MZB126- Formula- Sheet PDF

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For the final exam of MZB126 you are allowed 2 double sided sheets, this covers everything along with practice exam questions and answers....

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ODES Homogenous, Constant Coefficient ODEs (

Non-Homogenous, Constant Coefficient ODEs

yH ¿ 1. Assume y=C ert (RHS = 0) 2. Determine roots of the characteristic eqn. using quadratic formula (

r=

−b ± √ b2−4 ac 2a

1. Implement same steps to get 2. Make guess for

yP

( y p)

yH

1.

2.

3. Derive y P ' & y P ' ' and sub back into original expression 4. Solve unknown coefficients by equating t, cos, sin or at e terms 5. General soln.

y= y H + y p

Systems of ODES

Eigenvalues & Eigenvectors

y

'

, where

is a vector containing derivatives of the y-

vector, A is a coefficient matrix and b contains any non-y terms. 2. Create the variable w= y ’ , so that

w ’= y ’ ’ i.e. y '' + y ' + y =0 ' ' w + w + y=0→ w =− w− y [ 1 ] y '=w[ 2]

k1

&

Joint Probability (AND): 1. Mutually Exclusive:

P (C )=P ( A ∩ B )=0 2. Independent:

P (C )=P ( A ∩ B )=P ( A ) P(B) 3. Dependent:

|[

]|

factor:

][ ] [ ]

Example:

dy 1 + y=t +4 dt t R (t ) =e t =e =t 1 1 t3 y= ∫ ( t+ 4 ) t dt= ( +2t 2 +C1 ) t t 3 2 C1 t ∴ y = +2 t+ 3 t

i=C

dV dt

(take either row 1

][

]

Probability Bayes’ Rule:

∑

F=ma NSL: inertia & accel) Linear Motion:

di dt

,

1 ∫ v dt L KVL: ∑ V loop =0 KCL: ∑ i ¿ =∑ i OUT (series loop) Ohm’s Law: v =iR (resistor behaviour) i=

Spring:

ds dt

,

a=

ω=

dθ dt

,

v=

Rotational Motion:

∑ T =Iα

&

(Rotational

dv dt α=

dω dt

F s=−kx (acts down & sub straight) 2 F a=−d v (acts down & sub

Air resistance: straight) Damping:

Fd =−cv (acts down & sub straight) Fb (acts up & sub into NSL) Torque: T =F∗d perp , Rotational Damping: T =−cω

Buoyancy:

Distributions

P ( A| B ) P ( B ) P ( B| A )= P (A ) P ( A ) =∑ P ( A∨Bi) P (Bi) i

P(Y )¿

V L =L

dy cos ( t ) = dt sin ( y ) ∫ sin ( y ) dy=∫ cos (t ) dt −cos ( y ) =sin ( t ) +C 1 −sin ( t )−C 1 (¿) ∴ y =cos−1 ¿

Mechanical Modelling

1 V C = ∫ idt , C

Capacitor:

Inductor:

4000 c 1 +4,000,000 c 2=0 , c 1=−1000 c 2 −1000 c 2 c −1000 ~ c 1= 1 = = (do same 1 c2 c2 process for λ2 )

[][

(incl.

ln ( t)

) dt

Example:

+C!)

1

or row 2)

k2

integrating

Electrical Modelling

det|A−λI |=0 (apply quadratic formula) − λ 4,000,000 = ad−bc=0 i.e. −1 (−5000−λ ) Eigenvectors: ( A− λI )~ c =0 Eigenvalues:

[

2

4. Use initial conditions to solve

&

C2

λ1=−4000 4000 4,000,000 c 1 0 = −1 −1000 c 2 0

λ t λ t y = k 1~ c1 e + k2 ~ c2 e 1

C1

i.e. For

3. Calculate eigenvalues/vectors to get:

Calc.

∫ 1 dt

6. If given, use initial conditions to solve for

'

1.Convert ODE to the form: f ( y ) dy = g (t 2. Integrate both sides (+C on one side) 3. Rearrange to solve y

to:

P ( t ) dt

*RESONANCE PROBLEMS: Multiply guess for y p if y H is the same

y = Ay + b

ODE

R (t ) =e∫ 1 y= 3. ∫ Q ( t ) R ( t ) dt R(t)

i.e. r 2−6 r +5=0 3. Use the table below to determine general soln.

1. Write system in form:

Convert

Separation (1st order ODEs – non-linear)

dy + P ( t ) y=Q (t) dt

by looking at RHS and using table:

)

at e ( C1 cos ( bt ) +C2 s 4. If given, use initial conditions to solve C 1 & C2

Integrating Factor (1st order ODEs)

Binomial (Discrete) �~ �(�, �) Probability of a Bernoulli random variable (2 outcomes) occurring x times, over n trials. (Lizard question!

Poisson (Discrete) 𝑁 ~ �(�) Events occur at avg rate, and time between events is independent (# cars in a minute).

n! n− x x p ( 1− p ) x! ( n−x ) ! E(N )=np , Var (N )= np (1− p ) x −λ λ e P (N =x )= p ( x )= x! E(N )=λ , Var (N )=λ P(N=x )= p(x )=

P ( A ∩ B) = P( A) P ( A ∩ B ) =P ( B∨A ) P( A )

P (C )=P ( B∨ A ) Same as: OR Relationship:

P ( A ∪ B ) =P ( A )+P ( B )−P( A ∩ B)

( A B ) P (B) P ( B| A )= P | ∑ P( A∨ Bi )P( Bi )

P ( A ∪ B ∪ C )= P (A ) + P ( B ) +P(C ) Often simpler to calculate the complement:

´ P ( A ∪ B ) =1−P ( A´ ∩ B) i.e. P ( X ≥ 1 )=1−P(X =0)

x∈X

E ( X ) =∫ x∗f (x )

Var ( X )= E (( X−μ )2 )= ∑ ( x−μ )2 p(x ) x∈X

x

{

(Continuous)

Variance:

P ( X ≤ x )=F ( u)=∫ f ( u ) du

2

Var ( X )= E ( X ) −E (X ) =∫ ( x−μ ) f (x )

is the pdf)

2

2

1 p(x)= e 2σ √2 π σ X−μ Z= , E ( X )=μ , Var (X )=σ 2 σ [ Z N ( 0,1) ]

P(X ≤ 8) 2

}

− ( x−μ)

Gaussian/Normal (Continuous) 𝑁 ~ �(�, �2) Bell curved, the outcome is commonly close to the mean and rarely far away, utilise a standard normal table to evaluate.

(Discrete)

0

f (u )

(Discrete)

X

Cumulative Distribution Function (CDF):

p( x ) =λ e− λx P (T 2 (or use t n−2, 1 −α/2 (n-2 for linear) and T-table), then there is strong evidence to reject the nullhypothesis. Adjusted R-squared: indicates how much variation in y is explained by x in the linear model. In other words, it represents how well the line of best fits accounts for variance in the data. 95% Confidence Interval ( α =0.05 ¿

CI = ^β0 ± tn−2, 0.975∗S β CI = ^β ±t ∗S 1

Where,

n−2,0.975

0

β1

(SE Column)

Miscellaneous '

u

'

u

u ⋅v =u v +uv ' , Chain Rule: e =u e u u ' v −u v ' , Trigonometric: s  c -s  -c  s (deriving) = Quotient Rule: v v2 Integration by Parts: ∫ uv=u ∫ v−∫(u ' ∫ v) A 1 B = + Partial Fractions: P ( K−P ) P K−P Product Rule:

Inference (hypothesis testing and confidence intervals for sample data) Hypothesis testing: Sample mean and variance: 1 1. H 0 : μ=μ0 (Null Hypothesis)

´x =

N

∑ xi

2.

1 s= ∑ ( xi −´x )2 N−1

H A : μ ≠ μ0

(Alternative Hypothesis)

2

Test statistic:

CI = ´x ±

t n−1,1−α/ 2∗s √N

If

~ 95% sure that the true mean lies within range

(SE Column) P-value: ~ Determine the value of

|T test|=t n−1,1−α/ 2 If If

α=0.05 α =0.01

´x −μ0 s /√ N |T test| >t n−1,1−α/ 2

T test =

Confidence interval:

α

that makes:

some evidence strong evidence

Or, in general: T test 100) =1−0.7549 ≈ 0.24 Trains depart from the station every 60 mins, you arrive at a random time. What is the probability you wait more than 15 mins? Uniform, as area under curve is 1:

' t t ∴ y ( t )=e (C 1 cos ( √ 7 t ) +C2 sin ( √ 7 t ) )+ e ( − √ 7 C1 s ' ∴ y ( 0)=0=C 1 +√ 7 C2 , C2 =− 10 / √ 7 10 t sin ( √ 7 t ) ∴ y (t)=e 10cos (√ 7 t )− √7

(

Assume that at a factory, that 1 in 1000 of the parts produced are found to be defective. If the factory produces 100 parts per hour, what is the probability that there will be a day without defects? Average defects per day = λ=0.001∗100∗24 = 2.4 (Poisson)

2.4 (¿ ¿ 0 ∙ e−2.4 ) =0.091, 9.1% 0! P ( X=0 )=¿ Over 10 days, what is the probability that at least two of the days pass without defects. This can be considered as a binomial distribution with n=10, p=0.091 :

P ( X ≥ 2 )=1−(P ( X =0 ) + P ( X =1 ))

)

An unforced damped vibration system can be modelled by the 2 d y c dy k + + y =0 . Given 2 d t m dt m m=5 kg , c=1 Nms , k =10 N /m and the system is released from rest at a

ODE:

height of 10 cm, get a specific solution for the vibration. 2 k d y dy +0.2 + 2 y=0 2 dt m dt 2 r +0.2 r +2=0

PDF : f (t ) =1/60 60 CDF : F ( t ) =∫ f ( t ) dt

0 1 10−1 10! 0.091) (1−0.091) 10−0 + 0.091) (1−0.091 ) ) P ( X ≥ 2) 10 ! 1! ( 10 −1 ) ! ( =1−( 0 ! (10 −0 ) ! ( P ( X ≥ 2 )=1−( 0.3851579196 + 0.3855816357 ) ≈ 0.2292∨22.92 %

60 60 t F ( t ) =[ ] = − 15 =75 % 60 15 60 60

R2=1−

15

What is the probability you wait between 5 and 15 mins? 15

15 5 F ( t ) =[ t ] = − =16.6 60 5 60 60 What is the avg. wait time? 60

60

0

0

E ( T) =∫ t f ( t ) dt=∫

[ ] 2

E ( T) =

t 120

60

= 0

t dt 60

602 −0=30 120

2 ∑ ( y i −^y i) 2

∑ ( y i− ´y )

2

=1−

( n−2 )s ∑ y i2− ´y 2

• The number of girls in a family of 8 people = Binomial • The number of people who pass you by as you walk from Central Station to QUT = Poisson • The number of red lights you hit as you walk from Central Station to QUT = Binomial (Success or Failure) • The position a dart lands when thrown by at a dart board, by someone without skill = Uniform • The time between when you and your best friend enter the MZB126 lecture = Exponential • The height of the next person you see = Normal

2

0.2 ) −4∗1∗2  −0.2 ± √( 2∗1 r= r=−0.1± 1.4107 i(Complex roots) −0.1 t ∴ y (t)=e ( C 1 cos (1.4107 t ) +C 2 sin ( 1.4107 t ) ) −0.1∗0 y (0)=e ( C1 cos ( 1.4107∗0) +C 2 sin ( 1.4107∗0) ∴C 1=0.1 −0.1t ' ∴ y ( t )=−0.1 e (C 1 cos ( 1.4107 t ) +C2 sin ( 1.410 ' y ( 0 ) =−0.1 C1 +1.4107 C 2=0, 0.1∗0.1 ∴C 2= =0.007088 1.4107 −0.1 t ( 0.1 cos (1.4107 t ) +0.007089 sin(1.41 ∴ y ( t )=e −0.1 t ∴ y ( t )=e ( 10 cos( 1.4107 t ) +0.7089 sin ( 1.4107...