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Tom Obrien

HNC Level 4

Unit Number and title: Unit 1
Analytical Methods for Engineers
Assignment Title: Calculus
Assignment Number: 3
Date: November 2016

1

Tom Obrien

HNC Level 4

Contents Page:
Page 1:
Page 2:

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Title Page

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Contents Page

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Task 3
...
2
Question 1
Question 2
Question 3
Question 4

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Task 3
...
4
Question 1

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Task 3
...
4
Question 3

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Task 3
...
5
Question 5a
Question 5b
Question 5c
Question 5di

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Task 3
...
4
Question 6a
Question 6b
Question 6c

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Page 7 to 8:

Page 9:

Page 9 to 10:

Page 10:

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Page 14 to 15:

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2

Tom Obrien

HNC Level 4

3

Task 3
...
2
1)
dy
= 6cosec 3t
dx

dy
= 18cosec 3t
dt
substitute t =

π
4

3π
3π
dy
= 18 cosec ×
× cot
4
4
dt
cot

2)

3t
3π
= tan
= −1
4
4

36 × −1 = −36
y = 3√x

2

y = 3x sin2x

1
Lny = ln(3) + ln(x) + ln(sin2x)
2
1 2cos2x
1 dy
= 0 + +
2x
sin2x
y dx
dy
1 2cos2x
=y
+
dx
2x
sin2x

3)

dy
= 3√x sin
dx

1 2cos2x
+
2x
sin2x

2arctan2x
u=

u = 2x

2
×2
1 + (2x)

u=

4
1 + 4x

Tom Obrien

HNC Level 4

4)
arcosh4x
u=
u=

u = 4x
1

(4x) − 1
4

√16x − 1

6

Tom Obrien

HNC Level 4

7

Task 3
...
4
1)
y=

x=

2x
− 5x + 12x − 7
3

dy
= 2x − 10x + 12
dx

−b ± √b − 4ac 10 ± (10 − 4 × 2 × 12)
=
2a
2×2
x=

10 + 2 10 − 2 10 ± √4
=
=
4
4
2×2
+x = 3 and − x = 2

d y
=2
d x

− 10 + 12 = 4 − 10

x = 3 so 4 × 3 − 10 = 2 (MAX)

2)

x = 2 so 4 × 2 − 10 = −2 (MIN)
A(surface area) = πr + 2πrh
v = 1000000 = πr h
h=

10
πr

a = πr + 2 ×

10
πr

a = πr + 2 × 10 × r
du
= 2πr
dx

da
= 2πr − 2 × 10 r
dr
= 2πr −

2x10
=0
r

2πr =

2x10
r

=0

Tom Obrien

HNC Level 4

r =

v+

v=

v=

v=

π
(hr + 2r
3

r=

π
(r +
3

π
(r +
3

π
(r
3

rh + h
v=

r) × (h +

r) × (h +

r 2r

r +r

2πrh
×
3

v=

2x10
r

3√10
= 68
...
2

π
h) − r h
3

πr
×
3

2πrh du
=
3
dv

dv
×
dh

h)

π
h) − r h
3

r

h+

π

h

h

r = 8cm
h = 0
...
2) ( × 8 × −0
...
702cm

π
h) − r h
3

Tom Obrien

HNC Level 4

11

4)
H = KI Rt
R=

H
KI t

1
dr
=
dt KI t

dr −2 × H
=
KI t
di

dr =
=±

−H
dr
=
dy KI t

1
−2H
−H
dht
di +
dt
KI t
KI t
KI t

0
...
0025
0
...
6 − ±0
...
0025

= ±0
...
95%

)

Tom Obrien

HNC Level 4

5)
S = 2t − 13t + 24t + 10
v=

dx
= 6t − 26t + 24t
dt

A=

dv d x
=
= 12t − 26
dt dt

so t = 0 and v = 10ms

5a)

so t = 0 and a = −26ms
t = 4s

v = (6 × 4) − (26 × 4) + (24 × 4) = 16ms
5b)

a = (12 × 4) − (26) = 22ms

t=

v = 6t − 26t + 24t

26 ± √−26t − 4 × 6 × 24
2×6
=

5c)

64 ± √100 64 ±
=
12
122

+=

10

5di)

37
9
and −=
6
2
a = 10

= 12t − 26

10 + 26
= t = 3s
12

5

a=5

= 12t − 26

5 + 26
= t = 2
...
8s
12
5e)
v = (6(3) − 26 × 3 + 24) − (6(2) − 26 × 2 + 24) = 100ms

13

Tom Obrien

HNC Level 4

14

6a)

6b)
y = 9−x

y=x +3

y=

9 − x dx

y=

x + 3dx

A=

A=
=

9×2−2
2

A=
−

9−x
2

x
+ 3x
3

9 × −3 − 3
2

= 16 + 31
...
5

−3
2
+3×2 −
+ 3 × 3 = 8 − −18 = 26
...
5 − 26
...
83

Tom Obrien

HNC Level 4

15

6c)
y = 9−x

y=x +3

y = (9 − x)(9 − x) = 81 − 9x − 9x − x
y = 81 − 18x + x

y = (x + 3)(x + 3) = (x + 3x + 3x + 9)
=

v=π

+6

+9

y dx

81 − 18 +
81x −

x
9x
+
3
x



x + 6x + 9x

x
+ 2x + 9x
5

=

−3
2
− 9 × −3 + 81 − 3
− 9 × 2 + 81 × 2 −
3
3
−(

2
−3
+2×2 +9×2 −
+2 −3 +9 −3 )
3
3

2
2
= (π 2 − 36 + 162 − (−9 − 81 − 243)) − (π 6 + 16 + 18 − (−48
...
6
5
3
= π 461

2
− π(170)
3

1450
...
3 cubic units



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