QUESTION / ANSWER BOOKLET ENGSCI 211
THE UNIVERSITY OF AUCKLAND
SUMMER SCHOOL 2021 Campus: City
MATHEMATICAL MODELLING 2 Term Test
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(Time allowed: 90 minutes)
Answer all questions.
The answer for each question should be written in the space provided. If you
run out of space, extra space is available at the back of this booklet.
An appendices booklet containing information for the Data Analysis section and a formula sheet is provided.
This is a restricted book lite test. This means that you may bring in a single piece of A4 paper that is written or typed on both sides, NO attachments; glued, staples or otherwise attached. This will be checked prior to the start of your test.
Calculators are not permitted.
Surname: Forename(s): University of Auckland ID number:
Question 1 2 3 4 5 6 7 Total Outof 10 4 5 5 10 6 10 50
Page 1 of 11
QUESTION / ANSWER BOOKLET ENGSCI 211 ID:
SECTION A – Ordinary Differential Equations Question 1 (10 marks)
Consider the ordinary differential equation:
y′′ +4y′ +8y = 16t2 +4
subject to the initial conditions:
y(0)=1 , y′(0)=2
(a) Find the complementary function for the unforced response, yc. Write this in a trigono-
metric form if appropriate. (4 marks)
Homogeneous form of the ODE:
y′′ + 4y′ + 8y = 0 Trial solution for complementary function:
yc = Aeλt Characteristic equation by substituting in trial solution:
Roots of characteristic equation: Complementary function:
λ2 +4λ+8=0
λ = −2 ± 2i
yc = e−2t C1e2it + C2e−2it
Rewriting in a trigonometric form:
yc =e−2t(Acos(2t)+Bsin(2t))
(b) Find the particular integral for the forced response, yp. (3 marks)
Page 2 of 11
QUESTION / ANSWER BOOKLET ENGSCI 211 ID:
Trial solution of a similar form to f (t) of ODE:
yp = b0t2 + b1t + b2
No resonance as not of similar form to any term in yc. Derivatives of this: yp′ =2b0t+b1
yp′′ = 2b0
2b0 +8b0t+4b1 +8b0t2 +8b1t+8b2 = 16t2 +4
By equating like terms:
8b0 = 16 8b0 + 8b1 = 0 2b0 +4b1 +8b2 =4
Which can be solved for unknowns to give b0 = 2, b1 = −2 and b2 = 1. The particular intergral is given by:
yp = 2t2 − 2t + 1
Substituting into the ODE:
(c) Find the total solution to the ODE. (3 marks)
General solution equal to y = yc + yp: y=e−2t(Acos(2t)+Bsin(2t))+2t2 −2t+1
It’s first derivative:
y′ =−2e−2t(Acos(2t)+Bsin(2t))+e−2t(−2Asin(2t)+2Bsin(2t))+4t−2
Applying the initial conditions:
A+1=1 −2A + 2B − 2 = 2
Which can be solved for A = 0 and B = 2. The total solution is: y=2e−2tsin(2t)+2t2 −2t+1
Page 3 of 11
QUESTION / ANSWER BOOKLET ENGSCI 211
Question 2 (4 marks)
Consider the ordinary differential equation:
y′ =(t−y)2
subject to the initial condition y(0) = 0. Perform one iteration of the improved Euler
method for a time step of ∆t = 1.
Governing equations required for one iteration of the method: k1 = (tn − yn)2
yE =y +∆t·k n+1 n 1
k=t −yE 2 2 n+1 n+1
yIE =y+∆t(k+k) n+1 n 2 1 2
Solving each of these and entering results into a table:
n t y k yE k yIE n n 1 n+1 2 n+1
Page 4 of 11
QUESTION / ANSWER BOOKLET ENGSCI 211
Question 3 (5 marks)
Use the method of Laplace transforms to solve the ordinary differential equation: y′′ + 9y = 3 δ (t − 2)
y(0)=0 , y′(0)=0
Take Laplace transform of ODE and apply initial conditions: s2Y +9Y =3e−2s
Y=e−2s· 3 s2 + 9
Find inverse Laplace transform of F (s):
f(t)=L−1 3
Through the shift-in-t theorem, the ODE solution is: −1−2s 3
s2 + 9 f (t) = sin (3t)
y (t) = L e · s2 + 9
y (t) = u (t − 2) f (t − 2)
y (t) = u (t − 2) sin (3 (t − 2))
Page 5 of 11
QUESTION / ANSWER BOOKLET
ENGSCI 211
Question 4 (5 marks)
Identify the location of all five of the stationary points of:
f(x,y)= 2 −2x2y−xy2 +4xy+1
You do not need to classify these stationary points.
Both first-order partial derivatives:
∂f =xy2 −4xy−y2 +4y
∂f =x2y−2×2 −2xy+4x ∂y
Simplify one of these equations, ∂f , to find where it is equal to zero: ∂y
x (xy − 2x − 2y + 4) = 0 x (x − 2) (y − 2) = 0
This indicates that this partial derivative is equal to zero for x = 0, x = 2 and y = 2. Substituting each value into ∂ f to find where it is also equal to zero.
For x = 0:
−y2 + 4y = 0 y (4 − y) = 0
Both derivatives are zero when x = 0 and y = 4, or when x = 0 and y = 0. For x = 2:
2y2 − 8y − y2 + 4y = 0 y (4 − y) = 0
Both derivatives are zero when x = 2 and y = 4, or when x = 2 and y = 0. For y = 2:
4x − 8x − 4 + 8 = 0 x=1
Both derivatives are zero when y = 2 and y = 1.
To summarise, there are five stationary points: (0, 0), (0, 4), (2, 0), (2, 4) and (1, 2).
Page 6 of 11
QUESTION / ANSWER BOOKLET
ENGSCI 211
Question 5 (10 marks)
(a) Consider the double integral:
with the following region of integration, R:
sinx2 +y2 dxdy
Evaluate this double integral using polar coordinates.
Page 7 of 11
QUESTION / ANSWER BOOKLET ID:
ENGSCI 211
Integral limits in polar coordinates:
3π 3 2
I= f(r,θ)|J|drdθ
The Jacobian in polar coordinates is |J| = r. Re-writing the function from Cartesian to polar coordinates:
f (r,θ) = sinr2 The integral in polar coordinates is:
The inner integral can be solved via substitution, u = r2:
3π 3 2
rsin r drdθ
A (θ) = 12 [− cos (u)]94
A(θ)= cos(4)−cos(9) 2
sin (u) du
Solving the outer integral:
3π cos(4)−cos(9)
I = π(cos(4)−cos(9)) 2
(b) Consider the double integral:
Sketch the region of integration and re-write the integral after changing the order of
integration. You do not need to evaluate the integral. (3 marks)
Page 8 of 11
f (x, y) dx dy
QUESTION / ANSWER BOOKLET ENGSCI 211 ID:
Sketch of the integration region:
Re-writing the integral with changed order of integration:
2 2x 0 x2
f (x, y) dy dx
Page 9 of 11
QUESTION / ANSWER BOOKLET ENGSCI 211 ID:
SECTION C – Data Analysis Question 6 (6 marks)
(a) Comment on the validity of the following claim:
A two-sample t-test on the difference in average ride-share waiting times between two platforms returned a p-value of 0.82. This gives us certainty
that the null hypothesis must be true.
The following questions relate to the analysis given in Appendix A: Orange Juice – One-sample Analysis.
(b) Briefly comment on the exploratory analysis presented.
(c) Interpret the p-value resulting from the t.test() run on this dataset.
No, as there may be a small non-zero difference that might also not be inconsistent with the data we have observed. We cannot state for certain that there is no difference at all.
Center: about 250 mg/L
Spread: Between 100 and ∼ 400 (or 300 with 2 possible outliers) Skew: right skewed
No evidence against the hypothesis that the median pectin content in the orange juice is 240 mg/L
Page 10 of 11
QUESTION / ANSWER BOOKLET
ENGSCI 211
Question 7 (10 marks)
The following questions relate to the analysis given in Appendix B: Orange Juice – Regression Analysis.
(b) With reference to an appropriate plot, comment on whether oj.lm satisfies the equality of variance assumption. (1 mark)
(c) Assuming that the relevant assumptions are satisfied, write a brief Executive Sum- mary of the main conclusions you would draw from this analysis. (3 marks)
(a) Write down the regression equation fitted in oj.lm. iid
2 Sweetnessi = β0 + β1 × Pectini + εi where εi ∼ N (0, σ )
The residual plot shows broadly constant scatter, except for two points with very low sweetness – so EOV probably ok.
Our interest in the data is to estimate the relationship between the amount of pectin detected in orange juice, and its sweetness as determined by professional tasters.
We estimate that for each additional mg of pectin per litre in orange juice, the mean sweetness score decreases by between 0.00043 and 0.0042 points.
Our model explains 23% of the variation in sweetness scores.
(d) Give predictions for the sweetness of an individual batch of orange juice, where the pectin content was:
(i) 100mg/L (ii) 300mg/L
and comment briefly on the validity of each prediction. (2 marks)
(e) The orange juice manufacturer has collected more data on the chemical composition of batches of orange juice, and the corresponding subjective sweetness score as evaluated by professionals. A machine learning model trained on this data is to be used to predict how consumers would perceive the sweetness of each batch of orange juice. State a concern you might have with this model. (1 mark)
(f) A colleague suggests that fitting a quartic term (i.e. a x4 term) would improve the model fit and hence result in better predictions in the future. Evaluate briefly whether this is
100mg/L: not valid, outside range.
300mg/L: within range, so is valid / low R2 means perhaps invalid – score between 5.1 and 6.0.
Training set is biased as it uses professional scores, not that of consumers who may have different tastes / preferences.
a good idea.
Not a good idea; although this improves training quality, it would lead to increased variance in predictions on future data.
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