UNIVERSITY OF WARWICK Year 1
School of Engineering June 2015
ES183 Title of Examination:– Section 1
MATHEMATICS AND SYSTEMS MODELLING Page 1
Question 1 Mark
Allocation
Marks
******************************************* Begin i ************************************* ** Bi **
Answer to question (i):
Given : z1 2 4i z2 3 5i 1 mark
(a) z 3 z1 z 2 2 4i 3 5i 6 10i 12i 20 26 2i
2 2
(b) z3 26 2 676 4 26.08 1 mark
5 3i
(c) Given : z 4 .
3 2i
5 3i 3 2i 15 10i 9i 6 9 19i 9 19
From this z 4 i 0.6923 1.4615i
3 2i 3 2i 9 6i 6i 4 13 13 13 2 marks
******************************************* End i ************************************** **E i**
******************************************* Begin ii ************************************ **B ii**
Answer to question (ii):
The derivatives are :
dy
(a) for y 8 x 4 3 x 3 2 x 9 is
dx
dy 2 marks
32 x 3 9 x 2 2
dx
dy
(b) for y sin cos 3x use chain rule with u cos 3 x
dx
dy d sin u du 2 marks
dx du dx
du d cos 3 x
with 3 sin 3x one gets 1 mark
dx dx
dy d sin u du
cos u 3 sin 3 x
dx du dx
cos cos 3x 3 sin 3 x
3 sin 3x cos cos 3x 1 mark
dy 3x 3
2x 2 5
(c) for y is
dx cos x
Let u 3 x 3 2 x 2 5 and v cos x 1 mark
u dy vu ' uv'
such that y then 2 marks
v dx v2
One gets u ' 9 x 2 4x and v' sin x
2
dy vu ' uv' cos x 9 x 4x 3x 3 2x 2 5 sin x 1 mark
dx 2 2
v cos x
multiplying out to simplify further not required to obtain full marks.
******************************************* End ii ************************************* ** E ii**
, UNIVERSITY OF WARWICK Year 1
School of Engineering June 2015
ES183 Title of Examination:– Section 1
MATHEMATICS AND SYSTEMS MODELLING Page 2
Question 1 Mark
Allocation
******************************************* Begin iii *********************************** **B iii**
Answer to question (iii):
Given: x 2 y 3 x 2 3 y 3 0 . Differentiate through with respect to x .
d x2 y3 d (x 2 ) d (3 y ) d ( 3)
0
dx dx dx dx
d y3 d x2 3 d (x 2 ) d (3 y ) d ( 3)
x2 y 0
dx dx dx dx dx
d y3 d x2 3 d (3 y )
x2 y 2x 0
dx dx dx
d y 3 dy d (3 y ) dy
x2 2 xy 3 2x 0
dy dx dy dx
dy dy
x 2 3y 2 2 xy 3 2x 3 0
dx dx
dy
3x 2 y 2 3 2x y 3 1 0
dx
dy 2x y 3 1 2x 1 y 3
dx 3x 2 y 2 3 31 x2 y2 4 marks
******************************************* End iii ************************************ ** E iii**
******************************************* Begin iv *********************************** ** B iv **
Answer to question (iv):
Give vectors are: a 3i 2 j 4k and b 5i 1j 6k
3 5
(a) Dot (scalar) product: 2 1 3 5 2 1 4 6 15 2 24 11 3 marks
4 6
i j k 2 6 41 12 4 8
(b) Cross product: a b 3 2 4 4 5 3 6 20 18 38 3 marks
5 1 6 31 2 5 3 10 13
******************************************* End iv ************************************* **E iv**
School of Engineering June 2015
ES183 Title of Examination:– Section 1
MATHEMATICS AND SYSTEMS MODELLING Page 1
Question 1 Mark
Allocation
Marks
******************************************* Begin i ************************************* ** Bi **
Answer to question (i):
Given : z1 2 4i z2 3 5i 1 mark
(a) z 3 z1 z 2 2 4i 3 5i 6 10i 12i 20 26 2i
2 2
(b) z3 26 2 676 4 26.08 1 mark
5 3i
(c) Given : z 4 .
3 2i
5 3i 3 2i 15 10i 9i 6 9 19i 9 19
From this z 4 i 0.6923 1.4615i
3 2i 3 2i 9 6i 6i 4 13 13 13 2 marks
******************************************* End i ************************************** **E i**
******************************************* Begin ii ************************************ **B ii**
Answer to question (ii):
The derivatives are :
dy
(a) for y 8 x 4 3 x 3 2 x 9 is
dx
dy 2 marks
32 x 3 9 x 2 2
dx
dy
(b) for y sin cos 3x use chain rule with u cos 3 x
dx
dy d sin u du 2 marks
dx du dx
du d cos 3 x
with 3 sin 3x one gets 1 mark
dx dx
dy d sin u du
cos u 3 sin 3 x
dx du dx
cos cos 3x 3 sin 3 x
3 sin 3x cos cos 3x 1 mark
dy 3x 3
2x 2 5
(c) for y is
dx cos x
Let u 3 x 3 2 x 2 5 and v cos x 1 mark
u dy vu ' uv'
such that y then 2 marks
v dx v2
One gets u ' 9 x 2 4x and v' sin x
2
dy vu ' uv' cos x 9 x 4x 3x 3 2x 2 5 sin x 1 mark
dx 2 2
v cos x
multiplying out to simplify further not required to obtain full marks.
******************************************* End ii ************************************* ** E ii**
, UNIVERSITY OF WARWICK Year 1
School of Engineering June 2015
ES183 Title of Examination:– Section 1
MATHEMATICS AND SYSTEMS MODELLING Page 2
Question 1 Mark
Allocation
******************************************* Begin iii *********************************** **B iii**
Answer to question (iii):
Given: x 2 y 3 x 2 3 y 3 0 . Differentiate through with respect to x .
d x2 y3 d (x 2 ) d (3 y ) d ( 3)
0
dx dx dx dx
d y3 d x2 3 d (x 2 ) d (3 y ) d ( 3)
x2 y 0
dx dx dx dx dx
d y3 d x2 3 d (3 y )
x2 y 2x 0
dx dx dx
d y 3 dy d (3 y ) dy
x2 2 xy 3 2x 0
dy dx dy dx
dy dy
x 2 3y 2 2 xy 3 2x 3 0
dx dx
dy
3x 2 y 2 3 2x y 3 1 0
dx
dy 2x y 3 1 2x 1 y 3
dx 3x 2 y 2 3 31 x2 y2 4 marks
******************************************* End iii ************************************ ** E iii**
******************************************* Begin iv *********************************** ** B iv **
Answer to question (iv):
Give vectors are: a 3i 2 j 4k and b 5i 1j 6k
3 5
(a) Dot (scalar) product: 2 1 3 5 2 1 4 6 15 2 24 11 3 marks
4 6
i j k 2 6 41 12 4 8
(b) Cross product: a b 3 2 4 4 5 3 6 20 18 38 3 marks
5 1 6 31 2 5 3 10 13
******************************************* End iv ************************************* **E iv**
