Download Arc Length, Area, and the Arcsine Function Author(s): Andrew M

Arc Length, Area, and the Arcsine Function
Author(s): Andrew M. Rockett
Source: Mathematics Magazine, Vol. 56, No. 2 (Mar., 1983), pp. 104-110
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2690381
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Arc Length,Area, and the Arcsine Function
ANDREW M. RocKETrr
University
LonigJslanid
NY 11548
Greenivale,
of algebraicquantitiesprovidean essentialmotivation
forelemenGeometric
interpretations
titlesofmanyintroductory
textsalludeto the
tarycalculus.The "calculuswithanalyticgeometry"
of thederivative
While
relianceon pictorialimagesin thedevelopment
and thedefinite
integral.
and their
dominatemostpresentations
of the trigonometric
functions
considerations
geometric
and studyof
switchto algebraicmethodsfortheintroduction
thereis a nearuniversal
derivatives,
theinversetrigonometric
functions
(see [61,[3],and so on). This noteshowshow thegeometric
and their
ideas used in the definitions
of the definiteintegral,the trigonometric
functions,
inversefunctions.
derivatives
Reference
maybe continuedin thediscussionof thecorresponding
of thetheoryof elliptic
willbe made to therelationof thesegeometric
ideas to thedevelopment
and to Euler'smethodof findingalgebraicadditiontheorems
forcircular,
functions
hyperbolic,
and lemniscate
sines.
"Typicalcalculus"versusthearcsineas arc length
Taking[6] and [3] as ourmodels,we notethatin thetypicalmodemtextbook,
afterthedefinite
integral
has beendefined,
theapplications
includethearea betweentwocurvesand thearclength
formula.Sincefewintegration
techniques
are available,thearc lengthproblemsare restricted
to
"nice> curvesy = (x) such thattheintegralJf,'1+ f'( x)2 dx is particularly
simpleand sometimesan author'sapologyis offered
forthelack of interesting
applications
(see [3],p. 429).
The introduction
of the trigonometric
functions
followsa reviewof radianmeasureas arc
lengthmeasuredfromthepoint(1,0) on theunitcircleX2 + y2 = 1. The sineand cosineof a real
number0 are definedas thecoordinates
of thepoint(x, y) on theunitcircle0 radiansfrom(1,0)
of sin0 and cos 0 are derivedfromthesymmetries
of the
(see FIGURE I. Then theproperties
circleand the othertrigonometric
functions
are definedin termsof the sine and cosine.The
ofthesineand thecosinearefoundas consequences
derivatives
oflim0 0(sin0)/7 = 1. Thislimit
is established
by equatingarc lengthalongtheedgeof theunitcirclewiththearea of thesector
determined
by thearc (in FIGURE 2, 0 = 2 areaAOB) and thensqueezingthisarea betweentwo
triangular
regions.
Afterstudyingthe calculusof the six trigonometric
functions
("f(x)"), the corresponding
inversefunctions("f- 1(x)'") are soughtby reversingthe graphs("y = f(x)" becomes "x =
f`'(y)"),
makingarbitrary
choicesfor the "principalvalues" (see [6], pp. 295-6), and then
calculating
i(x)) = x.
Dx(f- l(x)) fromtheidentityf(f-
x2?y2
=I
(x, y)=(cos
,sinO)
B
(1,0)
0
o
FIGURE I
104
(0,0)
(1,0) A
FIGURE 2
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(/
1-
y)
-
(
s
-y2,y)
- --A(
-y 2, y)
0
(1,0)
(1,0)
FIGURE 4
FIGURE 3
In contrastto exploitingthe definitions
and propertiesof inversefunctions,
the arcsine
function
can be approachedin a moregeometric
way.Sincesin0 was definedas they-coordinate
of thepointon theunitcirclean arc lengthof 0 awayfrom(1,0), a geometric
attemptto invert
thisfunction
wouldask "giventhatthesecondcoordinateof thepointis y, fromwhatarclength
did it come?"Thereare two"small" answersto thisquestion(see FIGURE 3) and theninfinitely
moreseparatedfromeach otherby multiplesof 2 g. The "principalvalue" of theinversesine
as thesmallestdistancefromthestarting
function
maybe introduced
naturally
point,(1,0), that
willwork,and thisarc lengthmaybe called the"arc sineofy" and written
"arcsin(y)" or the
" sin- '(y)." We willuse the"arcsine"notationfortheremainder
" inversesineofy " and written
of our discussion.Thus the arcsinefunctionhas arcsin(y)=0 where -I/2 <0 < g/2 and
sinG = y.
Since arcsin(y)is an arc length,the arc lengthformulaf,511/
+f1'(t)dt can be applied to
f(t) = 21-t2 fromt = 0 to t =y (see FIGURE 4) to findthat
arcsin(y)=|/1
+f'(t )2dt
0
=
2
t
+
1
=j|
__
dt
(1)
dt
t2
and then,by theFundamental
Theoremof Calculus,it followsthat
D (arcsin(y))
1
=
Sincearcsin(l)= 7T/2,
we also have a simpleexampleof an improper
integral:
1
? /
2
dt=lmJ
y
I
/
1-t2 dt=-.
2
2
For theinversetangent,
a similarargument
yieldsan arctangent
function
arctan(w) 0 with
principalvalue -7T/2< 0 < 7T/2such thatarctan(w)is thedistancealong theunitcirclefrom
(1,0) to thepoint(x,y) withy/x=w. Since x= l-y2, we can solvey/x=w to findy=
w/ 1 + W2. Fromthearc lengthformula
(1) we have
arctan( w ) =
W/
o
W2
1-t2dt
dt.
VOL. 56, NO. 2, MARCH 1983
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105
Makingthechangeofvariablet=
u = w,we findthat
u/
arctan(W)
1 + u2, so thatt= O whenu= O and t= w/l1 + w2 when
=fV
du
+ U2
o
1?+U2(1?U2)
w
1
o
?2du,
101+
U2
and so
DW(arctan(w))=
2wFor theinversesecantof w (whereIwI> 1), if we seek thesmallestpositivearc lengthfrom
arcsec(w) = 0 with
(1,0) to thepointwithfirstcoordinate1/w,we obtainan arcsecantfunction
principalvalue0 < 0 < 7T/2and ST/2< 0 < ST.For w > 1, thearc length0 is ST/2- + (see FIGURE
5) so we have
- arcsin(1/w)
arcsec(w) =
and so
DW(arcsec(w))
1
-
__1
w2
21-(l/W)2
=
1
forw> 1.
W/W2_
that
For w < -1, we haveby symmetry
- arcsec(- w)
arcsec(w)=
and so
DW(arcsec(w))
=
D-D (arcsec(-w ))
and thus
DW(arcsec(w))
1wI
for > 1.
frw>1
1
dt forw) 1
IwI w2- 1
(We omitdiscussionof theformula
arcsec(w
)=w/w
0
o/1 _ t2
X_
=
x(t)=
j1-t2
_
t
X()
x(t)=(
(2
(
(I/W
(1,0)
FIGURE 5
106
W/
y
1-y2
rl
*
1 -y2,y)
W)
At
-(1,0)
FI
E
FIGURE 6
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in a subtleimproper
as it results
integral
calculation.)
The formulasfor the inversecosine,cotangent,and cosecantcan be obtainedby similar
arguments.
Area and the arcsine
Let us now studythe arcsineas an area. Let 0 < y < I be givenand let A denotethe area
boundedby theline x(t) =
-y2/y) t, the circlex(t)=
-tt2, and the x-axist = O (see
FIGURE 6). Sincearc lengthalongtheedgeof theunitcircleequals twicethearea of thesector
determined
bythearc,we havethat
arcsin(y) = 2 A
o
~~Y2
1y2
1-t2 dt- yl _-y2
=2f~'
2
(2)
Differentiating
(2), we have
y2
Dy(arcsin(y))= /l-y2?
'
21- y2
as before.
Alternatively,
our integralexpression
(2) forarcsin(y)providesa motivating
exampleforthe
1-t2 and dv= dt,we have
integration
by partsformulaJu-dv= u v - Jv du.Lettingu=
f1 - t2dt=t
-
-t2_vt
-t
t-
tV-t2-
-tv
dt
11_
-t2-f1-t22dtf
1
t2
dt
so
But then
arcsin(y)=2.f 1-VIt2dt-y ~I-y2
=Y +I-y+
-Y
J 1-t2
2 Y
"0
1-t2
1
dt-y I-y2
d
as before.
VOL. 56, NO. 2, MARCH 1983
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107
This development
of thearcsineas thearea boundedby thex-axis,thecirclex(t) = /1- t2
=
fromt 0 to t=y, and thelinefromtheoriginto thepoint( 1 -y2 , y) maybe adaptedto the
hyperbolax(t) = /1+ t2 to find the "inversehyperbolicsine" sinh-'(y) = lnj 1 +y2 +Yj.
However,unlikethe circle,thereis no simplerelationbetweenarc lengthand area for the
hyperbola
(see [4],? 6 and 7; fora representation
ofarclengthon thehyperbola,
see [1],? 60, and
in ?61). On the otherhand, a similarprocedureto
the corresponding
geometric
construction
sine"usingthelemniscate
insteadof thecirclemustbe carriedoutin terms
developa " lemniscate
of arclength(see [4],? 8).
Rationalization
of thearcsine
in (1) as a rationalfunction;
theintegrand
thatis,
We now turnto theproblemof expressing
we wishto expressI - t2 as thesquareof a rationalfunction.
If,in thetwo-variable
representationof Pythagorean
triples,
(a2
b2)2 + (2ab)2
+ b2)2
=(a2
(3)
we let a = I and b = u then
=
(1u2)2
(1
+ U2)2
_ (2U)2
or
(
2)
u)2
_(
t= 2u/(1 + u2) willresultin
and so thesubstitution
-
-t2
- U2
I U2
and
+
dt=d(1 u2)(2)(1+
(2u)(2u)d
2
u2
(1
u2)2
du.
+ u2)2
Then theindefinite
integral11/f1 - t2dtbecomes
I + I?u2
U2
____________I_
I-U2
(1 + u2)2
I___
udu=2f
du
1 + u2
and the integrandhas been rationalized.Solvingt= 2 u/( + u2) for u, we find that u=
foru with
(1 ? 1- t2)/t. Equatingt withsinO, we see that 1- t2= cos 0 and theexpression
theminussignbecomes
u = (I -cos 0 )/sin0 = tan(0/2)
for the rationalization
of trigonometric
and we have derivedthe " u = tan(0/2)" substitution
withthetypical"it has beendiscoveredthat..." treatment
integrals.
(Comparethisdevelopment
in [6],p. 368.)
Euler's sine sumformula
In the introductory
sectionof [5], Siegel describesFagnano's studyof arc lengthon the
of the arcsineon the circle)and speculatesthat
lemniscate(whichfollowsour development
to double arc lengthon thelemniscate
construction
Fagnano's 1718 discoveryof a geometric
resultedfromhis attemptto rationalizetheintegrandof thelemniscatesine. Thirty-five
years
later,Euler extendedFagnano's doublingtheoremto an algebraicadditiontheoremfor the
lemniscate
sineand he shortly
thereafter
his discovery
to ellipticintegrals.
generalized
Siegel([5],
of Euler'sresultfrom
p. 10) describestheaim of his firstchapterto be thefullerunderstanding
108
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in theirfulldomainof definition.
We concludeourdiscussion
theviewpoint
of analyticfunctions
forthearcsineadaptedfrom? 585 and 586
of thearcsinewithEuler'salgebraicadditiontheorem
of [2] (see also [2],CaputVI forEuler'sstudyof ellipticintegrals;[4], Chapter4; and [5],? 2).
Let - v7/2< T < v/2 be a fixedangle and let - 7y/2< 0, 4 < i/2 be any two angleswith
d(0 ? p) - 0, and ifwe set u = sinO and v = sin4, we
0 + k= T. Sincethesum0 + 4 is constant,
can rewrite
d(0 + p)= 0 as
(ju
(lo
VI
dt
1-dt)
2o4
_t2
equation
and so we havethedifferential
du
dv
?
-U2
t2
-v2
=0.
(4)
We seekan algebraicsolutionof (4) subjectto theconditionthat0 ? k- '.
second-order
equation
Euler'smainobservation
was thatifwe beginwiththesymmetric
u2 +
2auv+ v2= K2,
(5)
we can completethesquareon theleftsidein eitheru or v. In the
wherea and K are constants,
firstcase,we have
u 2+ 2auv
+ a2V2 = K2
+ (a2 -_ )V2,
so that
u+av=
(6)
VK2+(a2-_ )v2,
whilein thesecond,we have
v2+ 2auv + a2u2
=
K2 +
(a2 -)U2
so that
v+au=
(7)
K2+(a2-1)u2.
If we differentiate
equation(5), we find
2udu+ 2avdu+ 2audv+ 2vdv= 0,
termsand using(6) and (7),
whichbecomes,aftercollecting
?
dv
du
VK2+ (a2-_)u2
=0.
(8)
VK2+ (a2-_)v2
If we seta 2 1 = -K2 thenequation(8) is thesameas (4) and so a = 1 - K2. If we can solve
(5) forK usingthisvalueofa, we willfindan algebraicsolutionof(4) thatexpressestheconstant
value sin(0+ 4) in termsof sinO and sin4.
we have
ourvaluefora into(5) and rearranging,
Substituting
(u2 + v2)-K2=
21 -K2 uv
and squaringbothsidesgives
(U 2+ v2)2-2K2(U2+v2)+K4=4(l-K
a = u and b= v in equation(3) and solvingfor(U2
By setting
(u2 _ v2)2 - 2K2((U2
+ v2) - 2u2v2)
2)u
v2.
+ v2)2,
(9)
we can rewrite
(9) as
+ K4 = 0.
(10)
Since(10) is a quadraticin K2, we can completethesquarewithrespectto K2 to obtain
(K
(+ (2U2V2(U2
+ V2)))2 = ((U2 + V2)-2U2V2)2
_ (U2 _
VOL. 56, NO. 2, MARCH 1983
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11)
109
By expanding,collectingterms,and removingcommonfactors,the rightside of (11) may be
as 4u2v2(u2v2 - (u2 + v2) + 1). Takingsquareroots,we find
rewritten
K2 = (U2 + v2)-2u2v2
+ 2uv/ (1-u2)(1-v2)
(12)
theexpression
Regrouping
(U2 + v2) - 2u2v2 as (U2 _ u2v2) + (V2 _ v2u2), we see thattheright
square
and so
of
is
a
perfect
side (12)
K=u
1-v2
+ V
(13)
1-_2
Changingback to angles,K is theconstantsin(O+ k) while
VI- u2 = cos 0 so we have thesinesumformula
sin(O+?)
1-v2
=
1 - sin24
= sinOcosk + sin cos S.
=
cos4 and
(14)
we have established(14) foranyT.
Sincethefixedvalue of T playedno partin our calculations,
References
[1]
[2]
[3]
[4]
[5]
[6]
ArthurCayley,An Elementary
Treatiseon EllipticFunctions,2nd ed., 1895.Reprintedby Dover Pub., New
York, 1961.
Leonhard Euler, InstitutionesCalculi Integralis,Volumen Primum,1768. Reprintedas Leonhardi Euleri
Opera Omnia,SeriesI, Opera Mathematica,VolumenXI, Engel et Schlesinger,
eds., B. G. Teubneri,Leipzig,
1913.
Louis Leithold,The Calculus withAnalyticGeometry,4th ed., Harper& Row, New York, 1981.
A. I. Markushevich,
The RemarkableSine Functions(translatedby ScriptaTechnica,Inc.), Amer.Elsevier,
New York, 1966.
C. L. Siegel,Topics in Complex FunctionTheory,vol. I: Elliptic Functionsand Uniformization
Theory
(translatedby Shenitzerand Solitar),Wiley-Interscience,
New York, 1969.
GeorgeB. Thomas and Ross L. Finney,Calculus and AnalyticGeometry,5thed., Addison-Wesley,Reading,
1979.
Proof withoutwords:
Completing the square
X2 + AX= (X+ A/2)2-(A/2)2
x
+
A
+
LE
D. GALLANT
St. FrancisXavierUniversity
Nova Scotia
Antigonish,
-CHARLES
110
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