FSK 116 – Semester 1
Mathematics and Other
Essentials
Priorities
• Know how YOUR calculator works and
always have YOUR calculator with you.
• Always have a pencil (and an eraser) at
hand when doing Physics.
2
1
Angles
B
+
O
θ
A
• Angle θ is positive for anticlockwise rotation
• Angle θ is negative for
clockwise rotation
• 1 Rotation = 360°
• ½ Rotation = 180°
• ¼ Rotation = 90°
-
s1 s2 s3
= =
r1 r2 r3
s1
θ
r1
s2
r2
s3
r3
s
=θ
r
• Radians (rad)
⇒
3
Angles - 2
s1
θ
r1
r2
s2
s3
r3
s1 s2 s3
= = = constant
r1 r2 r3
⇒
s
=θ
r
• 1 Rotation = circ of circle = 2πr
r
r
∴
• 1 Rotation = 2π radians (360°)
• ½ Rotation = ½circ of circle = ½2πr
r
r
= π radians (180°)
• ¼ Rotation = ¼circ of circle = ¼2πr
r
r
= π/2 radians (90°)
∴
• 1° = 0.01745 rad
• 1rad = 57.3°
4
2
Calculator Skills
• Change angle setting between degrees – radian
– gradient: DGR
• Convert angle between degrees – radian –
gradient: DGR ►
• Set decimal and exponents: FSE
– Fixed – Scientific – Engineering
– Decimal – 10x – 103n
• Using 10x
– 3 × 102 press 3 then Exp then 2
– Exp ≡ × 10
5
Examples
•
•
Convert to Radians
– 1°
– 32°
– 175°
– 270°
– 315°
A circle has a radius of 3 cm
what is the arc length
subtended by an angle of 50°
•
Convert to degrees
– ½ rad
– ½ π rad
– 4/3 π rad
– ⅔ π rad
– 7 rad
– 6 rad
– 1 rad
6
3
Differentiation
Gradient of a straight line
•
•
•
y = mx + c
m, c – constants
Gradient – tempo at which
the y value changes with
increasing x.
–
•
Gradient = change in y
change in x.
2
y2
∆y
y1
1
θ
∆x
Co-ordinates of point 1 and
point 2.
Gradient =
=
∆y y2 − y1
=
∆x x2 − x1
Also:
( mx2 + c ) − ( mx1 + c )
tan θ =
=m
x2 − x1
x1
x2
∆y
= gradient
∆x
8
4
Gradient
of a straight
line
y
y
y
y2
y1
y
y1
y2
x1
x2
m
x
x1
x2
x
x1
m
y
x2
x
m
m
x
9
Gradient of a curve
•
•
•
Gradient changes at each
point
Also depends on the size
of interval ∆x.
Need new def. for gradient
–
•
The gradient of the tangent
to the curve at the point.
Can determine the
gradient of the tangent
using previous method.
–
y
x
Graphical method – not
always successful.
10
5
Gradient of a curve
•
•
•
Gradient changes at each
point i.e it’s a function of x
Can be expressed as a
function – the derivative.
Derivative notations:
d
( y)
dx
•
d-d-x of y
dy
or dy dx
dx
•
d-y-d-x
D ( y ) or Dx y
•
y
x
d of y or d-x of y
11
Handy Derivatives
y = xn
then
y = ax n
then
y = sin x
then
y = sin ax
then
y = cos x
then
dy
= nx n −1
dx
dy
= anx n −1
dx
dy
= cos x
dx
dy
= a cos ax
dx
dy
= − sin x
dx
y = cos ax
then
y = ln x
then
y = ax
then
y = ex
then
y = e f ( x)
then
dy
= −a sin ax
dx
dy 1
=
dx x
dy
= ( ln a ) a x
dx
dy
= ex
dx
dy
= f ' ( x ) e f ( x)
dx
12
6
Differentiation Rules
• Sum Rule
• Product Rule
y = f (x ) + g (x )
y = f ( x )g (x )
dy df dg
=
+
dx dx dx
dy
dg
df
= f
+g
dx
dx
dx
(
) (
• Quotient Rule
)
y = 2 x 2 + x + 3 + 3x 2 + 2 x + 1
f (x )
y=
g (x )
 df 
 dg 
g  − f  
dy
dx 

 dx 
=
dx
g2
y=
9x7
x2 + 1
(
(
)(
y = 3x 2 + 5 x 2 x 3 − 3
)
• Chain Rule
y = y (u ), and u = u (x )
dy dy du
=
dx du dx
( )
y = sin 4 x 2
)
13
Higher order Derivatives
y=3x^2+1
80
2
70
• If y = 3 x + 1
• dy/dx is gradient of y
• Where dy/dx = 0 - a
turning point.
•
d2y
=6
dx 2
• If d2y/dx2 <0 where dy/dx
= 0 – local maximum
• If d2y/dx2 >0 where dy/dx
= 0 – local minimum
50
40
Y
dy
= 6x
dx
30
20
10
0
-6
-4
-2
0
2
4
6
2
4
6
X
y' = 6x
40
30
20
10
Y
• Then
60
-6
-4
-2
0 76
0
-10
-20
-30
-40
X
14
7
Problems
• Determine the following:
– d y if y = 4 x + 2 x − 5 x − 2
2
4
2
dx 2
– d y if y = e
3
x2
dx 3
2
– ddx y if y = x x+ 1
2
–
d3y
if y = 2sin 3 x
dx 3
15
Example - 1
• An object is projected vertically upwards and the
height above ground, h is given by h = 20t – 5t2 meter,
in which the time, t, is measured in seconds. Calculate
the maximum height which it will reach and when it will
be reached.
16
8
Example - 2
• A flat metal plate (dimensions in cm) is bent in the form
of a gutter (see figure below). Find a formula for the
volume of the gutter in terms of the angle θ.
θ
θ
10
l
2
2
2
17
Example - 2
• The volume of a gutter was determined i.t.o. the angle of
the sides w.r.t. the bottom. Determine the angle at which
the gutter will contain a maximum volume.
θ
θ
10
2
b
2
l
2
2
θ h
θ
2
2
18
9
Example - 3
• A farmer possesses a rectangular sheet of polyethylene
plastic of 24 m by 9 m which he wishes to use as a
waterproof lining for a rectangular dam with constant
depth and vertical walls. To achieve this, equal squares
of side length x are cut from the corners and the edges
are glued together. Calculate the value of x for which the
volume will be a maximum.
x
x
19
Problems
• A sports field must be designed so that the
circumference is 440 m. The shape of the field is
2 semi-circles with a rectangular portion in
between. Determine the measurements so
that the area of the rectangular portion is a
maximum.
• Determine the were the maximum and minimum
values lie for the following function:
y = x 3 − 3 x 2 − 24 x + 1
20
10