UNIVERSITY OF DUBLIN
XMA1E02a
TRINITY COLLEGE
Faculty of Engineering, Mathematics
and Science
school of mathematics
JSF Engineering
JS MEMS
JF MSISS
Trinity Term 2015
MA1E02 — Engineering Mathematics II
May ??
Dr. M. Kulaxizi
Answer ALL questions
All questions have equal weight and all parts have equal weight
Page 2 of 11
1. (a) Find
dy
dx
XMA1E02a
if y = (arccos(x))2
Solution:
dy
arccos (x)
= 2 arccos (x)(arccos(x))0 = −2 √
dx
1 − x2
(b) Find the limits
i.
lim
x→0
sin(7x)
tan(3x)
Solution
lim
x→0
sin(7x)
7 cos(7x)
7 cos(7x) cos2 (3x)
7
= lim
=
lim
=
3
x→0
tan(3x) x→0 cos2 (3x)
3
3
ii.
lim+
x→π
sin(3x)
x−π
Solution
lim+
x→π
sin(3x)
3 cos(3x)
= lim+
= −3
x→π
x−π
1
iii.
(ln (x))2
x→∞
x
lim
Solution
2 ln (x)
(ln (x))2
lim
= lim
x→∞
x→∞
x
1
1
x
= lim
x→∞
2
x
1
=0
iv.
lim x7 e−7x
x→∞
Solution:
x7
x→∞ e7x
lim x7 e−7x = lim
x→∞
The exponential function goes to infinity faster than any polynomial, so the
limit is zero. Otherwise use the l’Hopital rule.
Page 3 of 11
XMA1E02a
(c) Assume that a population y decays according to the formula below for time t ≥ t0
y = ae−k(t−t0 ) + b
i. What is the population y equal to at time t = t0 ?
Solution: The polulation at time t = t0 is
y(t0 ) = a + b
ii. At what rate is the population y decreasing at time t = 5t0 ?
The rate is given by the derivative at t = 5t0 , therefore it is equal to
y 0 (t = 5t0 ) = −ake−k(5t0 −t0 ) = −ake−4kt0
iii. At what time t is the population equal to half its original value (original value
of the population is at time t = t0 )?
Solution: We need to solve the following algebraic equation for t:
a−b
a−b
1 a−b
a+b
= ae−k(t−t0 ) +b ⇔ e−k(t−t0 ) =
⇔ −k(t−t0 ) = ln
⇔ t = t0 − ln
2
2a
2a
k
2a
Page 4 of 11
XMA1E02a
2. (a) Evaluate the following integrals
i.
Z
sinh6 (x) cosh (x)dx
Solution:
Set u = sinh (x) ⇒ du = cosh (x)dx to rewrite the integral as follows
Z
Z
6
sinh7 (x)
u7
+C =
+C
7
7
u6 du =
sinh (x) cosh (x)dx =
ii.
Z
xe−x dx
Solution (by integration by parts)
Z
Z
−x
−x 0
−x
x(−e ) = −xe
xe dx =
Z
+
e−x = −xe−x − e−x + C
iii.
Z
π
cos2 xdx
0
Solution (using a trigonometric identity)
Z
π
2
π
Z
cos xdx =
0
0
1 + cos 2x
dx =
2
π
Z
0
Z π
Z
dx
cos 2x
1 π 1 2π
cos ydy =
+
dx = x|0 +
2
2
2
4 0
0
1
1
π 1
π
= π + sin x|2π
+ (sin 2π − sin 0) =
0 =
2
4
2 4
2
iv.
Z
x2
1
dx
−9
Solution
We use the method of partial fractions to rewrite
(x2
1
1
1 1
1 1
=
=
−
− 9)
(x − 3)(x + 3)
6x−3 6x+3
Page 5 of 11
XMA1E02a
so that
Z
Z Z
dx
1
dx
−
=
x−3 6
x+3
1
1
1 x − 3 = ln |x − 3| − ln |x + 3| + C = ln +C
6
6
6
x + 3
1
dx =
2
x −9
1 1
1 1
−
6x−3 6x+3
1
dx =
6
Z
v.
Z
+∞
e
1
dx
x ln3 x
Solution
Set u = ln x ⇒ du =
Z
e
+∞
dx
x
so that
ln b
Z ln b
dx
du
u−2 =
= lim
= lim
3
b→+∞ (−2) b→+∞ 1
u3
e x ln x
1
1
1
1
1
= lim −
− 1 = − (0 − 1) =
2
b→+∞
2 ln b
2
2
1
dx = lim
b→+∞
x ln3 x
Z
b
(b) A bullet of mass m, fired straight up with an initial velocity v0 , is slowed down
by the force of gravity mg and a drag force of air resistance kv 2 , where g is the
gravitational constant of acceleration and k a positive constant. As the bullet
moves upward, its velocity v satisfies the equation
m
dv
= −(kv 2 + mg)
dt
.
i. Show that if x = x(t) is the height of the bullet above the barrel opening at
time t, and v =
dx
,
dt
then
mv
dv
= −(kv 2 + mg)
dx
Solution
Using the chain rule, we can rewrite
dv
dv dx
dv
=
=v
dt
dx dt
dx
Page 6 of 11
XMA1E02a
Substituting to the given differential equation we find that
m
dv
dv
= −(kv 2 + mg) = mv
= −(kv 2 + mg)
dt
dx
ii. Solve the above differential equation to find x as a function of v, given that
the initial velocity of the bullet, for x = 0, is v = v0 .
Solution
dv
mvdv
mv
= −(kv 2 +mg) ⇔
= −dx ⇔ m
dx
(kv 2 + mg)
Z
vdv
=−
(kv 2 + mg)
Z
dx
The integration can be easily performed by substituting u = v 2 ⇒ du = 2vdv
m
2
Z
du
=−
ku + mg
Z
m
dx ⇔
2k
Z
du
=−
u + mg
k
Z
dx ⇔
m mg ln u +
= −x+C ⇔
2k
k
m 2 mg m 2 mg ⇔
ln v +
= −x + C ⇔ x = C −
ln v +
2k
k
2k
k
The final step is to fix the constant C by using the fact that when x = 0,
v = v0 . Substituting into the last equality we find that
0=C−
m 2 mg m 2 mg ln v0 +
⇒C=
ln v0 +
2k
k
2k
k
so that
m
x=
ln
2k
v02 +
v2 +
mg
k
mg
k
iii. Using your result in part (b) find the maximum height the bullet reaches. Use
the fact that the maximum height is attained when the velocity v vanishes.
Solution
xmax
m
=
ln
2k
v02 + mg
k
0 + mg
k
m
=
ln
2k
kv02 + mg
mg
Page 7 of 11
XMA1E02a
3. (a) Given the Maclaurin series for the function arctan x,
arctan x =
∞
X
(−1)k
k=0
x2k+1
2k + 1
show that the irrational number π is equal to:
∞
X
(−1)k
π=4
2k + 1
k=0
Solution
We know that arctan 1 =
π
4
therefore
∞
∞
∞
2k+1
X
X
(−1)k
(−1)k
π X
k 1
(−1)
=
=
⇔π=4
4
2k + 1 k=0 2k + 1
2k + 1
k=0
k=0
(b) Determine the value of the infinite sum
∞
X
3
4k
k=0
Solution
This is a geometric series
P∞
k=0
ark with geometric ratio r =
a = 4. We know that when the geometric ratio is r =
1
4
ark =
k=0
a
1−r
therefore
∞
X
3
3
=
k
4
1−
k=0
1
4
=
12
=4
3
(c) Show that the following series converges using the ratio test
∞
X
k
ek
k=1
Solution
and overall factor
< 1 the series converges
and the sum is equal to
∞
X
1
4
Page 8 of 11
XMA1E02a
The ratio test instructs us to the find the limit of the following ratio
ρ=
k+1
ek+1
lim
k→∞ kk
e
k+11
1
= <1
k→∞
k e
e
= lim
thereore the series converges.
(d)
i. Using the Maclaurin series for the function
∞
X
kxk =
k=0
1
,
1−x
show that for x ∈ (−1, 1)
x
(1 − x)2
.
Solution
We know that
∞
X
1
=
xk ,
1 − x k=0
, x ∈ (−1, 1)
Differentiating both sides of the equality we arrive at
∞
X
x
=
kxk ,
(1 − x)2
k=0
x ∈ (−1, 1)
ii. Use (i) to determine the value of the sum
∞
X
k
3k
k=0
Solution
Since the equality we proved in (i) is valid for all x ∈ (−1, 1), it is also valid
for x = 13 . So,
∞
1
X
k
3
=
1 2 =
k
3
(1
−
)
3
k=0
1
3
4
9
=
9
3
=
12
4
Page 9 of 11
XMA1E02a
4. (a) Find the inverse, if it exists, of the matrix

3 1 1




A=
1 3 1
1 1 3
Solution
To find out if an inverse exists, we need to compute the determinant of the matrix.

3 1 1




det A = 
1 3 1 = 3
1 1 3
3 1
!
−1
1 3
1 1
!
1 3
+1
1 3
!
=
1 1
= 3(9 − 1) − (3 − 1) + (1 − 3) = 20 6= 0
therefore A has an inverse. To find the invesre, we need to compute all the
cofactors of A and form the transpose of the matrix of cofactors from A. The
cofactors of A are:
C11 = 8, C12 = −2, C13 = −2,
C21 = −2, C22 = 8, C23 = −2,
C31 = −2, C32 = −2, C33 = 8
and the inverse is:

A−1 =
8
−2 −2


1 
 −2 8 −2 

20 
−2 −2 8
(b) Solve if possible, the following system of equations by finding the inverse of the
matrix of coefficients
3x + y + z = 1
x + 3y + z = 2
x + y + 3z = 3
Page 10 of 11
XMA1E02a
Solution
The matrix of the coefficients is

3 1 1



1 3 1


1 1 3
and is equal to A from (a) where we found its inverse to be

A−1 =
8
−2 −2


1 
 −2 8 −2 

20 
−2 −2 8
The solution to the system of equations can be found as follows
 1 
 

 
− 10
1
8 −2 −2
x






 
 y  = 1  −2 8 −2   2  ==  2 
 5 
 
  20 
9
3
−2 −2 8
z
10
1
which implies that the system has a unique solution, (x = − 10
, y = 25 , z =
9
).
10
(c) Solve the following system of equations
x + y + 3z = 6
2x + 2y + 6z = 12
2x + y + 6z = 11
How many solutions are there?
Solution
We observe that the second equation is just the first one multiplied by an overal
factor of 2. The system is therefore reduced to
x + y + 3z = 6
2x + y + 6z = 11
Page 11 of 11
XMA1E02a
Solving for x = 6−y −3z from the first and substituting into the second we obtain
12 − 2y − 6z + y + 6z = 11 ⇔ y = 1
Substituting back into the solution for x we obtain the solution of the system
(x = 5 − 3s, y = 1, z = s)
Alternatively, the system can be solved with Gauss-Jordan elimination or any other
method. The number of solutions is infinite.
(d) Assuming that all matrices are n × n and invertible, solve for the matrix D:
ABC T DBAT C = AB T
Solution
Since all the matrices have an inverse, we can simply multiply in the order written
to obtain
D = (C T )−1 B −1 A−1 AB T C −1 (AT )−1 B −1
c UNIVERSITY OF DUBLIN 2015