Curve Straightening and Formula Determination

Curve Straightening and Formula Determination

Ronald EdwardsWesleyan College, Macon, Georgia

Much has been said and done about integrating mathematics andscience programs at the high school level. In a more informal waythis can be carried on at the junior high school level for students whohave had algebra or are in the process of studying elementary algebra.The illustrations in this paper will be used as examples for unifyingthe study of mathematics and science at the high school and juniorhigh school levels.These mathematical treatments of simple laboratory experiments

or classroom demonstrations serve many purposes. They illustrate afunction of mathematics. They increase the students practice inarithmetic and other mathematical computations, putting a natural(or necessary) emphasis on discovery and intuition. The results causescience to be more meaningful and develop a comprehension of thescientific method and the spirit of discovery.The process of curve straightening and formula determination dis-

cussed can be started on the junior high school level and be developedthroughout the students high school career on a more sophisticatedplane as the mathematics develops. The student needs a minimum ofbasic algebra and some knowledge of graphing including a goodknowledge of linear graphs. The considerations in this paper strike amiddle ground, but can be lowered for the junior high or upgradedfor the calculus student.While conducting a standard experiment to determine the relation-

ship between the length of a spring and the force stretching thespring, the data in Table 1 and subsequent graph in Fig. 1 were pro-duced.

TABLE 1. RELATIONSHIP BETWEEN THE LENGTH OF A SPRING AND THEFORCE STRETCHING THE SPRING

Mass suspended from Force on the spring, F, Length of the spring,the spring in gm-weight in dynes L, in cm

20040060080010001200

1.96 X1053.5.7.9.1,

.92 X105,88 X108,84 X105.80 X105.176X106

56.5559.61.64.66.69.

,00,55.15.70.25

It is of interest to discover an algebraic relation between the force,F, acting on the spring and the length, L, of the spring. This ismuch more of a challenge and more fun than being given the related

149

150School Science and Mathematics

FIG. 1. The graph of the length, Z, of a spring as a function of the force, F,stretching the spring.

formula and showing that the experimental data agrees with theformula. The graph in Fig. 1 indicates the relation is nearly linearand hence the equation representing this relation is of the formL==m-F-{-b for some constants m and 6. The slope, w, of the straightline is

Li -Z2��

^F\ � F^

i.e., m equals the difference in the values of L for two points on thegraph divided by the difference in the values of F for these samepoints. We express this as the difference of L divided by the differenceof F, written AL/AF. Here

AL 69.25 cm � 56.55 cm�� = �� == 1.29 X 10-5 cm/dyne.A7? 1.176 + 106 dynes - 1.96 X 105 dynesAF

The y-intercept, b, is read from the graph extended to intersect they-axis. Here b ==54.1 cm. This should be the original length of thespring.Now, we have £=(54.1 cm)+(1.29XlO-5 cm/dyne)-F. We may

check the accuracy of this formula by considering F=5.88X105dynes. From the formula £==(54.1 cm)+(1.29X10-5 cm/dyne)�(5.88X105 dynes) =(54.1 cm)+(7.58 cm) =61.68 cm and from thecollected data in Table 1, £==61.55 cm. The percentage of errorbetween these values is about .2 percent.The value of this quantitative relationship (between L and F) is

that mathematical implications drawn from the formula may lead toworthwhile predictions or conclusions concerning the physical experi-ment.

Curve Straightening 151

Returning to the graph in Fig. 1 and the subsequently derived for-mula, we consider five significant mathematical observations.

(1) The ^-intercept of the graph is 54.1. Physically speaking wewould say the spring has length 54.1 cm when a force of 0 dynes isapplied to the spring. Hence, this is the original length of the spring.

(2) The slope of the line is w=L29XlO-5 cm/dyne. This repre-sents the change in length of the spring per unit change in force act-ing on the spring. At a given point on the graph, m represents the rateat which the length is changing with respect to the force acting onthe spring. The slope is constant in this experiment, i.e. for the springunder study. This observation may suggest that we determine thisrate of change experimentally for other springs. We may even predictthat a ^stiffer^ spring would have a smaller constant, m, i.e. the rateat which the length is changing with respect to the force acting onthe spring would be less.The idea of the rate of change of one variable with respect to

another can be generalized mathematically. For the calculus studentwe express this symbolically as dL/dF==m and say "the derivativeof L with respect to F is m^\

(3) The simplest prediction given the formula, 2L==54.1 cm

+(L29X 10~5 cm/dyne) -F, is predicting the length of the spring fora given force, F, in dynes or predicting a given force acting on thespring given the length, L, in centimeters. Although we have no datain our table giving the length of the spring when F== 5X 105 dynes, wemay substitute this value of F in our derived formula and obtain areasonable value for L. The student may return to the realityof the physical experiment and check the theoretical result whenF==5X 105 dynes by applying this force to the spring.

(4) The area, A, between the linear graph in Fig. 1 and the hori-zontal axis computed from the point where F==0 dynes to the pointwhere F=b dynes (6>0, a constant) is A==j. b -(54.1 cm+1.29X10~5 cm/dyne-6), i.e. the area of the triangle formed by the hori-zontal axis, the graph and the vertical line through the point (6, 0)is ^ its base times its height. Note the dimension of A is dyne-centi-meter. This is a unit measuring work. For a force of b dynes, A repre-sents the work done when the force stretches the spring from its neu-tral position (original length of 54.1 cm) to the length (54.1+1.29XIO-5-^) centimeters.

If the region whose area is desired is formed by a geometric figurewhose area formula is known, the student will have little trouble. Ifthe region is irregular as the shaded area shown in Fig. 2, the student(including the junior high student) may approximate the area. Thiscan be done using a carefully drawn graph on metric graph paper,


FIG. 2. The area between the graph of/ and the o;-axis betweenx==a and x==b.

counting the squares and estimating fractions of squares in theregion under study.We can generalize the idea of the area between the graph and the

horizontal axis in certain cases and write

A == f}wd^�/ n

read ^the integral of f(x)dx evaluated from a to b.^ Now, A definedin this manner will represent the area between the graph of / andthe horizontal axis from x==0 to x==b, if / is a positive function be-tween 0 and 5. In Fig. 2 we represent the area of the shaded region as

r/<x)dx.

This may be computed for a wide range of functions with which thestudent may be familiar.

(5) Given two points jP==(Zi, Fi) and Q=(L^ F^) on our graphwe may compute the length of the line segment PQ, written ~PQ.

~PQ equals V(Z2 - -Li)2 + (^2 - WHowever, in this case we cannot attach a dimension to the result orbeneficially interpret the result in our physical experiment. Thus, to


every mathematical computation resulting from the derived formulathere may not be a meaningful physical interpretation.

Again, with aid of elementary calculus the idea of length may beextended to non-linear functions or a method of approximation oflength may be devised by the precalculus student.As we collect data in the laboratory, we soon discover that not all

data is nearly as obliging as that in the above illustration. Data mayproduce graphs of great variety and varying degrees of complexity.The problem confronting us is to determine an equation describingthe graph and hence an equation which approximates the collecteddata in an experiment. In this problem our intuition plays a lead role;but, we may rely upon mathematical techniques to reach this goal.For the illustrations discussed here, it is assumed that the student isfamiliar with the basic shapes of graphs of polynomial functions,positive root functions and the basic trigonometric functions (sineand cosine).

Consider an experiment where we are to determine an equationexpressing the period, t, of a simple pendulum in seconds as a func-tion of the length, jL, of the pendulum in centimeters. We collect thedata given in Table 2 and represent it graphically in Fig. 3. Even thejunior high school student is with us here.

FIG. 3. The graph of the period, t, of a simple pendulum as a functionof the length, £, of the pendulum.

154 School Science and Mathematics

TABLE 2. RELATIONSHIP OF THE PERIOD, t, IN SECONDS TO THE LENGTH,I/, IN CENTIMETERS FOR A SIMPLE PENDULUM

Period of the pendulum in secLength of the pendulum in cm

3.510 3113.239 2592.910 2112.694 1802.453 150.52.204 121.31.724 751.393 50.980 25

Although the graph does not represent a linear relationship, wemight guess the graph is of the form t== k’ -^/L, t== k’ ^/L or t== k � ^/L{k a constant and n a positive integer). The graph of a positive rootfunction has this general shape. Now, we construct a table of valuesfor t and -\/L (using the data from Table 2) in Table 3. A table ofsquare roots or slide rule would be useful here. We hope our juniorhigh school student is with us. Complete understanding of the "whyîs not necessary at this point.

TABLE 3. RELATIONSHIP OF THE PERIOD, ^ IN SECONDS TO THE SQUAREROOT OF THE LENGTH IN Vcm OF A SIMPLE PENDULUM

-r» . i ,. ,, , , . Square root of the length of thePeriod of the pendulum in sec 4

pendulum in Vein

3.5103.2392.9102.6942.4532.2041.7241.393.980

17.6016.0914.5213.4112.2611.008.667.075.00

When we graph the data from Table 3, we note an interestingchange in the geometry of our graph; see Fig. 4. A straight line ap^proximates the data of Table 3 quite well, i.e. t as a function of -v/Lis nearly linear or t==m’\/L-[-b. Here

M 3.75 sec,

/��

^ -=. ��^ == ��: == .197 sec/ycm.AVL 19Vcm

Also, 6==0 sec as seen from Fig. 4, b being the ^-intercept. Hence,^==.197 sec/Vcm-VZ.


Thus, if our intuition leads us correctly, we may by the abovemethod ^straighten57 our curve to a linear graph and thereby arriveat the equation we desire. Shortly, we shall investigate why thismethod works.

Again, this formula, ^==.197 sec/Vein-V£, has predictive value.Given a period, t, in seconds we may predict the length, £, in centi-meters and given a length, L, in centimeters we can compute theperiod, f, in seconds. The rate of change in t with respect to ^/L maybe expressed as the slope of the graph in Fig. 4. However, the rate ofchange of t with respect to L is not constant. We can see from Table2 or Fig. 3 that this rate of change decreases as L increases.The derivative of t with respect to £, dt/dL, is given by dt/dL^=

. 192/2-1/VL. (the units being sec/cm) With this expression at anypoint on the curve we may find the rate of change of the period withrespect to the length. When L== 100 cm, dt/dL== .192/2 � 1/VIOO sec/cm==.0098 sec/cm.Although we may compute the area under the curve in Fig. 3 by

a graphical approximation or by computing the proper definiteintegral, we arrive at a value whose unit of measure is centimeterseconds. Since we attach no physical significance to this product ofL and I, this mathematical computation is of little significance.

Before we consider why this method works, perhaps a few wordsshould be directed to the junior high school student. Today greateremphasis in mathematics and science has been placed upon under-standing why this computation or that process works. However,practically speaking quite the opposite is taking place in our society.Most automobile drivers, laboratory technicians and computer pro-grammers do not know the "why55 of their machinery. The machineryis easy to operate and is quite dependable. The junior high schoolstudent should learn the "why^s^ he is able and use other machinerythat is easy to operate, available and dependable, e.g. log tables, trigtables, slide rule, computers, etc. . . .

Before we try to interpret data of greater complexity and perhapsof more interest, let us examine the method we have employed. Sup-pose ^ is a function of t and intuitively we guess s=<j>{t), then we plota graph of the ordered pairs (t, <f>(t)). If by our choice we "straighten"the curve to a linear graph, then we arrive at the result: s == m’ <^(f)+ 6.For example, consider the following table for the function /:

x [ 0 2 5 12

f(x) | 0 20 125 720

Now, consider the function g: x^�^fix) and its corresponding table


x2 |0 4 25 144

/�O) | 0 20 125 720

In this latter table,

Af(x) 20-0 125-20 720 - 20== 5.

Ax2~

4 - 0~~

25-4 "144-4Thus, the graph of g is linear. This due simply to the fact thatf(x) == 5x2 was the original rule of the function /. Now that this isknown, we may compute the slope in general. We have, f(x^) == 5x^2and f(xi) �== Sxi2 implies

/(^2) -f^l) 5[^2-^2J== 5.

x^2 � xi2 X22 � xr

Thus, if/were of the form/(^) = k ’x2, then g would be of the formg(x2)::=::m’x2-{-b, the graph of g being a straight line with slope m andy-intercept 6. If by a first choice for the function g we do not com-pletely ^straighten" our curve, we may try again until we find a goodchoice. If we have succeeded in "straightening" / to any degree by afunction g we may work on this new function g without returning tothe original data.The questions we must ask concerning this method in the general

case are: If we guess a suitable equation (assuming one exists), willthe curve be "straightened"? If wefind a function that will "straighten"our curve will the resulting equation reproduce our given data?To answer the first of these questions consider the following table:

X\ d\ 02 Os ’ ’ ’ CLn�� ,V [ bf b2 b3 ’ � ’ bn

the a/s and 6/s are real numbers and the <z/s are distinct. Now, sup-pose/(^) is a non-linear function and in the above table y= k -f(x) -{-L,for constants k and L. If our intuition suggests that the equation re-lating x and y in the table of original data is y==f(x), we construct thefollowing table of values:

/(^J/W fW ’ ’ -f^n)y | bi 62 � � * bn

Now, we compute the following:

b2 - b, __ (^/(a2) + L) - (k’f(az) + L)

/(î) - /(oi)~

/(^) -/(oi)k(fW -/(ai))

= kW -fW


This is true since the function represented in the original table ofdata is y==k’f(x)-}-L and bi=k’f(ai)-^-L for z=l, 2, �

� � , n. Also,

^3 � ^2 _ ^n � 6»-1 _

f(a3) - /(as)~

?" *

?

^n) - /(a.-i)

and in general if i^j,

bi - b,== k.

W -/(^)

Since these ratios are constant, f(x) is linear with respect to y and wehave the equation y = k - (x) 4- P for some constant p.To determine ? we substitute/(oi) ==/(^) and bi==y (in the equa-

tion y==k’f(x)-}-p) and obtain b^,==k-f(ai)-{-p or p=bi�k’f(ai).However, from ^v==^-/(^)+^/ we also have L^bi�k’fâi). Thus,p^L and we have reproduced the equation y==k’f(x)-\-L.Thus, if we guess the basic relation involved in relating the values

Oi and bi, we will both "straighten" the curve and arrive at the for-mula y==k’f(x)-{-L.To answer our second question, consider the table of data:

X | d\ (22 � � � CLn

y \ bi b^ � ’ ’ bn

Assume we guess that the function/ relates x to y above. We considerthe tables of values

f(x)\f(a,) /(^) . . ./(^)

V | bi b’2 ’ � ’ bn

Let us suppose that y and/(^) are related linearly, or the differencequotients are constant (or very nearly so), i.e.

bi - b,

^ == �� fQF ,^ ^ ./(a,) -f(a,)

Now, we have the equation (A) y==k’f(x)+L where L==bi�k-f(ai).To check that this derived equation (which has straightened ourcurve) will reproduce the original data, we substitute ai for x in equa-tion (A) and

y == k’f(az) + L = k’f(a,) + [b, - k’f(a,)i == b,.

Thus, if the solution is exactly linear, we reproduce the original dataexactly. If it is very nearly linear, we will approximate the data veryclosely.Let us employ this method once again on a collection of data from


TABLE 4. RELATIONSHIP OF THE LIGHT INTENSITY, /, IN FOOTCANDLESPASSING THROUGH TWO POLAROID FILTERS AND THE ANGLE,

<^», IN DEGREES BETWEEN THE FILTERS

Light intensity, 7, in footcandlesAngle, <^>, between polaroids

0 5810 5720 5330 4640 3750 2860 1970 1180 590 2100 3110 7120 14130 23140 32150 42160 49170 54

5 L

FIG. 4. The graph of the period, t, of a simple pendulum as a function of thesquare root of the length of the pendulum.


an experiment dealing with polarized light. Table 4 relates the lightintensity, 7, in footcandles of the light passing through two polaroidfilters and the angle, 0, in degrees between a polarizing polaroid andanalyzing polaroid.

TABLE 5. RELATIONSHIP OF THE LIGHT INTENSITY, 7, IN FOOTCANDLESPASSING THROUGH TWO POLAROID FILTERS AND THE COSINE

OF THE ANGLE BETWEEN THE FILTERS

Intensity, ICosine of the angle, <f>

585753463728191152

1.0000.9848.9396.8660.7660.6427.5000.3420.1736.0000

FIG. 5. The graph of the light intensity, Z, passing through the polaroidfilters as a function of the angle, <^, between a polarizing polaroid and analyzingpolaroid.

160 School Science and Mathematics

TABLE 6. RELATIONSHIP OF THE LIGHT INTENSITY, Z, IN FOOTCANDLESPASSING THROUGH TWO POLAROID FILTERS AND THE SQUARE

OF THE COSINE OF THE ANGLE BETWEEN THE FILTERS

Intensity, I Square of the cosine of <j>

58 1.00057 .97053 .88046 .75037 .59028 .41519 .25011 .1175 .0312 .000

50

50

10

FIG. 6. The graph of the light intensity, J, passing through the polaroid niters asa function of the cosine of the angle, 0.

If we guess that the basic shape of the curve is that of 7= cos <^,by examining the graph in Fig. 5, we construct Table 5. Plotting thevalues in Fig. 6, we see our curve is not fully "straightened.77 Wemust try again. Since we have straightened the curve somewhat andthe resulting curve may suggest the graph of y=x2, we constructTable 6 and the subsequent graph in Fig. 7. Now, we see that the


0 .1 .2 .5 .4 .5 .6 .7 .8 ,9 1,0

FIG. 7. The graph of the light intensity, J, passing through the polaroid niters asa function of the square of the cosine of the angle, <f>.

graph in Fig. 7 is nearly linear and hence 7= k{cos(|>)2-}-b, We find thenumerical values of the constants to be fe=54 and 5=5.5. Hence,

I == [(54) cos2 <t> + 5.5] footcandles.

Now, checking the value I when 0=30 degrees, we obtain

I = [54-cos2 (30°) + 5.5] footcandles == [54(.75) + 5.5] footcandles

=46 footcandles,

as it appears in Table 4. Note, as you approach angles near 0° or 90°the error increases rapidly.

MUSTARD CAUSES HYPERTENSION, HEART DISEASE,DOCTOR SA ^S

Mustard contains an ingredient that is the major cause of coronary disease andatherosclerosis (hardening of the arteries), an Ohio doctor believes. Together withtwo other condiments common to the American table, he says, it is also responsi-ble for most unexplained hypertension.The key ingredients, reported Dr. Jackson Blair of Lakewood, Ohio, are esters

of isothiocyanic acid, though he admitted that he does not have definite proof oftheir relationship to coronary disease.

Large amounts of pepper, ginger, and mustard were present in the diets ofhypertensive patients studied by Dr. Blair in 1948. Several years later, he re-ports, he produced "statistically significant hypertension77 in laboratory rats byadding the condiments to their diets.

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Curve Straightening and Formula Determination