201MS Coursework 2 (2015–2016) Page 1 of 15
COURSEWORK 2
Multivariable Calculus
This is an individual assignment which should be handed in to the ECB Assessment
Office before 16:00, 10 December 2015.
Estimated time: 12–15 hours.
Percentage of coursework component: 70
Intended Learning Outcomes:
1. solve selected second order linear, constant coefficient differential equations;
2. use techniques from multivariable calculus;
3. apply mathematical techniques and modern mathematical software to
engineering problems;
4. find Fourier series in trigonometric form and apply this to the solution of
partial differential equations;
Solution of the wave equation: 20 marks
MATLAB simulation: 10 marks
Maxima and minima: 25 marks
Total: 55 marks
For full marks, all working must be shown. All program files should be submitted
on paper as part of your working.
201MS Coursework 2 (2015–2016) Page 2 of 15
Part 1: TheWave Equation
The problem is concerned with a uniform string stretched along the x-axis between
x = 0 and x = L, where L is determined by the last digit of student ID, lSID, as
L = 1 lSID < 5,
2 lSID ≥ 5.
Thus, lSID = and L = (fill in boxes). For the rest of
the assignment use this value for L rather than variable L.
The one-dimensional wave equation is given by
1
c2
¶2u
¶t2 =
¶2u
¶x2 , 0 ≤ x ≤ L, t ≥ 0, (1)
where u(x, t) represents the displacement of the string at position x and time t. It is
assumed that the string is held fixed at both ends and so we have
fill in boxes (1 mark)
u( , t) = u( , t) = 0, t ≥ 0. (2)
Now assume the string is raised a distance μ at x = b and then released from rest at
t = 0. Zero initial velocity implies that
fill in boxes (1 mark)
¶u
¶t
( , ) = 0, 0 ≤ x ≤ L. (3)
The initial displacement (plotted in figure 1) is given by
fill in formula for b ≤ x ≤ L (2 marks)
u(x, 0) = u0(x) =

μx/b 0 ≤ x ≤ b
b ≤ x ≤ L.
(4)
We begin the assignment with some hand-calculations. Most of the calculations are
provided, you need only fill in the missing details. Space is provided at the end for
this. If you wish to word process your solutions, simply create your own template so
that it is clear which part of your solution refers to which part of the assignment.
Step 1: Use the method of Separation of Variables to derive solutions to (1)–(3)
having the form
u(x, t) =
¥
å
n=1
An sin
npx
L
cos
npct
L
, (5)
201MS Coursework 2 (2015–2016) Page 3 of 15
0 b L
0
μ
Figure 1: Initial displacement of string.
where An, n = 1, 2, . . . are constants that need to be determined in order that u(x, t)
also satisfies the constraint of (4). (Note: You may assume that the constant value
referred to by k in lecture notes or by a in the handout ‘PDE’s I: TheWave Equation’
is negative.)
Solution:
We begin by looking for solutions to Equations (1)–(3), having the form
u(x, t) = X(x)T(t). (6)
Substituting this into the wave equation gives
put result into box below (1 mark)
Dividing through by u(x, t) = XT then gives
1
c2
T′′
T
=
X′′
X
. (7)
The left-hand side is now a function of t only and the right-hand side is a function of
x only. It follows that both sides must equal the same constant value k say. Let us
first analyse the equation
X′′
X
= k ⇐⇒ X′′ − kX = 0. (8)
We only consider the case where k < 0. Let us first note that Equation (2) implies
that X(0)T(t) = X(L)T(t) = 0. Since we are not interested in the trivial solution
where T(t) ≡ 0, we deduce that
fill in boxes (1 mark)
X(0) = X(L) = .
201MS Coursework 2 (2015–2016) Page 4 of 15
Since k < 0, we set k = −p2, : Equation (8) has the general solution
put general solution into box below (2 marks)
for arbitrary constants A and B. The boundary conditions X(0) = X(L) = 0 give rise
to fill in working in box below for nontrivial solutions of X(x) (4 marks)
for any constant (fill in A or B as appropriate) and any positive integer n.
Returning to (7), we also have
1
c2
T′′
T
= k = −p2 = − np
L 2
,
where L is as setup on page 1.
Rearranging this produces
T′′ + npc
L 2
T = 0. (9)
The general solution to this is given by T(t) = C cos
npct
L
+ Dsin
npct
L
where C and
D are arbitrary constants. However, Equation (3) leads to D = 0.
show working for this in the box below (3 marks)
201MS Coursework 2 (2015–2016) Page 5 of 15
Therefore
T(t) = C cos
npct
L
.
for any constant C and any positive integer n. Combining our expression for X(x)
and T(t), we arrive at the following solutions to equations (1)–(3) from the
assignment:
u(x, t) = BC sin
npx
L
cos
npct
L
. (10)
Since the equation to be solved is linear we can add such solutions together to arrive
at the desired expression (Equation (5)):
u(x, t) =
¥
å
n=1
An sin
npx
L
cos
npct
L
. (11)
where An, n = 1, 2, . . . are arbitrary constants until we add the constraint of
Equation (4).
It follows from (11) that (1 mark)
u(x, 0) = u0(x) = (12)
Assuming An can be found to satisfy (12), we must have
An =
2
L Z L
0
u0(x) sin
npx
L
dx (13)
201MS Coursework 2 (2015–2016) Page 6 of 15
Step 3: Use this formula (in Matlab or Maple or by hand), together with the
expression for u0(x), as given by (4), to obtain (4 marks)
An = (14)
If using Matlab, you will need to set up μ, b, x and n as symbolic variables. Just use
mu to represent μ. Split the integral of (13) into two parts, given the change of
formula for u0(x) at x = b, as given in (4). After adding the two parts together,
simplify the resulting formula using the simplify command. In Maple, it is possible
to specify that n is integer valued, but I don’t know if you can do this in Matlab. You
will therefore need to simplify sin(np) as zero yourself.
Putting this expression for An into (5) gives
u(x, t) =
2μL2
p2b(L − b)
¥
å
n=1
1
n2 sin
npb
L
sin
npx
L
cos
npct
L
(15)
Step 4:
We finish the assignment by constructing an animation showing the vibration of the
string. Animation in Matlab and Maple was introduced in Question 3 in the
‘Introduction to Fourier Series’ handout.
If you are using Matlab, then you can fill in the missing details in the following
m-file. (4 marks)
% m-file for simulating one-dimensional wave equation.
% initialise constants
L= ; mu = ; N = ; b = ; c = 100;
% set up x-vector for plots
x = L*linspace(0,1,500);
% set up limits for x and y axes
v = [0 L -1.1*mu 1.1*mu];
% set up value for time step T
% T is the period of oscillation divided by 15
201MS Coursework 2 (2015–2016) Page 7 of 15
T = ;
% set up for loop for incrementing t in 16
% equal steps of length T
for m = 1:16
t = (m-1)*T;
% initialise u to be zero then sum first N terms
u = 0;
for n = 1:N
u = u+2*mu*L^2/(pi^2*b*(L-b)*n^2)
*sin(n*pi*b/L)*sin(n*pi*x/L)*cos(n*pi*c*t/L);
end
% plot this sum and store for movie using getframe
figure(1),plot(x,u),axis(v)
M(m) = getframe;
% replot in a 4X4 array for displaying frames
figure(2),subplot(4,4,m),plot(x,u),axis(v)
end
% show movie with 10 periods
figure(1),movie(M,10)
% display the individual frames
figure(2)
Attach your printouts of the Figure 2 window for three different numbers of the terms of
Fourier series given by N = 5, 10, 50. Comment on the effect of N on the accuracy of
solution by comparing the initial deflection of string (4) with solution (15) at t = 0 for
different values of N. (6 marks)
Use the following steps (or your own variation on them). No further guidance on
Maple is given here, but you can ask if you are interested.
1. Assign a meaningful value for μ. Do not copy a friend’s values, use your own!
201MS Coursework 2 (2015–2016) Page 8 of 15
2. Choose your value of b as b = L(lSID + 1)/11 using the last digit of your
student ID , lSID.
3. Create a suitable vector of x-values for simulating the vibrating string.
4. Create a vector to be used in an axis command to specify the limits on the x
and y axes for the simulation.
5. Set up a FOR loop, for example using the command
>> for m = 1:16
The idea is that each iterate of the loop will plot and store the string profile for
a specific time instant. These time instants should start at t = 0 and be equally
spaced, say T seconds apart. Assuming you are going to produce an animation
consisting of 16 time instants, the last plot will be for t = 15T. You should then
pick T such that 15T is one complete period. By this means, looping the
animation will produce a sensible animation for as many periods as you desire.
You should calculate what the period is, which will depend on your chosen
value for c. You may wish to assign this value for T in Matlab, before the FOR
loop (since it doesn’t change, putting such a command inside a loop is poor
programming).
6. Define the time instant. For example using
t = (m-1)*T;
where T has been constructed as suggested earlier.
7. Set up another FOR loop (inside the current one). The idea here is that this
loop will construct the sum of the first N terms of the Fourier sum given by
(15) at the time instant t. This is similar in nature to that given in the Fourier
series handout and so no further details are given.
8. Plot this truncated sum, using prescribed axes limits, into Figure 1 and store
the result using getframe. Also store the result into an array of plots in
Figure 2. This then ends the outer FOR loop.
9. Use the movie command to display the animation for 10 periods.
201MS Coursework 2 (2015–2016) Page 9 of 15
Part 2: Maxima and Minima
A tent has volume V which is determined by the last digit of student ID, lSID, as
V = =

4 m3 if lSID = < 5,
6 m3 if lSID = ≥ 5.
(fill in boxes). For the rest of the
assignment, use this value of V where required. The tent has the shape shown in the
figure with ends but no floor. It is desired to make the tent with the least possible
material.
✂
✂
✂
✂
✂
✂
✂
❆
❆
❆
❆
❆
❆
❆
✏✏✏✏✏✏✏✏✏✏✏✏✏✏
✏✏✏✏✏✏✏✏✏✏✏✏✏✏
❆
❆
❆
❆
❆
❆
❆
2w
q
L
q
1. Derive a formula for the volume, V, in terms of w, L and q. Then rewrite this
formula with L as the subject.
Solution: The area of each triangular end is given by Ae = wh, where the tent
height is given by h = . (1 mark)
It follows that the volume is given by V = . (1 mark)
Rewriting with L as the subject gives (1 mark)
L = . (16)
2. Derive a formula for the area of material, A, in terms of w, L and q and then
substitute for L as given by Eq. (16).
201MS Coursework 2 (2015–2016) Page 10 of 15
Solution: The area of each end is Ae and the area of each side is As = LH,
where H is the length of each upper side of the triangular ends, i.e.,
H = . (1 mark)
The total area is therefore given by A = 2Ae + 2As, which simplifies down to
A = 2w2 tan q +
2V
w
csc q
(where csc q = (sin q)−1).
Show the working for this in the box below (3 marks)
Given that V is a constant, this gives A as a function of the 2 variables, w and q.
3. Using either hand-calculation or Matlab (with Symbolic Toolbox), find the two
partial derivatives,
¶A
¶w
and
¶A
¶q
. (Aside: You can useMaple if you prefer, but
you may have trouble using Maple for part (d)). You will find Matlab
commands to answer question 5 from the Maxima and Minima handout (but
without any explanatory comments) at the end of the assignment. You can
modify these appropriately.)
201MS Coursework 2 (2015–2016) Page 11 of 15
Solution: InsertMatlab commands or hand calculations into the box below
with the final results in the following boxes (4 marks)
This gives
¶A
¶w
=
¶A
¶q
=
4. Find the stationary point (subject to the constraints that w > 0 and 0 < q <
p
2
).
If using Matlab, use the solve command. By default, Matlab will also find
stationary points that do not obey the above 2 constraints (even complex
solutions). You will have to pick out the relevant one.
Substitute the values w = w0, q = q0 say, as given by the stationary point for
the volume of the tent specified above, into the formula for L in part (a) to find
the corresponding value L = L0.
201MS Coursework 2 (2015–2016) Page 12 of 15
Solution: Attach Matlab commands and/or hand-calculations under heading
Question 1(d). (6 marks)
q0 =
w0 =
L0 =
5. Finally, use the second derivatives test (or otherwise) to show that the
stationary point of part (d) is a local minimum. Again, you may useMatlab (or
Maple) if you wish.
Solution: First we calculate the second partial derivatives evaluated at the
stationary point . Put theMatlab commands or the hand calculations for
¶2A
¶w2 ,
¶2A
¶q2 , and
¶2A
¶w¶q
at (w0, q0) into the box below. Solution: Put explanation in the
box below (6 marks)
Explain how the above can be used to show the stationary point is indeed a
local minimum.
201MS Coursework 2 (2015–2016) Page 13 of 15
Solution: Put explanation in the box below (2 marks)
TotalMarks 55
201MS Coursework 2 (2015–2016) Page 14 of 15
Possible Matlab commands for Part 2.
Please note that the order of solutions returned by the Matlab ‘solve’ command
matches the order of the unknowns specified in the command. The order of
solutions is not affected by the names of variable on the left hand side. For example,
the command ‘[a b]= solve(eq1,eq2,x,y)’ returns solution of two simultaneous
equations ‘eq1’ and ‘eq2’ first for the unknown ‘x’ and then for the unknown ‘y’,
which are stored as ‘a’ and ‘b’, respectively.
>> syms w d V
>> A = 3*w*d + 2*V/w + 2*V/d;
>> Aw = diff(A,w); pretty(Aw)
V
3 d – 2 —-
2
w
>> Ad = diff(A,d); pretty(Ad)
V
3 w – 2 —-
2
d
>> [wsol dsol] = solve(Aw,Ad,w,d)
wsol =
[ 1/3*18^(1/3)*V^(1/3)]
[ -1/6*18^(1/3)*V^(1/3)+1/6*i*3^(1/2)*18^(1/3)*V^(1/3)]
[ -1/6*18^(1/3)*V^(1/3)-1/6*i*3^(1/2)*18^(1/3)*V^(1/3)]
dsol =
[ 1/3*18^(1/3)*V^(1/3)]
[ -1/6*18^(1/3)*V^(1/3)+1/6*i*3^(1/2)*18^(1/3)*V^(1/3)]
[ -1/6*18^(1/3)*V^(1/3)-1/6*i*3^(1/2)*18^(1/3)*V^(1/3)]
>> w0 = wsol(1); pretty(w0)
201MS Coursework 2 (2015–2016) Page 15 of 15
1/3 1/3
1/3 18 V
>> d0 = dsol(1); pretty(d0)
1/3 1/3
1/3 18 V
>> h0 = V/(d0*w0); pretty(h0)
1/3 1/3
1/2 18 V
>> Aww = subs(diff(Aw,w),{w,d},{w0,d0})
Aww =
6
>> Add = subs(diff(Ad,d),{w,d},{w0,d0})
Add =
6
>> Adw = subs(diff(Ad,w),{w,d},{w0,d0})
Adw =
3
>> Delta = Aww*Add – Adw^2
Delta =
27

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