interesting way to remember how to calculate surfaces area in calculus that relies on a very old geometry idea.

We look at the ideas of Pappus (300 AD, Egypt)

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Theorems of Pappus:Area and Volume … Arclength and Surface Area

Theorem 1: If a region is rotated about an axis that does not

intersect with the region, then the volume generated equals the product of the area of the cross section of the region and the distance that the center of that cross section travels.

V = So dV = or These are the shell methods of volumes of revolution.

2 *r areaydxr *2xdyr *2

65


-2

-4

-5 5

Rotate the yellow region about x = -1.

Pappus says that the volume generated by the black strip is dV =

1 22 ( 1)*( )x y y dx

66


Theorem 2: If a arc is rotated about an axis that does not

intersect with the arc, then the surface area generated equals the product of the length of the arc and the distance that the center of that arc travels.

SA = =

The radius is determined by the axis about which the arc is rotated (and there are the other two arclength formulas mentioned earlier)

arclengthr *2 dxdx

dyr 2)(1*2

67


It seems that this second theorem works perfectly well as a geometry formula …

2

-2

-5 5

Since we get the lateral area of a cylindrical prism.

68


…but it seems a bit suspicious for an arc that is not perpendicular to the axis of rotation…

2

-2

-5 5

f x = 0.125-x2 A

B

But remember that arc AB is very small.

69

Surface Area (examples)

Calculate the surface area when the curve f(x) = x2 on [0 , 4] is rotated about the y axis.

SA = 4

2 2

0

2 * 1 ( ) 2 1 (2 )b

a

dyr dx x x dx

dx

whose value can be done with a u du substitution or a fnInt on a calculator.

70


Calculate the surface area when the curve f(x) = x2 on [0 , 4] is rotated about the x axis.

SA = 4

2 2

0

2 * 1 ( ) 2 1 (2 )b

a

dyr dx y x dx

dx

42 2

0

2 1 (2 )x x dx

71


Calculate the surface area when the curve f(y) = sin(y) on [0 , ] is rotated about the x axis.

SA =

2

22 2

0

2 * 1 ( ) 2 1 (cos( ))b

a

dxr dy y y dy

dx

The numerical integral is needed to calculate this value.

fnInt( , y, 0 , )22 1 (cos( ))y y 2

72


Calculate the surface area when the curve defined by x = cos(t) and y = sin(t) is rotated about the x-axis. Use 0 t

SA =

2 2 2 2

0

2 * ( ) ( ) 2 ( sin( ) (cos( ))b

a

dx dyr dt y t t dt

dt dt

=

0

(2 (sin( ))*1) 4t dt

73


Note that we have just calculated the surface area of a sphere.

V =

SA =

34

3r

24 r

74

Gaussian Quadrature …How the TI calculators evaluate numerical definite integrals

Now, we are in the mid to late 1980’s with technology.

The Texas Instruments family of calculators has a fancy method to evaluate the definite integrals which is fast and usually quite accurate.Their website describes it as a version of

Gaussian quadrature.

75


TI-S2 FAQ Tl-83 f AO TI-8S FAQ TI-86 FAQNumeric integration on the TI-82, TI-83, TI-85, TI-86 - how does it work?

The method used in the TI-82, 83, 85, and 86 is known as Gauss-Kronrod integration. It is a good deal more sophisticated than commonly known methods like Simpson's rule or even Romberg integration. In fact, the widely used mainframe library program QUADPACK uses this technique. A detailed discussion is much better left to the literature on this subject, which is voluminous. A few comments, however, may help you in utilizing this function.

The concept begins with Gaussian quadrature rules, which have the property that by sampling a function at "n" points, an integral can be estimated that would be exact if the function were a polynomial of degree "2n-l". For comparison, the trapezoid rule is of polynomial degree one and Simpson's rule is of polynomial degree 3. But while Simpson's rule gives an exact integral for cubic polynomials based on three points, a Gauss quadrature rule with three points will give the exact integral of polynomials up to order 5. The next concept is to use a pair of Gauss quadrature rules of different order to provide both an integral result and a corresponding error estimate. The development of optimal extension pairs (pairs for which the lower order sample points are a subset of the higher order ones so that extra function evaluations are avoided) was done by A.S. Kronrod in 1965. The use of these pairs allow the method to be "adaptive". The integral is developed by beginning with the whole interval, computing the integral and the error estimate. If the error is too large, the interval is bisected and the process repeated on each half. Anytime a segment passes with regard to its "share" of the total error budget, that integral is added to the total integral. Other segments that don't pass are further divided. Note that this is quite different than some "adaptive" methods that iterate with smaller intervals until the result changes by less than some tolerance. Our method computes error from two integral estimates and only takes new function samples in rapidly changing regions that require it.

Formerly on http://www.ti.com/

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From a user standpoint, this means that you need to specify an error "tolerance" that you want to achieve for the integral. In our implementation this is an "absolute" error bound. This means the algorithm will quit when the absolute value of the difference in its two integrals is less than the tolerance you specify. Hence, you may not want to ask for a tolerance of .00001 if you compute an integral that has a value of order 10,000 because you are asking for 10 accurate digits. Another thing to watch out for is the fact the function is sampled a finite number of times on the first integration attempt. These points are not equally spaced and are in fact clustered toward the end points. However, it is not uncommon to have a function that is virtually constant or perhaps zero throughout much of the range for integration. In this case, the integrator may quit early and fail to detect the existence of function behavior in other regions. An example of this is:

fn!nt(eA(-t/le-6),t,0,l). This may seem appropriate to an electrical engineer who wants to compute all the "energy" in this waveform with a microsecond order fall time. However, just over a value of 0.002 in the interval 0 to 1, the value of this expression becomes less than le-999 and underflows to zero. So the algorithm returns a value of zero for this integral. A range of (0 , 0001) would be more appropriate for the problem and also because the integral is small, a tol [tolerance] of le-10 is not excessive.

References: "Numerical Methods and Software", David Kahaner, Cleve Moler and Stephen Nash,

Prentice Hall, 1989, Chapter 5.

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It works something like this:The interval along the x-axis is divided into two

equal pieces.The area under each half is calculated.The half-intervals are each cut in half again, and

the area is calculated again. If the area originally calculated does not change

when the interval is subdivided, then it is assumed that the area is close to correct.

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If the area is significantly different (whatever “significantly different” is), then it is further subdivided again and again until one calculation is not significantly different than the preceding one.Thus, parts of a function are divided more than

other parts. (Which parts?)

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However, in exchange for the accuracy (and often the speed) of the calculations, there are occasional examples for which the calculators make BIG errors, or for which answers are not available. Ex: fnInt(1/x , x , 1 , 1 x 1020) Ex: fnInt(sin(x) , x , 0 , 1000) is really slow. Ex: fnInt(sin(x) , x , 0 , )

gives 2.718 x 10-9 for an answer, which is only slightly different from the exact answer of 0.

500

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Ex: fnInt(e^(-10x), x, 0, 10000)

gives 0, when the actual answer is 0.1

There is a similar strategy similar to the coastline measurement being used, but the entire function is not sampled if changes are not detected earlier enough, so major errors in calculation can occur.

100001000010 10 100000 00

0

1 1( ) 0.1

10 10x xe dx e e e

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{ 1 x < 2

f(x) = 1001 2 < x < 3

1 x > 3

Calculators can accept these piecewise defined functions:

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The area under the curve on [0, 1000] ought to be 2000, but fnInt (Y1 , x , 0 , 1000) gives 1000.

1 x < 2f(x) = 1001 -2 < x < 3

1 x > 3{

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In conclusion …

Arclength and surface area may not be the most crucial topics in an AP Calculus class.

…but there are some connections with ideas and the historical development that make the topic fascinating and valuable.

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In conclusion …

We followed the ideas of measuring a straight line from the methods of Pythagoras, which the students learned in algebra and geometry classes. So the new calculus methods have pedagogical ties to previous knowledge.

We connected the geometry of the Pythagorean theorem to the algebra of the coordinate plane.

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In conclusion …

We help students to make the leap to the measurement of an arc as a series of line segments, because of the local linearity of most functions.

This develops the “official” methods of measuring arclength.

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In conclusion …

But….the anecdotes from history bring some authentic relevance to why these lengths are valuable: Geography in the 20th century use calculus ideas

to measure coastlines.Calculator technology adapt issues that plague

coastline measurement, developing highly effective ways to evaluate definite integral.

Calculus improves the methods of algebra and geometry to do effective calculations.

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In conclusion …

Students often cannot help but be eager to learn, especially when the intellectual

twists and turns are authentic and unexpected.

1 Arc Length and Surface Area Calculus Techniques Meet History David W. Stephens The Bryn Mawr School Baltimore, MD NCTM – Baltimore 2004 15 October 2004.

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