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 …show more content…
Firstly the base was designed.
Functions used for the base of the wine glass were Function | Domain | Range | y=4 | 0≤x≤1 | | y=-4 | 0≤x≤1 | | x=0 | | -4≤y≤4 | x=1 | | -4≤y≤4 |
Then the stem was then designed. A trigonometric function was used for the stem of the graph.
Firstly, y=sin(πx) and y=-sin(πx) were entered as functions. The two functions were given domains of 1≤x≤5 and ranges from -1≤y≤1 to make sure that the total length of the wine glass did not exceed 15cm.The result is shown below.
These functions were then changed from sin to cos. With this change, the domains and ranges for both functions stayed consistent. The result is shown below.
As shown above, the function y=±cos(πx) has an opening at the end of the stem. The domain of the both of these functions was changed to 1≤x≤3.5 to solve this problem. The result is shown below.
Functions used for the stem of the wine glass were Function | Domain | Range | y=cos(πx) | 1≤x≤3.5 | -1≤y≤1 | y=-cos(πx) | 1≤x≤3.5 | -1≤y≤1 |
Lastly, the bowl of the wine glass was designed. The bowl of the wine glass must have an x intercept of 3.5 to meet with the stem. First of all the functions y=lnx and