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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: jasoN- on October 07, 2010, 10:50:58 pm

Title: (Perhaps) Useful Formulas + basic revision
Post by: jasoN- on October 07, 2010, 10:50:58 pm: This is a revision of the basics and formulas that may be useful, enjoy.

Things that we may need to know that may or may not supplied (from what I've seen used in the VCAA exams) include:
PLEASE FEEL FREE TO ADD THINGS + CORRECT ANY ERRORS!

Area of a segment(radians): $A = \frac{R^2}{2}\left(\theta-\sin\theta\right)$ may be helpful when a line intersects a circle 'find the shaded area'
Area of a sector(radians): $A = \frac{1}{2} R^2 \theta$
Area of a sector(degrees): $A = \pi R^2 \cdot \frac{\theta}{360}$ (R = radius)

Supplementary angles (made this one up from some other formula): $cos^{-1}{\frac{A}{B}} + cos^{-1}{\frac{-A}{B}} = \pi$ where $A<B A,B \in R+$
Don't know what this is may be used for but: $sin^{-1}{\frac{A}{B}} + sin^{-1}{\frac{-A}{B}} = 0$ where $A<B A,B \in R+$

Complex numbers:
Typical complex number circle formulas: (good to know, as these questions pop up in multiple choice 'which one does not represent a circle in an Argand diagram')
$|z-a-bi|=R$ , centre $(a,b)$ radius $R$
$(z-a-bi)(\bar{z}-a+bi)=R^2$ , centre $(a,b)$ radius $R$

'Arg(z)' laws:
$-\pi<Arg(z)\leq\pi$
$Arg(\frac{z_1}{z_2}) = Arg(z_1) - Arg(z_2)$ where $Arg(z_1), Arg(z_2) \in Arg(z)$
$Arg(z_1z_2) = Arg(z_1) + Arg(z_2)$

If a polynomial has all real coefficients, then roots contain even numbers of Imaginary (eg. if $z^3+2z^2+z+8=0, z=x+yi, z=x-yi, z=w$ ) (Conjugate root theorem) else does not apply.

Linear:
Midpoint $(x_1, y_1), (x_2, y_2): (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
Bisector: $|z-a-bi|=|z-x-yi|$ i.e. line perpendicular of the midpoint between points $(a,b)$ and $(x,y)$
$m=\tan{\theta}$ from positive x-axis
Cubic factorisation:
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$

Vectors:
Scalar resolute of $\tilde{a} \parallel \tilde{b} =\tilde{a} \cdot \tilde{b}$
Vector resolute of $\tilde{a} \parallel \tilde{b} =(\tilde{a} \cdot \Hat{\tilde{b}})\Hat{\tilde{b}}$
Vector resolute of $\tilde{a} \perp \tilde{b} =\tilde{a} - (\tilde{a} \cdot \Hat{\tilde{b}})\Hat{\tilde{b}}$
Angle between two vectors: $\theta=\cos^{-1}\frac{\tilde{a}\cdot\tilde{b}}{|\tilde{a}||\tilde{b}|}$
Speed = $|\dot{r}|$ where r is the position vector

Tip: In questions that have 3-Dimensional vectors, $a\tilde{k}$ being the altitude, let the component of $\tilde{k}=0, ie. a=0$ to find when the particle 'hits' the ground

Linear dependence:
If one vector can be expressed by a scalar multiple of the others, then the vectors are linearly dependent.
ie. $\tilde{c}=m\tilde{a}+n\tilde{b}$
$If k_1\tilde{a}+k_2\tilde{b}+k_3\tilde{c}\neq0$ and all 'k' values equal 0 then the vectors are linearly independent

Calculus:
If $\frac{dP}{dt} \varpropto P$ then $\frac{dP}{dt} = kP$
Newton's Law of Cooling: $\frac{dT}{dt}-k(T-T_s), T_s$ = Temperature of surroundings

Partial fractions:
Type 1: Linear Factors $\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$
Type 2: Repeated Factor $\frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}$
Type 3: Irreducible quadratic factors $\frac{1}{ax^2+bx+c} = \frac{Ax+B}{ax^2+bx+c}$

Inverse circular functions domain and range restricting:
eg. 'State the domain and range of $\frac{a}{\pi}cos^{-1}(b(x-\frac{\pi}{c}))+d$
Start with the 'original' domain and range
Domain: $x\in[-1,1]$
Substitute x for whatever's inside the inverse function, i.e. $b(x-\frac{\pi}{c})$
$\Rightarrow -1\leq b(x-\frac{\pi}{c}) \leq 1$
Simple rearrangement to find domain:
$\Rightarrow \frac{-1}{b}+\frac{\pi}{c}\leq x \leq \frac{1}{b}+\frac{\pi}{c}$

Range(slightly different approach): $y\in[0,\pi]$ , Let $b(x-\frac{\pi}{c})$ = x for simplicity
$\Rightarrow 0\leq cos^{-1}(x) \leq \pi$
Observe the given equation; first multiply by $\frac{a}{\pi}$ then add $d$
$\Rightarrow 0\cdot\frac{a}{\pi}\leq cos^{-1}(x) \leq \pix\frac{a}{\pi}$
$\Rightarrow 0\cdot\frac{a}{\pi}+d\leq cos^{-1}(x) \leq \pix\frac{a}{\pi}+d$
$\Rightarrow d\leq cos^{-1}(x) \leq a+d$

Kinematics:
(only for constant acceleration)
If the sign (i.e. + or -) of the acceleration = the sign of the velocity, the particle is speeding up
If the sign (i.e. + or -) of the acceleration $\neq$ the sign of the velocity, the particle is slowing down up

Distance = speed x time
Average speed = total distance travelled / time (only positive)
Average velocity = displacement / time (can be positive or negative)
Instantaneous velocity: $\frac{dx}{dt}, \dot{x}, x'(t), v$

In a velocity-time graph the DISTANCE = the area under the curve (if under x-axis then take the negative)
   DISPLACEMENT = Add positive areas, subtract negative areas
eg.
velocity
|
|
|\
| \
| \
| \ /¯¯| Distance = A1 + A2 + A3 (magnitude)
| A1\ /A3| Displacement = A1 - A2 + A3
-----\-------/---------------time
| \_A2_/
|
|

Mechanics
Changing momentum $\Delta{P}=m\Delta{v}$ , m is mass in kg, v is velocity
Friction: Friction is greatest when sliding is about to occur. When the body is about to slide down, its state is known as 'limiting equilibrium'.
Limiting friction force: $F_{max}=\mu R$
Friction opposes motion, however it cannot exceed $F_{max}$ .
When motion is not about to occur $0\leq F<\mu R$

Recognising which method for integral calculus:
1) Divide first if possible $\frac{x^4+3x}{x}$
2) If the numerator is the derivative of the denominator -> log recognition $\frac{2x^3}{1+2x^4}$
3) Can you factorise the denominator? Use partial fractions $\frac{3x}{6x+2x^2}$
4) Inverse circular functions $\frac{2}{\sqrt{8+4x^2}}$
5) Using substitution (change of variable) $\frac{e^{tan^{-1}x}}{1+x^2}$

Slope fields:
If $\frac{dy}{dx}$ is in terms of x, the gradients follow the y-axis (i.e. on a point x, there is only one gradient)
If $\frac{dy}{dx}$ is in terms of y, the gradients follow the x-axis (i.e. on a point x, there is more than one gradient)

Nature of a graph f(x)=y, some questions ask for 'show that point P is a maximum' etc. (imagination required)
$-----------------------------------------------------------$
$----- \frac{d^2y}{dx^2}>0 -------- \frac{d^2y}{dx^2}<0 -------- \frac{d^2y}{dx^2}=0$

$\frac{dy}{dx}>0$ increasing concave up increasing concave down positive P.O.I

   | /¯¯ /¯¯
   __/ | __/ notice: P.O.I is has the steepest gradient

$\frac{dy}{dx}<0$ decreasing concave up decreasing concave down negative P.O.I

   | ¯¯\ ¯¯\
   \__ | \__

$\frac{dy}{dx}=0$ local minimum local maximum stationary point of inflection

   \ / /¯¯\ __/ \__
   \__/ / \ / or \

$-----------------------------------------------------------$
Title: Re: (Perhaps) Useful Formulas + basic revision
Post by: aznanthony on October 07, 2010, 10:58:06 pm: Thats a nice list, thanks for posting it :)
Title: Re: (Perhaps) Useful Formulas + basic revision
Post by: TrueTears on October 07, 2010, 11:01:33 pm: Quote from: jasoN- on October 07, 2010, 10:50:58 pm
'Arg(z)' laws:
$-\pi<Arg(z)\leq\pi$
$Arg(\frac{z_1}{z_2}) = Arg(z_1) - Arg(z_2)$
$Arg(z_1z_2) = Arg(z_1) + Arg(z_2)$

Not true in general.

Think about what you can do to generalise it :)
Title: Re: (Perhaps) Useful Formulas + basic revision
Post by: jasoN- on October 07, 2010, 11:06:56 pm: edit:nvm this is complex lmao
tell me please :D
Title: Re: (Perhaps) Useful Formulas + basic revision
Post by: TrueTears on October 07, 2010, 11:10:08 pm: Think about what happens when you use the above to simplify expressions involving Arg(z), what if the final result is not between -pi and pi? Think about how you would normally correct that so it IS between -pi and pi and then generalise it with your expressions above.
Title: Re: (Perhaps) Useful Formulas + basic revision
Post by: jasoN- on October 07, 2010, 11:18:39 pm: $Arg(z_1), Arg(z_2) \in Arg(z)$ and hence in the domain?