Math
s | Mathematics B Assignment | Linear, periodic and exponential functions | | HARBECK, Henry | Due: 2nd November |
Contents Contents 1 PART A – KNOWLEDGE AND PROCEDURES 2 KAP Q1 2 KAP Q2 3 KAP Q3 5 KAP Q4 6 PART B – MODELLING AND PROBLEM SOLVING 8 MAPS Q1 8 MAPS Q2 9 MAPS Q3 11 MAPS Q4 18 References 26
PART A – KNOWLEDGE AND PROCEDURES
All working for Part A is done in centimetres unless otherwise stated.
KAP Q1 Function | Domain | Range | Equation | AB | 0 ≤x≤1 | y=4 | y=4 | CD | 1 ≤x ≤10 | y=1 | y=1 | DE | 10 ≤x ≤17 | 1 ≤y ≤4.5 | y= 0.5x-4 |
The functions AB and CD were found to be constants.
AB was found to be y=4 from 0 ≤x ≥1
CD was found to be y=1 from 1 ≤x≥10
DE was found to be y=0.5x-4 from 1≤x≥4.5
Working for DE
Points are (10, 1) and (17, 4.5) m= y2-y1x2-x1 m= 4.5-117-10 m= 3.57 m=.5 y= .5x+c
Substitute in (10, 1)
1= .5 ×10+c
1= 5+c
-4=c
c=-4 y= .5x-4
KAP Q2
Functions used for the base of the wine glass were y=4 y= -4 x=0 x=1
Functions used for the stem of the wine glass were x=1 x= -1
Functions used for the bowl of the wine glass were y= .5x-4 y= -.5x+4 x=17 Functions used for the first initial were x=3 x=5 y= -6
Functions used for the second initial were x=7 x=9 y= -6
KAP Q3
The value assigned for this question is 10.0
Since the allocated value is the same as the value in the graph of KAP Q2 no changes have been made, except the initials have been removed.
The volume of a truncated cone is V (cm3)=13πr12+r1r2+r22h
Where V= volume, r1= the larger radius, r2= the smaller radius and h= height r1=4.5, r2=1, h=7
V= 13π4.52+4.5*1+127
V= 13π20.25+4.5+17
V=13π25.757
V= 13π(180.25)
V=188.7573586cm3
V≈188.76cm3
KAP Q4
The trapezoidal rule is abfxdx= 12wE+2M
Where E=fa0+f(an) is the sum of the end values
And M=fa1+fa2+ . . .+f(an-1) is the sum of the middle values
For this question 7 sections will be made.
The width (w) of each
Contents Contents 1 PART A – KNOWLEDGE AND PROCEDURES 2 KAP Q1 2 KAP Q2 3 KAP Q3 5 KAP Q4 6 PART B – MODELLING AND PROBLEM SOLVING 8 MAPS Q1 8 MAPS Q2 9 MAPS Q3 11 MAPS Q4 18 References 26
PART A – KNOWLEDGE AND PROCEDURES
All working for Part A is done in centimetres unless otherwise stated.
KAP Q1 Function | Domain | Range | Equation | AB | 0 ≤x≤1 | y=4 | y=4 | CD | 1 ≤x ≤10 | y=1 | y=1 | DE | 10 ≤x ≤17 | 1 ≤y ≤4.5 | y= 0.5x-4 |
The functions AB and CD were found to be constants.
AB was found to be y=4 from 0 ≤x ≥1
CD was found to be y=1 from 1 ≤x≥10
DE was found to be y=0.5x-4 from 1≤x≥4.5
Working for DE
Points are (10, 1) and (17, 4.5) m= y2-y1x2-x1 m= 4.5-117-10 m= 3.57 m=.5 y= .5x+c
Substitute in (10, 1)
1= .5 ×10+c
1= 5+c
-4=c
c=-4 y= .5x-4
KAP Q2
Functions used for the base of the wine glass were y=4 y= -4 x=0 x=1
Functions used for the stem of the wine glass were x=1 x= -1
Functions used for the bowl of the wine glass were y= .5x-4 y= -.5x+4 x=17 Functions used for the first initial were x=3 x=5 y= -6
Functions used for the second initial were x=7 x=9 y= -6
KAP Q3
The value assigned for this question is 10.0
Since the allocated value is the same as the value in the graph of KAP Q2 no changes have been made, except the initials have been removed.
The volume of a truncated cone is V (cm3)=13πr12+r1r2+r22h
Where V= volume, r1= the larger radius, r2= the smaller radius and h= height r1=4.5, r2=1, h=7
V= 13π4.52+4.5*1+127
V= 13π20.25+4.5+17
V=13π25.757
V= 13π(180.25)
V=188.7573586cm3
V≈188.76cm3
KAP Q4
The trapezoidal rule is abfxdx= 12wE+2M
Where E=fa0+f(an) is the sum of the end values
And M=fa1+fa2+ . . .+f(an-1) is the sum of the middle values
For this question 7 sections will be made.
The width (w) of each
References: http://keisan.casio.com/has10/SpecExec.cgi?id=system/2006/1223372110 - Used for KAP 3 New Q Maths by R. Brodie and S. Swift P258 http://www.physicsforums.com/showthread.php?t=638735 – Mark 44 post #5 – “Here’s why…” – used for MAPS 3 and 4 TO DO Re-do graphs - maps 2, 3, 4 Label graphs - maps 3 Printing money