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MSE2183 Assignment 3 of 2022

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MSE2183 Teaching Geometry Mathematics 218 Assignment THREE of 2022 solutions.

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  • August 25, 2022
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  • 2022/2023
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MSE2183 ASSIGNMENT 3 2022

SECTION A

Question 1

1.1

Activity 2.1

Number 1

1.4




1.5




Number 2

2.1

,3.1




1.2

Activity 2.2

Number 1




Number 2

,Number 3




Activity 2.3

‘Software for mathematics Visualisation – review by John’

1. The uses of sketchpad as a geometry teaching tool.

The Geometer’s Sketchpad is a computer software that may be used to draw graphs and make
constructions in Mathematics. The software can also be specifically used in a Geometry lesson. Arcs,
polygons, and coordinates may be drawn/plotted using the geometer’s sketchpad. The software can
analyse and interpret the drawn figures. Images of drawings may be imported into the sketchpad.
These imported images may be edited by changing orientation or size.

2. An outline of similar programs that are related to a sketchpad.

Graphmatica

This is a graphing software that plots graphs of equations and inequalities including polar, parametric
and ordinary differential equations.

RedCrab

This is an advanced calculator that handles complex calculations. It can also read notation from
uploaded images.

Advanced Trigonometry Calculator

This is an advanced calculator that handles not only trigonometric calculations. It also solves
quadratic equations, linear equations, problems in geometry and those in linear programming.

Graphing Calculator 3D

This is a graphing software that plots 2D and 3D graphs, including implicit equations, parametric
equations, inequalities and contour plots. The graphs may also be transformed and viewed using
different graphics and colours.

xFunc

This is an advanced calculator that handles complex calculations and shows the working or steps. The
solutions are displayed using relevant mathematical notation and my be copied from the software.

, Activity 4.7

𝑉 = 350 ml

𝑉 = 350 cm3

𝑉 = 𝜋𝑟 2 ℎ
350 = 𝜋𝑟 2 ℎ

350
=ℎ
𝜋𝑟 2

350
ℎ=
𝜋𝑟 2

𝑇𝑆𝐴 = 2𝜋𝑟 2 + 2𝜋𝑟ℎ

350
𝑇𝑆𝐴 = 2𝜋𝑟 2 + 2𝜋𝑟 ( )
𝜋𝑟 2

350
𝑇𝑆𝐴 = 2𝜋𝑟 2 + 2𝑟 ( )
𝑟2

350
𝑇𝑆𝐴 = 2𝜋𝑟 2 + 2 ( )
𝑟

2 × 350
𝑇𝑆𝐴 = 2𝜋𝑟 2 +
𝑟

700
𝑇𝑆𝐴 = 2𝜋𝑟 2 +
𝑟
700
Let 𝑓(𝑟) = 2𝜋𝑟 2 + 𝑟


𝑓 (𝑟) = 2𝜋𝑟 2 + 700𝑟 −1

𝑓 ′ (𝑟) = 2 × 2𝜋𝑟 2−1 + (−1) × 700𝑟 −1−1

𝑓 ′ (𝑟) = 4𝜋𝑟1 − 700𝑟 −2

𝑓 ′ (𝑟) = 4𝜋𝑟 − 700𝑟 −2

When 𝑓 attains its minimum, 𝑓 ′ (𝑟) = 0

4𝜋𝑟 − 700𝑟 −2 = 0

4𝜋𝑟1 × 𝑟 2 − 700𝑟 −2 × 𝑟 2 = 0 × 𝑟 2

4𝜋𝑟1+2 − 700𝑟 −2+2 = 0𝑟 2

4𝜋𝑟 3 − 700𝑟 0 = 0

4𝜋𝑟 3 − 700 (1) = 0

4𝜋𝑟 3 − 700 = 0

4𝜋𝑟 3 = 700

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