Math Review: Circles, Spheres, Trig, Integration, Diff. Equations, Polar Coords, Volume, Slides of Mathematics

Mathematical formulas and solutions for various topics including circles and spheres, trigonometry, integration, differential equations, and complex numbers. It covers the circumference, area, and surface area of circles and spheres, the sin and cos functions, integration in cartesian and spherical polar coordinates, and complex numbers. It also includes solutions for differential equations and equations in spherical polar coordinates.

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2021/2022

Uploaded on 09/27/2022

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MATHEMATICS REVISION NOTES:
1 Circles and Spheres
Circumference of a circle of radius
R
C
= 2
๎˜™ R
Area of a circle of radius
R
A
=
๎˜™ R
2
Surface Area of a sphere of radius
R
A
= 4
๎˜™ R
2
V
olume of a sphere of radius
R
V
=
4
3
๎˜™ R
3
2 Trigonometry
sin
๎˜’
= 2 sin
๎˜’
2
!
cos
๎˜’
2
!
cos
๎˜’
= cos
2
๎˜’
2
!
?
sin
2
๎˜’
2
!
= 2 cos
2
๎˜’
2
!
?
1= 1
?
2 sin
2
๎˜’
2
!
d
d๎˜’
tan
๎˜’
= sec
2
๎˜’
=
1
cos
2
๎˜’
d
d๎˜’
cot
๎˜’
=
?
csc
2
๎˜’
=
?
1
sin
2
๎˜’
3Integration
Z
a
0
x
sin
xdx
= sin
a
?
a
cos
a
1
pf3
pf4

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MATHEMATICS REVISION NOTES:

1 Circles and Spheres

Circumference of a circle of radius R

C = 2  R

Area of a circle of radius R A =  R^2

Surface Area of a sphere of radius R

A = 4  R^2

Volume of a sphere of radius R

V =

 R^3

2 Trigonometry

sin  = 2 sin

cos

cos  = cos^2

sin^2

= 2 cos 2

1 = 1 2 sin 2

d d tan  = sec^2  =

cos^2  d d cot  = csc^2  =

sin^2 

3 Integration

Z (^) a 0

x sin xdx = sin a a cos a

4 Spherical Polar Co ordinates

A p oint whose cartesian co ordinates are (x; y ; z ) can b e describ ed in terms of spherical p olar co ordinates (r;  ; ) where x = r sin  cos  y = r sin  sin  z = r cos  ;

or inverting, we have

r =

q x^2 + y 2 + z 2

 = tan ^1

p x^2 + y 2 z

 = tan ^1

 (^) y x

5 Volume Integrals

In Cartesian Co ordinates (x; y ; z )

dV = dx dy dz

In spherical p olar co ordinates (r;  ; )

dV = r 2 dr sin  d d:

6 Di erential Equations

dy dx = ay

Solution y = y (0) exp(ax)

dy dx = cebx^ ay

Particular solution: Search for a solution of the form

y = Aebx^ :

x dy dx = by 2 We rewrite this as dy y 2 = b dx x Integrating b oth sides and adding the constant of integration, we get 1 y = b ln(x) + C

Taking the inverse y =

(C + b ln(x)) C is determined from the b oundary condition, y = y 0 when x = x 0 - we can write the solution as y =

1 =y 0 ln(x 0 ) + b ln(x)

This may b e rewritten in a simpler form as

y = y 0 1 + by 0 ln(x=x 0 )

7 Complex Numb ers

z = x + i y = jz j ei^ ;

where jz j =

q x^2 + y 2 ; tan  = y x w =

z

x + i y

x i y x^2 + y 2 = jw j ei^ ;

where jw j =

p x^2 + y 2