Mathematical Modelling in the Sciences
(ACM 10070) – Mock Midterm test
Monday 17th October (50 minutes), 12:00
This is a mock midterm test and is designed to give you and me feedback on how
the module is going. It does not count towards your final grade. However, you can use
it to obtain extra credit1 that can be used to top the marks you have received from
the other assignments. You will be handed back your corrected exam in a few days.
1. Solve the following initial-value problems:
dy
+ 3y = 7,
dx
y(0) = 2
and
dy
+ y = cos(2x),
y(0) = 3.
dx
Answer clue: Both questions involve using the integrating-factor technique. In the
second question, you will need to show that
Z
ex cos(2x) dx = 51 ex [2 sin(2x) + cos(2x)] .
2. Solve the equation
dy
+ y = cos2 (x).
dx
Hint: Take as given the following trigonometric identity:
cos2 A = 21 (1 + cos 2A) .
3. Solve the following initial-value problem:
du
= u1/2 t,
dt
Hint: Separation of variables!
u(0) = 0.
4. Let T (t) be the temperature of a hot object at time t. The temperature cools to
the background temperature TB < T (t) according to Newton’s Law of cooling:
dT
= −k (T − TB ) ,
k ∈ R+ .
dt
Solve for T (t). Leave your answer in terms of T0 , the initial temperature of the
object (i.e. T (0) = T0 ).
1
A maximum of 10 bonus points, equivalent to 2.5% of the final grade for the module
1
Mathematical Modelling in the Sciences
Mock Midterm test
Formulæ in the Differential and Integral Calculus
Derivatives
y
xn
sin x
cos x
tan x
cot x
csc x
y./x.
nxn−1
y
sec x
cos x
− sin x
sec2 x
− csc2 x
− cot x csc x
= − cos x/ sin2 x
sin−1 xa
cos−1 xa
tan−1 xa
cot−1 xa
csc−1 xa
y./x.
tan x sec x
= sin x/ cos2 x
y
sec−1
√ 1
a2 −x2
− √a21−x2
a
a2 +x2
a
− a2 +x
2
− x√xa2 −a2
x
a
ex
eax
ax
ln x
y./x.
a
√
x x2 −a2
ex
aeax
ax ln a
1
x
Integrals
y
xn
sin x
R
cos x
sin x
sec2 x
tan x
csc2 x
− cot x
tan x sec x
yx.
xn+1
n+1
n 6= −1
− cos x
sec x
R
y
yx.
cot x csc x − csc x
√ 1
sin−1 xa
a2 −x2
or
− cos−1 xa
a
tan−1 xa
a2 +x2
or
− cot−1 xa
√ a
sec−1 xa
x x2 −a2
or
− csc−1 xa
eax
eax
a
ax
ax
ln a
y
R
sinh x
yx.
ln x
cosh x
sech2 x
tanh x
√ 1
x2 +a2
sinh−1 xa√
2
2
= ln x+ xa +a
√ 1
x2 −a2
cosh−1 x√
a
2
2
= ln x+ xa −a
1
tanh−1 xa
a
1
= 2a
ln a+x
a−x
1
x
1
a2 −x2
2