28127goat.htm updated 29jul2008

MY 11 TRISECTION APPROXIMATION METHODS Pg. 127

THE PLANIMETER SOLVES GOAT'S PROBLEM

R = 2cosa || a = cos^-1 R/2 || Rsin2Æ = sin2a || 2Æ = p - a || pi = p =

FIG. 28

PROBLEM:

1.	In FIG. 28.a a goat is tethered to a leash in yellow area A and eats

	the equivalent of area *A = pir² = p1 = 3.14159265359* square

	units of grass per day. How much longer does the leash, R units,
	Pg. 128
	have to be in order for the goat to eat the same amount when

	tethered at the circumference of the old circle, in order to optimize

	its eating area; so that yellow area A equals green area B?

CONSTRUCTION:

1.	In FIG. 28.b above: draw any suitable straight line CDEOFG.

2.	At pt O draw a suitable circle with radius r = OE = OF = 1 unit.

3.	At pt E draw another unit circle with radius r = OE = CE = 1 unit,

	intersecting the previous circle at pt A, as shown in FIG 28.b.

4.	Assuming the correct radius R > r = 1 draw another circle with

	centre O intersecting the circle CBAO at pt B and line CDEOFG

	at pts D and F. [IDEALLY: area A = area B = p*r² = p]

5.	Let OB = OD = OG = R units. [2 > R > 1 is approximation]

6.	From FIG 28.b one can easily draw FIG 28.c to represent the
	Pg. 129
	areas in question. Thus area E + F + C = C + G = G + D = p

	and thus area C = area D which is solveable, explained below.

7.	Each section FIG 28.c, FIG 28.d, FIG 28.e and FIG 28.f

	represents FIG 28.b in FIG 28 above.

Pg. 130

FIG. 29

THE LONG SOLUTION:

1.	Using CALCULUS and LA GRANGE for each of the coloured

	areas in FIG. 29 above (representing FIG. 28.b and FIG. 28.c

	above) one gets the following integral formula for the radius R of

	the new leash.


	Pg. 131


	Since R = 2cosa [from FIG. 28.e]

	and since sin2a = 2sinacosa [from CRC TABLES]

	therefore R²(p - a) + 2a - sin2a - 2p = 0

	or R²(p - a) - Rsina + 2a - 2p = 0
	Pg. 132
	Since a = p - 2Æ thus sina = sin(p - 2Æ) = sin2Æ [see FIG. 28.d]

	therefore 2ÆR² + Rsin2Æ - 4Æ = 0

	Thus the roots of the quadratic equation are:

	R = - sin2Æ/4Æ +/- [(sin2Æ)² + 32Æ]^½/4Æ

	For one solution the irrational part equals zero:

	therefore sin²2Æ + 32Æ² = 0

	and therefore sin2Æ = +/- 4Æ(- 2)^½

	thus R = - sin2Æ/4Æ = (- 2)^½ = 2^½i

THE SHORTEST SOLUTION IN LAGRANGE:

FIG. 30

Pg. 133

	From which one also gets: 2ÆR² + Rsin2Æ - 4Æ = 0

	which also gives: R = - sin2Æ/4Æ = (- 2)^½ = 2^½i.

THE SIMPLER SOLUTION:

1.	Again in FIG. 28.c areas (E + F + C) = (C + G) = (G + D) = pi = p.

2.	Thus area C = D and area (E + F) = G.

3.	Areas F + C = F + D = pR² - pr² = p(R² - 1).

	[2 > R > 1 is approximation & r = 1]

4.	From FIG 28.b/d isosceles DDBO: [CRC TABLES]

	a.	The 2 sides OD = OB = R and let base ÐODB = ÐOBD = Æ.

	b.	The base lenght DB = 2RcosÆ and Height = Rsine2Æ.

	c.	The area *DDBO = ODHeight/2 = .5R²sine2Æ**.

	d.	Let ÐDOB = a. [ sinÆ = Rsin2Æ/2RcosÆ ]

5.	From FIG 28.b/d/e isoscele DEBO: [CRC TABLES]
		Pg. 134
	a.	The 2 sides EO = EB = r = 1 and base ÐEOB = ÐEBO = a.

	b.	The base length OB = R = 2cosa and Height = sine2a.

	c.	The area *DEBO = OEHeight/2 = .5sine2a = .5Rsinea.**

6.	Also in FIG. 28.a/b/c/f area A = p = area B = pR² - 2(x + y + z) = pi.

7.	[The area of a slice of pie of a circle is the radius squared times half the angle - theorem]

	Thus twice area *(X + Y) = r²(4Æ - p) = 4Æ - p**

	and twice area DEBO = area X = sin2a = Rsin2Æ

	and thus twice area Y = 4Æ - p - Rsin2Æ

	also twice area (Z + X) = R²a = R²(p - 2Æ).

	Therefore twice area (X + Y + Z) = area (C + F) = p(R² - 1) =

	R²(p - 2Æ) + 4Æ - p - Rsin2Æ

	and again 2ÆR² + Rsin2Æ - 4Æ = 0

	which again gives R = - sin2Æ/4Æ = (- 2)^½ = 2^½i

Pg. 135

IMAGINARY EQUATIONS:

Imaginary formulas always facinated me and here I had one right in my

own lap. It facinated me so much so that I had to try to use it. So here

it is with the help of sinh x = (e^x - e^-x)/2; knowing that the

planimeter uses e^x and that my limits are 1 < R < 2; I played

FIG. 31

Pg. 136

with my diagrams (see FIG. 31 above) and arrived at the formula below,

giving R = 1.24311673443 - even the planimeter agrees with me!

GENERATED FORMULAS:

1. Starting with R = x + yi = 0 + 2^½i

2. Using limits 1 < R < 2

3. Investigating Hyperbolic formulas which are:

sinhy = (e^y - e^-y)/2 & coshy = (e^y + e^-y)/2

4. I noticed that:

(1 + e^y)/e^y > 1 with y=¥ & (1 + e^y)/e^y < 2 with y=0

ie. 1 £ (1 + e^y)/e^y £ 2 or R (1 + e^y)/e^y

5. Thus creating R = x + yi = 0 + 2^½i =

= [(e^x + e^y)/e^{(x + y)} ](y² - 1)^½ =
Pg. 137
= [(e⁰ + e^{2^{^½}})/e^{(0 + 2^{^½}}) ](2 - 1)^½ = (1 + e^{2^{^½}})/e^{2^{^½}} =

=1.24311673433

The planimeter agrees with this value rather than with

NEWTON'S approximation method value of 1.2451215 which is a

differance of 8 thousands - every computer gave a different value!

NOTE:

The formula could also have been (e^x + e^y)/e^{(x + y)}

= (e⁰ + e^{2^{^½}})/e^{(0 + 2^{^½}}) = (1 + e^{2^{^½}})/e^{2^{^½}} = 1.24311673433

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MY 11 TRISECTION APPROXIMATION METHODS Pg. 127

THE PLANIMETER SOLVES GOAT'S PROBLEM

R = 2cosa || a = cos-1 R/2 || Rsin2Æ = sin2a || 2Æ = p - a || pi = p = FIG. 28

R = 2cosa || a = cos-1 R/2 || Rsin2Æ = sin2a || 2Æ = p - a || pi = p =

PROBLEM:

CONSTRUCTION:

FIG. 29

THE LONG SOLUTION:

THE SHORTEST SOLUTION IN LAGRANGE:

FIG. 30

THE SIMPLER SOLUTION:

IMAGINARY EQUATIONS:

FIG. 31

GENERATED FORMULAS:

NOTE:

R = 2cosa || a = cos^-1 R/2 || Rsin2Æ = sin2a || 2Æ = p - a || pi = p =

FIG. 28

R = 2cosa || a = cos^-1 R/2 || Rsin2Æ = sin2a || 2Æ = p - a || pi = p =