Math Portfolio Finished SL TYLE II CIRCLES

Download Math Portfolio Finished SL TYLE II CIRCLES

Post on 01-Nov-2014

103 views

Category:

Documents

1 download

DESCRIPTION

MATH PORTFOLIO SL TYPE II CIRCLES 2012

TRANSCRIPT

IB SL Math Type 1 Circles Math Portfolio Sharat Ramamani

Ms. Alsop Due: June 4, 2012 MPM-3UW

TABLE OF CONTENTS Introduction ___________________________________________________________ # Method _____________________________________________________________ # Case 1 ______________________________________________________________ # Case 2 ______________________________________________________________ # Validity of General Statement _____________________________________________ # Conclusion __________________________________________________________ #

Introduction

Method To find chord , a variety of circular and trigonometric properties shall be applied, to arrive at the general statement representing chord . Figure 2: C2, with AGO as a circumscribed right angle triangle within the circle, APO as an isosceles triangle within AGO, and AOO and APO as right angle triangles within APO, generated by Wolfram Mathematica.

Several claims must be proven in order to derive the general statement representing chord : Claim 1: Side OG represents the diameter of C3, or twice the length of radius OP. The radius of C2 is stated in the problem as OP. Side OG represents 2OP, or the diameter of C 2 (see Figure 2). Equation 1: Claim 2: Side OA represents the radius of C3. Figure 1 depicts the intersection of the three circles. The center of C3, point A, is connected to point O by side OA. Point O is on the circumference of C3, therefore Side OA represents the radius of C3. The radius of C1 has the same value as r. Equation 2: In this investigation, lowercase r will represent the radius of C3.

Claim 3: Chord segment OO is half the length of chord . Figure 3: C1, C2 and C3 with side AP drawn on C3

O

If a line segment is drawn, attaching point A to point P, then chord becomes a part of APO, which is within C3 (see Figure 3). Since both side AO and side AP extend from the center of C3 out to any point on C3s circumference, they have the same side length, as they both represent the radius of C3. Due to the fact that AO and AP have the same side length, APO is an isosceles triangle. Furthermore, if a line segment AO is drawn from the center of C 3, and bisects APO to a point O on chord , then two right angle triangles are created; AOO and APO. O on both right angle triangles contain the 90 right angle. Since APO is bisected, that means point O bisects chord , which also means that chord segment OO is half the length of chord . Equation 3: 2

Claim 4: AGO is a right angle triangle To prove this claim, Thales Theorem must be applied. Thales Theorem states that the diameter of a circle always subtends a right angle to any point on the circle. Figure 2 contains C2, which circumscribes AGO. Side OG represents the diameter of the circle, as proven by Claim 1, and point A represents a point on the circumference on C2. Therefore, according to Thales Theorem, AGO is a right angle triangle, and A is a right angle. Using these four proven claims, it is possible to derive the general statement that represents chord . Equation 4: ( )

The initial equation of cos(o) =

is used to define cos(o), as

and will be

substituted into other equations later on. Now that cos(o) has been defined, it is possible to work with the triangles inside AGO. It should be noted that r cannot be equal to or greater than 2OP, or else O will not exist. If r is equal to 2OP, then O is 0, which cannot exist with the geometric figure given, so it is a limiting case. If r is greater than 2OP, then O does not exist, as there is no intersection point between C1 and C2, where the normal intersection point is Point A, which is non-existent. Figure 4 and 5* compare O when 2OP>r and 2OP r

Figure 5: Case when OG (or 2OP) < r

*Created using the geometrical program Geogebra

Equation 5:

( ) ( ( ( ))( ) )( )

In these set of equations, AOO is being analyzed, in Figure 3. Chord segment OO is being solved for, by using the trigonometric ratio of cosine. Cos(o) is then substituted with as it is defined like such in Equation 4, and thus resulting in chord segment OO = Equation 6: Equation 6 combines the proven claims and simplifies the other equations to produce the general statement for chord , that is, Case 1 In the first case, r=1 , and OP must be solved for when OP = 2, 3 and 4. The general statement that models OP can be used, but can also be modified to better fit the case. Instead of Equation 7: when r=1 and OP=1 when r=1 and OP=2 when r=1 and OP=3 when r=1 and OP=4 ,substitute r for 1, as it represents a constant in this case. . ( ) ,

In case 1, is the inverse of OP, due to the fact that r stays at a constant 1. There is no maximum value of OP, but the minimum value of OP is . This is because when OP becomes larger , chord becomes smaller; but when OP < ,then C3 which contains the point P ceases to exist, meaning that chord will not exist. Figure 5 models this limiting case, where C3 does not exist.Figure 6: Graph of OP vs. OP when r=1

Length of OP' vs. Length of OP when r=12.5

Value of OP'

2 1.5 1 0.5 0 0

y=02 4 6 8 10

Value of OP Figure 6 is a representation of values of OP ranging from 1-10, and the consequent values of OP when r=1. It can be used to justify the modified general statement of when r=1, that is, . This graph represents an inverse relationship between and OP. There is a clear horizontal asymptote at y=0, depicting how as the value of OP gets larger, the value of decreases but never reaches 0. The graph also shows the minimum value of OP and the maximum value of OP, which is 0.5 and 2 respectively. The domain and range for Figure 6 is {xR||x>0.5} and {yR||2>y>0}, where x and y represent the value of OP and the value of , respectively Case 2 In the second case, OP is set as a constant, where OP = 2, and OP must be solved for when r=2, 3 and 4. Another modification to the general statement of statement to fit Case 2 is . must be made to model this case better. OP is substituted for 2, as it is a constant. Thus the modified general

Equation 8: when r=1 and OP=2 when r=2 and OP=2 when r=3 and OP=2 when r=4 and OP =2

In Case 2, is larger than r for every value of r except for when r=1. The same limitation of r < 2OP is evident in Case 2, as C3 becomes non-existent when r>2OP. If C3 is non-existent, then is also non-existent, because point P lies on C3.Figure 7: Graph of OP vs. r when OP=2

Length of OP' vs. Length of r when OP=29 8 7 6 5 4 3 2 1 0

Value of OP'

0

1

2

3

4

Value of r Figure 7 depicts the relationship between r and when OP = 2. The maximum value of r is 2OP because point A ceases to exist when the limit of 2OP is surpassed; therefore C3 also ceases to exist. Since OP is held at a constant 2, 2OP=4, thus the maximum value of r in case 2 is 4. A minimum value of r does exist. The value of r must be greater than 0 due to the fact that a radius of 0 does not produce a value of ,and also C1 and C3 are not formed. The minimum and maximum values of are related to that of the minimum and maximum values of r. The minimum value of is 0, and the maximum value of is half the square of the maximum value of r, that is, = 8.

Validity of General Statement To test the validity of the general statement, different values of r and OP will be put into the general statement, to produce a value of ; and the same values of r and OP will be applied to C1, C2, and C3 on Geogebry graphing software, to virtually compare the lengths of OP. Figure 8: C1, C2, and C3 of when OP=4 and r=3 Example 1: OP=4, r=3

Figure 8 depicts C1, C2, and C3, with the radius of C1 and C3 being 3, and the radius (OP) of C3 being 4. The general statement correctly determined the value of ,as validated by the measurement tool on Geogebra. Both the general statement and the software determined the length of to be 2.25 when r=3 and OP=4.Figure 9: C1, C2, and C3 and OP of when OP=2 and r=1.5

Example 2: OP=2, r=1.5 Yet again the general statement produced a value of OP identical to the value measured by Geogebra, using the Conclusion same r and OP values in both mechanisms. It Chord is represented by the general statement ,where r is the radius of C1 and is reasonable to say C3, OP is the radius of C2general is a chord on C3 and on the radius of C2. The general that the , and statement was derived by utilizing basic trigonometric and chord properties, and also using statement Thames Theorem. The general statement was validated by inputting different values of r and OP into the general formula to generate a value for ; and graphing the three circles along is valid. with the same values of r and OP, and comparing the values of outputted by both mechanisms. The resulting outcome was that both the software and the general statement came up with the same values for , meaning that the general statement was proven to be valid. Chord however, only exists when 0

Recommended

View more >