ES 183 June 2017 Exam Solutions

UNIVERSITY OF Warwick Year 1
School of Engineering June 2017
ES183 Title of Examination: -
ENGINEERING MATHEMATICS AND SYSTEMS
MODELLING Page 1

Question 1-3 Mark
Allocation
Marks
***************************************** Begin i ******************************************* *** Bi ***
The diagonal can be expressed as: (2 marks)

****************************************** End i ******************************************* *** Ei ***

***************************************** Begin ii ****************************************** *** Bii ***
Given: 2 3 , 3 2 4 and 5 7 6

i j k 1 4 3 2 4 6 2
(3 marks)
(i) a b 2 1 3 3 3 2 4 9 8 1
3 2 4 2 2 13 4 3 1
2 5 (3 marks)
(ii) ∙ 1 ∙ 7 2∙5 1∙7 1∙ 6 9
1 6
Note: When vector of part (i) was incorrect but, if then, the scalar product in part (ii) is calculated
correctly, based on the wrong result of part (i), then full marks will be given for part (ii).
****************************************** End ii ******************************************* *** Eii ***

***************************************** Begin iii ****************************************** *** Biii ***
3
Vector form is: ∙ 3 14 (2 marks)
1
****************************************** End iii ******************************************* *** Eiii ***

***************************************** Begin iv ****************************************** *** Biv ***
(a) ln : Apply product rule :
Use: , ln → 2 ln 2 ln (3 marks)

(b) √ 1 : Apply chain rule :
√ 1 1 → 1 ∙2 1


1 1 (3 marks)

(c) 3 : Use implicit differentiation :

2 , 1, 3 , 3 3


Hence 2 1 3 3


Rearranging yields (3 marks)
(Note that ⁄ is given in terms of and .)
****************************************** End iv ******************************************* *** Eiv ***

, ***************************************** Begin v ****************************************** *** Bv ***
(a) z1 3 4 i

r 32 42 25 5
4
and tan 0.9273 or 53.13 (2 marks)
3
1 2i
(b) Given is z 2
4i
1 2i 4i 4i 8 1 1
z2 z2 i (2 marks)
4i 4i 16 2 4
n
(c) De Moivre’s theorem is : cos i sin cos n i sin n
5
Given tan 68.1986 1.190 rad
2
8
Hence, z 38 5.385 cos 68.1986 i sin 68.1986
5.3858 cos 8 68.1986 i sin 8 68.1986
707107.9 cos 545.589 i sin 545.589 (3 marks)
707107.9 0.99525 0.097387 i 703749.1 68863.14 i
(Equivalent working in radians is of course accepted as correct too)
****************************************** End v ******************************************* *** Ev ***


***************************************** Begin vi ****************************************** *** Bvi ***
3 2 6 1
Given with 2 3 .
8 5 2 5


18 (3 marks)
Multiplying out for gives: 18 17 →
17
After multiplying out left-hand-side and equating:
3 2 18
8 5 17
(3 marks)
Solving system of simultaneous linear equations yields 4, 3
****************************************** End vi ******************************************* *** Evi ***


*** Bvii ***
***************************************** Begin vii ******************************************
x 3 3x 5
Given function is: y
x2 3
2
3
x 3x 5 x x 3 5 x x2 3 5 5
“Divide out”: y 2 2 2 2
x 2
x 3 x 3 x 3 x 3 x 3
5
For x , then 2 0
x 3
(3 marks)
Hence, the oblique asymptote is: y x
*** Evii ***
****************************************** End vii ******************************************


***************************************** Begin viii ***************************************** *** Bviii ***
To show that 6 -5x is odd we must show that the condition is satisfied.
Now, 6 5 6 5 since 1
6 5
But this is . Therefore we have shown and, consequently is an odd function. (3 marks)


****************************************** End viii ****************************************** *** Eviii ***