Prep mathematics for Tech
Table of Contents
- Equations
- Quadratic equation
- System of two linear equations
- Example 2
- Example 3
- Example 4
- Geometry 2d
- Triangles
- Right-angled triangle
- Basic trigonometry
- Trigonometry - example
- Quadrilaterals
- Geometry 3d
- Cylinder
- Cone
- Sphere
Quadratic equation
If the highest power of a variable is 2, then we talk about quadratic equations. The solution is found by substituting multipliers into the formula.[br][br]First, the equation must be rewritten in the standard form[br][br] [math]\large \textcolor{blue}{ax^2+bx+c=0.}[/math][br] [br]Solutions are[br][br] [math]\large \textcolor{blue}{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.}[/math][br][br]If the discriminant (expression inside the square root) [math] b^2-4ac[/math] [br][br][list][*]> 0, there are two different real solutions. [/*][*] = 0, there is only one real solution. [/*][*] < 0, there is no real solution but the complex solution can be found.[/*][/list]
[br][color=#0000ff]Example 1. [color=#000000]Solve [math] 2x^2-7x-5=0.[/math] [/color][/color][br][br]The equations is in a standard form, so [color=#0000ff][color=#000000][math] a=2,\, b=-7 [/math][/color][/color] and [color=#0000ff][color=#000000][math] c=-5.[/math][/color][/color] By substituting these to the formula, we get the solutions:[br][br][math] \begin{array}{rcl}[br]x&=&\frac{-(-7)\pm\sqrt{(-7)^2-4\cdot 2\cdot(-5)}}{2\cdot 2}\\[br]&=&\frac{7\pm \sqrt{49+40}}{4}\\[br]&=&\frac{7\pm \sqrt{89}}{4}[br]\end{array}[/math][br][br][math] x=\frac{7+\sqrt{89}}{4}\approx 4.1 \;\; \vee (=\text{or})\;\; x=\frac{7-\sqrt{89}}{4}\approx -0.6 [/math][br][br][br][color=#0000ff]Example 2. [color=#000000]Solve [math] x(2x-3)-3x(1-x)=-1.[/math] [/color][/color][br][br]Let us simplify the equation first by removing brackets and combining like terms:[br][br][math]\begin{array}{rcll}[br]x(2x-3)-3x(1-x)&=&-1\\[br]2x^2-3x-3x+3x^2&=&-1\\[br]5x^2-6x&=&-1\\[br]5x^2-6x+1&=&0&|\text{Standard form!}[br]\end{array}[br][/math][br][br]As the equations is of degree two (the highest power of the variable), it must be given in the standard form. [color=#0000ff]Values of parameters in a formula are always looked from a standard form.[/color][br][br][math] a=5,\; b=-6,\; c=1[/math][br][br][math]x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 5\cdot 1}}{2\cdot 5}=\frac{6\pm 4}{10}\\ \vspace{15mm}[br] x=\frac{6+4}{10}=1 \;\;\text{or}\;\; x=\frac{6-4}{10}=\frac{2}{10}=\frac{1}{5}[/math][br] [br]
Triangles
Sum of all angles is 180°. The area of a triangle can be solved with the formula[br][br] [math]\LARGE A=\frac{1}{2}ah,[/math][br][br]where [i]a[/i] is the base of the triangle and [i]h[/i] is the perpendicular height of the triangle.[br][br]There are three special cases of triangles and all of them have features simplifying solving:[br][list][*][color=#0000ff]equilateral triangle[/color]: all sides are equal and all angles are 60°. [br] [/*][*][color=#0000ff]isosceles triangle[/color]: two sides of a triangle equal. Thus, base angles are also equal. [br] [/*][*][color=#0000ff]right-angled triangle[/color]: one angle is 90°.[br] [/*][/list]The height of an isosceles or an equilateral triangle bisects the base into two equal parts. [br][br]Identifying the parts of a triangle must be done in a systematic way: a vertex is marked with uppercase letters (e.g. A), the corresponding lowercase letter for the side opposite the vertex (a), and the angle at the vertex is marked with a corresponding greek letter ([math]\Large\alpha[/math] ). Identifying is done anticlockwise.[br][br]
Cylinder
Basic objects in space are [color=#0000ff]cylinder, cone [/color]and [color=#0000ff]sphere[/color]. [br][br][br]In the below graph, you can see two diffenrent cylinders.[br]
[br][br]In technology, special cases of cylinders have got their own "nicknames".[br][br][list][*][color=#0000ff]cylinder[/color] refers to circular ends[br][/*][*][color=#0000ff]prism[/color] refers to polygon ends[/*][/list][br]These names are also used when spoken about cylinders, so mathematical definition is quite unknown: [br] [quote][br][br] [color=#0000ff]Cylinder is formed by using two identical simple closed plane curves that are parallel to each other and joining up corresponding points on each of the curves with straight lines[/color][color=#0000ff].[/color][br][br] [/quote][br][br] [br]A cylinder is called [color=#0000ff]a right circular cylinder[/color] if its ends are circles and the line joining their centres is an axis of symmetry of the cylinder. A cylinder is called [color=#0000ff]a regular cylinder[/color] if its ends are polygons and the line joining their midpoints is an axis of symmetry of the cylinder. [br][br]The basic idea in calculating the volume of a cylinder[br][br] [math]\Large \textcolor{blue}{V=Ah}[/math],[br][br]where [i]A [/i]is the area of the base and [i]h[/i] is the perpendicular height of the cylinder.[br]
Example 1.
[br][br]There was a tubular steel in a classroom. The cross-section of this tubular steel was square with a outer side length of 3 cm and the steel was 1mm thick at every side. Define the weight of this tubular steel, if it has a length of 40 cm.[br][br]The tubular steel is in a shape of cylinder, where the inner part is empty. The area of the cross-section is a subtraction of outer area and empty part.[br][br] [math]A_u=(3\text{ cm})^2=9\text{ cm}^2\\[br]A_s=(3\text{ cm}-2\,\cdot\,0.1\text{ cm})^2=7.84\text{ cm}^2\\[br] A=A_u-A_s=1.16\text{ cm}^2[/math][br] [br][br]Now, the volume can be solved with the basic formula:[br][br] [math] V=Ah=1.16\text{ cm}^2\,\cdot\,40\text{ cm}=46.40\text{ cm}^3.[/math][br] [br]The density of a steel is 7850 kg/m[sup]3[/sup]. The weight of the tubular steel can be solved from the formula[br][br] [math] \rho =\frac{m}{V}\;\;\Leftrightarrow \;\;m=\rho V.[/math][br] [br]As the density is in different unit than area, we must change the unit for cubic centimetres:[br][br] [math]\rho =\frac{7850\text{ kg}}{\text{m^3}}=\frac{7850\text{ kg}}{(100\text{cm})^3}=\frac{7850\cdot 1000\text{ g}}{10^6\text{ cm}^3}=7.85\text{ g/cm}^3.[/math][br] [br]The weight is[br][br] [math]\ m=7.85\text{g/cm}^3\,\cdot \,46.40\text{ cm}^3=364.24\text{ g}\approx 360 \text{ g}.[/math][br] [br][br]
Example 2.
A lamp is hanging in the middle of a room of 6m × 4m. The lamp can be at most 4.5 meters from corners on the floor. How high from the floor can the lamp be?
Solution.
With Pythagoras' theorem is space, we get quadratic inequality:[br][br][math] \begin{array}{rrl}&\sqrt{l^2+p^2+h^2}&\leq 4.5\\[br]\Rightarrow& 3^2+2^2+h^2&\leq 4.5^2\\[br]\Leftrightarrow& h^2&\leq \frac{29}{4}\\[br]\Leftrightarrow& 0\leq h&\leq 2.7 \\[br] \end{array}[/math][br][br]The lamp can be at most 2.7 metres from the floor.
Example 3. Volume of a hole
[br]
Define the volume of the hole.[br][br]When solving the volume of a cylinder, we need the cross-section area perpendicular to the height. Now, the circular hole is not perpendicular to the height of the beam but the length of the hole. So, we should think the volume in parts.
The length of the hole in this case is [br][br][math]\ l=\sqrt{40^2+50^2}\approx 64.0[/math] [br][br]The volume of the hole is the sum of right circular cylinder and the ends. The ends together form a circular cylinder with the height [i]y[/i]. [br][br]Because [i]y[/i] would have been subtracted from the original length, but it would be used in volume of the ends, the volume can be solved with the basic formula: [br][br][math] V=\pi r^2 l= \pi \cdot 10^2\cdot 64.0\approx 20116 \text{ mm}^3 = 20.1 \text{ cm}^3[/math][br][br]