Other Coordinates System
Polar coordinates.
When you have to work curves defined by a Point that is moving around the origin using Cartesian
Coordinates not efficient since the equations that describe the curve is not a function. To solve this
reason Bernoulli revive an old system used by the Greek astronomer Hipparchus (190-120 BC).
The polar coordinates use an origin that call Pole and an axis and a horizontal semi-line oriented
positive to the right. A point has coordinates are Theta and r that is the distance from the origin to
the point P an angle Theta is the angle of sides OP and the polar axis
(r, 𝜃)
Pole
O
(origin)
Polar Axis
Ex 1: Graph r=sin( )
𝜋/2+ 𝜋/3
Let do a table for (r, ) 𝜋/2+𝜋/4
x
x
𝜋 − 𝜋/3
𝜋/2
x
𝜋/3
𝜋/4
x
𝜋/6
x
x
𝜋
𝜃
x
x
0
Type equation here.
r = sin 𝜃
0
0.0
𝜋/6
𝜋/4
1/2 = 0.5
𝜋/3
3/2= 0.9
𝜋/2 + 0
1.0
2/2= 0.7
𝜋/2 +𝜋 /6
3/2= 0.9
+ 𝜋/4
2/2= 0.7
+𝜋/3
3/2= 0.9
𝜋/2 +𝜋/2
+
0.0
𝜃
0
Ex 1: Graph r=sin(𝜃 )
Let do a table for (r, 𝜃 )
𝜋/2 + 𝜋/3
x
x
𝜋/2+𝜋/4
𝜋 − 𝜋/3
x
Type equation here.
Type equation here.
𝜋/6
x
x
𝜋/4
2/2= 0.7
𝜋/2 + 0
𝜋/4
x
x
𝜋
𝜋/3
x
𝜋/6
0
1/2 = 0.5
𝜋/3
𝜋/2
0
r = sin 𝜃
3/2= 0.9
1.0
𝜋/2 +𝜋 /6
+ 𝜋/4
3/2= 0.9
+𝜋/3
3/2= 0.9
𝜋/2 +𝜋/2
+
0.0
2/2= 0.7
Trick #3: PARTIAL FRACTIONS.
Let P(x)/Q(x) a polynomial fraction where Q(x) is completely factored and deg P(x) < deg Q(x) then the fraction can be
written as sum of simples fraction where the denominators are the factors involved in the factorization of Q(x) using the
following two rules.
𝐴1
𝐴2
𝐴𝑛
1) For each linear factor of type (ax+b)𝑛 we add n simple fractions (𝑎𝑥+𝑏)1 + (𝑎𝑥+𝑏)2 + … . + (𝑎𝑥+𝑏)𝑛
𝑛
2) For each linear factor of type (a𝑥 2 + 𝑏𝑥 + 𝑐) we add n simple fractions
𝐴1𝑥 +𝐵1
(
a𝑥 2 +𝑏𝑥+𝑐
Ex 1.
Ex
)
1 +
𝐴2 𝑥+𝐵2
(
a𝑥 2 +𝑏𝑥+𝑐
3𝑥 2 −𝑥+5
𝑥−1 𝑥+4 ( 2𝑥−1)2
𝐸𝑥+𝐹 𝐴𝑥+𝐵
2. 2
𝑥 +4
𝑥 2 +1
+
)
=
2 + ⋯+
𝐴
𝑥−1
𝐶𝑥+𝐷
(𝑥 2 +1)2
+
+
𝐴𝑛𝑥+𝐵𝑛
(a𝑥 2+𝑏𝑥+𝑐)
𝐵
𝑥+4
𝐸𝑥+𝐹
𝑥 2 +4
+
𝐶
2𝑥−1
𝑛
+
𝐷
(2𝑥−1)2
Ex 3. Integrate
2𝑥 2 +7𝑥+15
(x+2)2 (x+3)
dx , first decompose
2𝑥 2 +7𝑥+15
(x+2)2 (x+3)
=
𝐴
(𝑥+2)
+
𝐵
𝑥+2 2
+
𝐶
(𝑥+3)
Multiplying both sides by (x + 2)2 (x + 3) we get
2𝑥 2 − 7𝑥 + 5 = 𝐴 𝑥 + 2 + 𝐵( 𝑥 + 3 + 𝐶 𝑥 + 2 2
Take x= - 2
8-14+15= 9
B(1) = 9  B = 9
Take x= -3
A( -1) + C(-1) = - 3
 - A - C= -3  A + C= 3
Take x=0
A(2) + B(3) + C( 4) =5  2A +3B =5 and since B=9
5
So
Ex. 4
Ex. 5
𝟐𝒙𝟐 +𝟕𝒙+𝟏𝟓
(𝐱+𝟐)𝟐 (𝐱+𝟑)
4𝑥 + 5
𝑑𝑦 =
𝑥2 + 𝑥 − 2
𝐝𝐱 =
−𝟏𝟏
𝒙+𝟐
dx +
4𝑥 + 5
𝑑𝑥 =
𝑥+2 𝑥−1
𝟗
𝒅𝒙
𝒙+𝟐 𝟐
−
𝟏𝟒
dx
(𝒙+𝟑)
1
3
+
𝑑𝑥 =
𝑥+2 𝑥−1
4𝑥 2 + 17𝑥 − 4
𝐴
𝐵𝑥 + 𝐶
𝑑𝑥 =
+
𝑥 + 5 𝑥2 + 1
𝑥+5
𝑥−1
Type equation here.
A +C = 3  C = 14
2A + 27 = 5  A = -11
= - 11ln|x+2|-
𝟏
(𝑥+2)2
+ 14 ln|x+3| + C
𝑙𝑛 𝑥 + 2 + 3𝑙𝑛 𝑥 − 1 + 𝐶
PARAMETRIC EQUATIONS
When “the equation in Cartesian Coordinates is to complex” mathematician use an extra
variable t called “parameter” such that both x and y are functions of t like x= x(t) and
y= y(t) . We say then that the original function is now defined by x=x(t) , y= y(t), y(t)
Ex 1. Find the Cartesian equation corresponding to the curve defined by x = 3t
y = 𝑡2 + 1
𝑥
From the first parametric equations we get t = and replacing on the second we get
y =(
𝑥
3
)2 + 1=
𝑥2
9
3
+1 and we recognize that is a parabola.
Ex 2. Find the length of the graph defined by x=4cos(t) and y =4sin(t)
It is clear that 𝑥 2 + 𝑦 2 = 4 1 that is a circle with radio 2. So the length is 2 pi
Type equation here.
Type equation here.Type equation here.
LENGH OF AN ARC IN PARAMETRIC
Let have a curve defined in parametric form =x(t) ; y=y(t) , we get that the arc length
𝑡2
𝑡1
of an arc is given by s =
ds =
𝑑𝑥
𝑑𝜃
2
𝑑𝑦
+
𝑑𝜃
𝑑𝑥 2
𝑑𝜃
1+(
𝑥′ 𝑡 2
)
(𝑦 ′ (𝑡)
𝑑𝑡
2
𝑑𝑥 =
𝑑𝑥 2
𝑑𝜃
+
𝑑𝑦 2 𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑥
=
𝑑𝑥 2
𝑑𝜃
+
Ex 1: Find the length of the curve defined by x = 2t-1 and y=𝑡 2 +1 from t = 0 to t= pi/6
Let find dx/dt = 2 and dy/dt =2t …
𝑝𝑖/6
ds= 0
2 2 + 2𝑡 2 𝑑𝑡 = 4 1 + 𝑡 2 𝑑𝑡 =4 sec3 𝜃 d𝜃 =
Use t =tan𝜃
dt=sec2 𝜃 d𝜃
1
(sec
2
𝑑𝑦 2
𝑑𝜃
𝑑𝜃
Type equation here. .
𝜃 𝑡𝑎𝑛𝜃 + ln |sec𝜃+tan𝜃|=
𝐸𝑥. 3
𝑥2 𝑦2
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔ℎ𝑡 𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑖𝑝𝑠𝑒 2 + 2 = 1
𝑎
𝑏
2
𝑑𝑦
𝑑𝑠 = 1 + 𝑦′
2
=
1+
𝑑𝜃
𝑑𝑥
𝑑𝑥 =
𝑑𝜃
𝑑𝑥
𝑑𝜃
2
(𝑢𝑠𝑒 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑑𝑦
+
𝑑𝜃
2
𝑑𝜃 =
=
𝑆 = 4𝑎
𝜋
2
0
1+
𝑘 2 𝑡𝑎𝑛2 𝜃
𝑥 = 𝑎 sin 𝜃
𝑦 = 𝑏 cos 𝜃
𝑑𝜃 , where
𝑘2
𝑎
=
𝑏
𝑎 cos 𝜃
1+
2
<1
2
+ −𝑏 sin 𝜃
−𝑏 sin 𝜃 2
𝑎 cos 𝜃 2
𝑑𝜃
2 𝑑𝜃
Type equation here.
This integral cannot be solved since
the there are not exist a function
whose derivative is
REMARK:
Niels Abel (1813-1829) study 3 types of family functions that he call Elliptic Integrals and
prove that there have not anti-derivative and integral is one them ,this means that the
ellipsis length can not be calculate by integration.
There are tables for Elliptic integrals. where you find he value.
Ex 1. A function Y =f(x) is defined by the parametric equations x= 2t+1 and y = t+3. Find y = f( x )
In other words we need to eliminate t using the given equations.
From the first
2t = x-1
From the second
t=y-3
1
3
1
3

X - 1 = 2(y-3)  x – 1 = 2y – 6  y = 2 x − 2 y= f(x)= = 2 x − 2 . It is a line
Ex 2. Get y=f(x) for x(t)= a cos(t) and y(t) = b sin(t)., where a,b are positive numbers.
Since
cos2
𝑡 +
𝑠𝑖𝑛2 (t)
=1

𝑥(𝑡) 2 𝑦(𝑡) 2
(
) +(
) = 1 It is an ellipse.
𝑎
𝑏
ARC LENGTH IN PARAMETRICS EQUATIONS.
Since ds =
1 + 𝑦′
2
dx and 𝑦 ′ =
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
 ds =
1+
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
2
𝑑𝑥 =
𝑑𝑥 2
𝑑𝑡
+
𝑑𝑦 2
𝑑𝑡
dt
Ex 1. A function Y =f(x) is defined by the parametric equations x= 2t+1 and y = t+3. Find y = f( x )
In other words we need to eliminate t using the given equations.
From the first
2t = x-1
From the second
t=y-3
1
3
1
3

X - 1 = 2(y-3)  x – 1 = 2y – 6  y = 2 x − 2 y= f(x)= = 2 x − 2 . It is a line
Ex 2. Get y=f(x) for x(t)= a cos(t) and y(t) = b sin(t)., where a,b are positive numbers.
𝑥(𝑡) 2 𝑦(𝑡) 2
 (
) +(
) = 1 It is an ellipse.
𝑎
𝑏
Since cos2 𝑡 + 𝑠𝑖𝑛2 (t) = 1
ARC LENGTH IN PARAMETRICS EQUATIONS.
Since ds = 1 + 𝑦 ′
2
𝑑𝑦/𝑑𝑡
′
and 𝑦 =  ds = 1 +
𝑑𝑥/𝑑𝑡
2
dx =
𝑑𝑥2 𝑑𝑦2
+
𝑑𝑡
𝑑𝑡
2
𝑑𝑥
𝑑𝑡
dx =
ds =
𝑑𝑥
𝑑𝑡
2
𝑑𝑦
+
𝑑𝑡
𝑑𝑥 2
𝑑𝑡
+
2
𝑑𝑥
𝑑𝑥
𝑑𝑡
𝑑𝑦 2
dt
𝑑𝑡
𝐸𝑥. 3
𝑥2 𝑦2
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔ℎ𝑡 𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑖𝑝𝑠𝑒 2 + 2 = 1
𝑎
𝑏
𝑑𝑦
𝑑𝑠 =
2
1 + 𝑦′
1+
=
𝑑𝑥
2
𝑑𝜃
𝑑𝜃 =
𝑑𝜃
𝑑𝑥
𝑑𝜃
2
𝑑𝑦
+
𝑑𝜃
𝑑𝑥 2
𝑑𝜃
(𝑢𝑠𝑒 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑥 = 𝑎 sin 𝜃
𝑦 = 𝑏 cos 𝜃
2
Remark. Mathematician Niels Abel (1810- 1828) study for
do not exist anti-derivatives
𝑑𝜃 =
𝑅𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑑𝑥, 𝑑𝑦, 𝑑𝜃 𝑤𝑒 ℎ𝑎𝑣𝑒:
𝑑𝑠 =
𝑆 =4
𝑎 cos 𝜃
𝜋
2
0
2
+ −𝑏 sin 𝜃
𝑎𝑐𝑜𝑠𝜃 2
𝑑𝑠 = 4
𝜋
2
0
2
2
2
2
𝑑𝜃 = 𝑎 𝑐𝑜𝑠 𝜃 + 𝑏 𝑑𝜃 =
𝑎2 𝑐𝑜𝑠 2 𝜃
𝑏
1+
𝑎
𝑏
1+
𝑎
2
𝑡𝑎𝑛2 𝜃
𝑑𝜃 = 𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 !
2
𝑡𝑎𝑛2 𝜃
𝑑𝜃
There do not exist a function whose derivative
is integral function if k=
𝑏 2
𝑎
< 1.
REMARK:
Niels Abel (1813 - ?) study 3 types of family functions that he call Elliptic Integral and prove that there have not
anti-derivative. He also publish a table giving the approximated value the for different limits.
integral is one them, this means that the ellipsis length can not be calculate by integration.
Sequences and Series
A sequence { 𝑎𝑛 } is a function whose domain is the non negative integer 𝑍 + U {0} and the
range a subset of the real numbers, meaning a(n) : Z+ U {0} ∶ −−→ R.
The functional values : 𝑎1 , 𝑎2 , 𝑎3 , ……. are the n components and a(n) is called the
general term of sequence.
𝑛
1 2 3 4
𝑛
Ex 1: {
} = { ,
, , ,.. . ,
, . ...}
𝑛+1
2 3 4 5
𝑛+1
Let’s graph the sequence ….
1.0
The graph says
lim of seq = 1, e.g. Converge!
Ex 2. Test for convergence
𝑛3 +2𝑛2 −2
{ 2
}
𝑛 +4
dividing num. and denom. by 𝑛3 we get
0.5
1
2
1+𝑛+−𝑛^2−
1
+4/𝑛
𝑛
Taking lim we get infinity, so the sequence Diverge!
𝑛→∞
1
2
3
4
5
6
7 8
Definition: Let { 𝑎𝑛 } be a sequence we will call Series the sum adding of the terms of the
sequence.
Definition: Let { 𝑎𝑛 } be a sequence we will call Series the sum adding of the terms of the
sequence.
Ex 1 a𝑛 ={ 1, x ,𝑥 2 , 𝑥 3 , 𝑥 4 , ……… } then
Series.
Definition : A polynomial Series is
∞
𝑛=1 𝑎𝑛
∞
𝑛=0 𝑎𝑛
= {1 + x + 𝑥 2 + 𝑥 3 + 𝑥 4 + ……… } is a
is convergent if lim
𝑛→∞
∞
𝑛=1 𝑎𝑛
exist.
Ex 2. Test for convergence the series = 1-1+1-1+1-1 …… This series is Divergen , since the
sum is alternating from -1 to +1
Remark: This sequence converge to different values in concordance with the way we
regroups the terms of the series
A) (1-1)+(1-1)+(1-1) …….. = 0
B) 1- ( (1-1)+(1-1) + ……... ) = 1
A) (1-1)+(1-1)+(1-1) …….. = 0
B) 1- ( (1-1)+(1-1) + ……... ) = 1
C) (-1+(1-1)+(1-1)+(1-1)….. =-1
This series is Divergent , since the sum is alternating from -1 to +1 This example shows us that we
need a more strong definition for convergence of series that assures us that the limit be if exist be
a unique real number. To get this we are going add that the series of the absolute values of the
terms should be convergence, to assures that that the new adding be associative and
commutative.
Type equation here.
GEOMETRIC SERIES
Definition: The series
series.
∞
𝑛=0
𝑎𝑟 𝑛 where r is a fix real number are called a geometric
Theorem: geometrics series converge if |r|<1
divergent if |r|≥1
2
Proof: Call S(n) = a + ar + a𝑟 + 𝑎𝑟 3 +
+𝑎𝑟 𝑛
multiplying by r
r S(n )=
ar + a𝑟 2 + 𝑎𝑟 3 +
+𝑎𝑟 𝑛 + 𝑎𝑟 𝑛+1 + ……..
𝑎𝑟 𝑛+1
Substracting … S(n) (1-r) = So if |r | < 1 the series Converges
if |r |≥ 1 Diverges.

𝑎−𝑎𝑟 𝑛+1
S(n)= 1−𝑟
, so lim S(n )= 𝑓 𝑥 =
𝑎 −0 𝑎
=
1−𝑟 1−𝑟
1
Given ∆ ABC area = 2 bh where h = CH = √ 𝐴𝐶 2 − ()AH 2 = =
3
2
a
C
So area Type equation here.
So area= A =
3
4
𝑎2
For second triangle a is replacing by
So the series is A+
1
4
𝐴+
1
42
𝑎
2
1
4
and Area by = (first area)
A+....
So the sum of all areas… is S =
𝐴
1
1−( )
4
a
Geom. Series with r=1/4
=
4𝐴
3
=
4
3
3
4
𝑎2 =
3
3
𝑎2
A
0.5 a
H
B
Definition:
∞
𝑛=1 𝑎𝑛
𝑥 is absolute convergent if
∞
𝑛=1
|𝑎𝑛 𝑥 | is convergent.
Remark: The absolute convergent assures us that adding of a series is commutative and
associative.
Ratio Test : Let
∞
𝑛=1 𝑎𝑛
𝑥 a series and such that lim |
𝑛→∞
𝑎𝑛+1 𝑥
𝑎𝑛 𝑥
| = r.
1) The series converges
if r < 1
2) The series diverges
if r > 1
3) There are no information if r =1
Proof: 1) The limit assures that exist an integer N such that |
𝑎𝑛+1 𝑥
𝑎𝑛 𝑥
| < 1 for every n>N
So |𝑎𝑁 𝑥 |>r |𝑎𝑁+1 𝑥 |>𝑟 2 |𝑎𝑁+2 𝑥 |> …. Is a GEOMETRIC SERIES OF 0< r<1 is convergent
𝑁
This implies that ∞
𝑛=1 |𝑎𝑛 𝑥 | =
𝑛=1 |𝑎𝑛 𝑥 | + 𝑡ℎ𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑟𝑖𝑒 is convergent.
2) and 3) is the similar argument.
𝐸𝑥. 1
𝑥𝑛
𝑛=1 𝑛!
∞
𝑅 = lim
𝑛→∞
𝑥 𝑛+1
𝑥𝑛𝑥
𝑛!
𝑥 𝑛+1
𝑛!
𝑛+1 !
∗
∗
= lim
= lim
𝑥𝑛
𝑛→∞ 𝑛 + 1 𝑛! 𝑥 𝑛
𝑛→∞ 𝑛 + 1 ! 𝑥 𝑛
𝑛!
0 < 1 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑟𝑒𝑎𝑙 𝑥.
lim
𝑛→∞
𝑥
𝑛+1
=0
𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑜𝑓 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑜𝑓 𝑆𝑒𝑟𝑖𝑒𝑠
𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 …
∞
𝑎𝑛 (𝑥 − 𝑎)𝑛
𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡
𝑛=0
𝐷𝑒𝑚: 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑎
𝑎𝑛+1 (𝑥 − 𝑎)𝑛+1
lim
=
𝑛→∞
𝑎𝑛 (𝑥 − 𝑎)𝑛
𝑆𝑜
𝑟 (𝑥 − 𝑎) < 1 
𝑎𝑛+1 (𝑥 − 𝑎)
𝑎𝑛+1 𝑥 − 𝑎 𝑛 (𝑥 − 𝑎)
lim
= 𝑟|𝑥 − 𝑎| < 1
lim
= 𝑛→∞
𝑛
𝑎
𝑛→∞
𝑛
𝑎𝑛 (𝑥 − 𝑎)
−1
𝑟
1
< 𝑥−𝑎 <r
1
1
 ( a -- 𝑟 < x < a + 𝑟 )
1
Remark : The lector should check if for the series converge or diverge for x= a ± 𝑟 and see if it is need to
improve to (a , b) to [a , b) or (a , b ] or [a , b] by checking for convergence at x=a and x=b.
𝐸2. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑜𝑓𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑓𝑜𝑟𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒
𝑥 𝑥3 𝑥5 𝑥7
𝑆 𝑥 =
−
+ − +⋯
1! 3! 5! 7!
Using 𝑟𝑎𝑡𝑖𝑜 𝑡𝑒𝑠𝑡 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑥 𝑛+1
𝑥
𝑥𝑛𝑥
𝑛!
𝑛+1 !
lim
=
lim
∗ 𝑛 =
𝑟 = lim
= 𝑛→∞
𝑛→∞
𝑛
+
1
𝑛
𝑛 + 1 𝑛! 𝑥
𝑥
𝑛→∞
𝑛!
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑟𝑒𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑥 lim
𝑛→∞
1
𝑛+1
=0
𝐸3. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑦.
∞
(𝑥 − 2)𝑛
𝑛=0
𝑥 − 2 𝑛 (𝑥 − 2)
𝑥−2 𝑛
(𝑥 − 2)𝑛+1
= 𝑥 − 2 lim
= 𝑥−2 <1
𝑟 = lim
= lim
𝑛→∞
𝑛→∞ (𝑥 − 2)𝑛
𝑛→∞ (𝑥 − 2)𝑛
(𝑥 − 2)𝑛
−1 < 𝑥 − 2 < 1 , 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑖𝑛 1 < 𝑥 < 3
𝐿𝑒𝑡 𝑢𝑠 𝑠𝑒𝑒 𝑖𝑓 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑥𝑡𝑟𝑒𝑚𝑒𝑠 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 3
𝑊ℎ𝑒𝑛 𝑥 = 1 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒
∞
(𝑥 −
2)𝑛
∞
=
𝑛=0
(1 −
2)𝑛
∞
=
𝑛=0
(−1)𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑛=0
𝑊ℎ𝑒𝑛 𝑥 = 3 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒
∞
(𝑥 −
2)𝑛
∞
=
(3 −
𝑛=0
𝑛=0
∞
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒
2)𝑛
∞
=
(1)𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑛=0
(𝑥 − 2)𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑓𝑜𝑟 𝑥 ∈ (1,3)
𝑛=0
𝑈𝑠𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑠𝑒𝑟𝑖𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
∞
∞
𝑆𝑖𝑛𝑐𝑒 𝑙𝑖𝑚
𝑓(𝑥𝑖 )∆x = =
𝑛=0
𝑓 𝑥𝑖 𝑑𝑥
0
𝑇ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑥𝑖𝑠𝑡𝑠. 𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠.
∞
𝐸1.
1
𝑝 =
𝑥
𝑥=1
∞
1
𝑥 −𝑝+1
∞
1
𝑑𝑥 =
=
𝑥𝑝
−𝑝 + 1 1
𝑓𝑜𝑟 𝑝 > 1, 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
𝑆𝑜 𝑓𝑜𝑟
𝑓𝑜𝑟 𝑝 < 1, 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑓𝑜𝑟 𝑝 = 1, 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑓𝑜𝑟 𝑝 > 1, 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑥𝑖𝑠𝑡𝑠
𝑓𝑜𝑟 𝑝 < 1, 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
𝑓𝑜𝑟 𝑝 = 1, 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
𝑃𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑓𝑢𝑛𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑎
∞
𝑎𝑛 (𝑥 − 𝑎)𝑛
𝑓 𝑥 =
𝑛=0
1𝑠𝑡 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑎𝑛
𝑓 𝑥 = 𝑎0 + 𝑎1 𝑥 − 𝑎 + 𝑎2 (𝑥 − 𝑎)2 +𝑎3 𝑥 − 𝑎
𝑓 𝑎 = 𝑎0 + 𝑎1 0 + 𝑎2 (0)2 + ⋯ = 𝑎0
3
𝑓′ 𝑥 =
+ 3 ∗ 4𝑎4 𝑥 − 𝑎
1 𝑎1 + 𝑎2 𝑥 − 𝑎 + 2 ∗ 3𝑎3 𝑥 − 𝑎
𝑓 ′′ 𝑥 =
0+
𝑓 (3) 𝑥 =
0+
𝑓 (4) 𝑥 =
0+
0
0
𝑛=0
+
+
3! 𝑎3
0
𝑓 (𝑛) 𝑥
(𝑥 − 𝑎)𝑛
𝑛!
4
+ 4! 𝑎4 𝑥 − 𝑎
+
4
+⋯
0! 𝑎0 = 𝑓 𝑎
2! 𝑎2 + 3! 𝑎3 𝑥 − 𝑎 + 4! 𝑥 − 𝑎
∞
f(x) =
2
+ 𝑎4 𝑥 − 𝑎
4! 𝑎4
3
+⋯
𝑎4
+ …
+...
+⋯
1! 𝑎1
=
𝑓 ′( 𝑎)
2! 𝑎2 = 𝑓 ′′ 𝑎
3! 𝑎3 = 𝑓 (3) (𝑎)
4! 𝑎4 = 𝑓 (4) (𝑎)
𝑎𝑛
=
𝑓 (𝑛) 𝑥
𝑛!
in the same year Taylor publish the series using . Later in the same year Brooke Taylor
publish the extension series in the vicinity of x=a with a≠ 0.
Ex 1. Find the polynomial series of McLauring for f(x)= 𝑒 𝑥
1
𝑥
0
f(x)= 𝑒  f(0) = 𝑒 = 1 => 𝑎0 = =1
f’’(x)= 𝑒 𝑥
 f(‘’0)=
We have
𝑒𝑥=
1+
𝑒0
1
1!
+
=1 => 𝑎1 =
1
2!
+
1
3!
+
0!
1
=1
1!
1
+ ...…
4!
and since => 𝑎𝑖 =
.
1
𝑛!
+
1
=1
𝑖!
for all i=0,1,2,3…..
1
+ .....
(𝑛+1)!
Let find the interval of convergence (using the Ratio Test ) 𝑟 = lim
𝑛→∞
lim
𝑛→∞
1
𝑛+1 !
1
𝑛!
=
1
lim | |
𝑛→∞ 𝑛+1
𝑥𝑛+1
𝑛+1 !
𝑥𝑛
𝑛!
=
=0
So, The series converges when r =|x-0|=|x|0 < 1 meaning always, so interval of
convergence is (-∞, ∞).
Ex 1. Find Mclauring series for sinx
Getting the f’ , f” …
1) f(x) = sin x = 0 +
1
1!
x
0
+2! 𝑥 2
f’(x) = cosx
f’’(x)= sinx
f’’’(x) = cosx start reppeting !
f(0) =
f’(x)= cosx
f”(x)=- -sinx
f’’’(x)= -cosx
f’’’’(x)= sinx
+
−1
3!

.... = x
𝑥3
−
3!
Sin(x) = x
𝑥3
−
3!
+
𝑥5
5!
+
𝑥5
5!
……
= x
𝑥3
−
3!
+
𝑥5
5!
……
f’( 0) =. 1
f’’’(0 )=0
f’’’( 0)= -1
f””(0)=0 start reppeting !. . .
Remark: Since the set of series convergent is closed for + , - , * , / , including * by a real number and more Plus like the
set of polynomials, we can see the convergent series as an extension of set of the polynonmial, e.g. polynomials of
infinite degree . Further more if f(x) = ak-6∑_(k=0)^∞
Ex 2. Find Mcauling series for cosx , since we known the series for sinx then 𝑤𝑒 𝑐𝑎𝑛 𝑔𝑒𝑡 𝑐𝑜𝑠𝑥as
′
𝑠𝑖𝑛𝑥 = (x
𝑥3
− 3!
+
𝑥5
5!
……)’ = 1
𝑥2
− 2!
+
𝑥4
4!
……) ‘ =
−1 𝑥 𝑘−1
𝑛
𝑘+1
𝑘=1 (−1)
2𝑘+1 !
sin 𝑥
𝑥
Ex 3. The function f(x) =sin(x)/x do not has antiderivative. Solve
3
𝑠𝑖𝑛 𝑥
𝑥
𝑑𝑥 =
𝑛
=
𝑘1
(−1)
𝑘=1
5
x − 𝑥3! + 𝑥5! ……
𝑥
𝑥 𝑘−1
2𝑘 + 1 !
dx =
(1 −
𝑥2
3!
+
𝑥4
5!
……) dx = 1-
𝑥3
(3!)3
+
𝑥5
5!.(3)
𝑑𝑥 𝑢𝑠𝑖𝑛𝑔 𝑠𝑒𝑟𝑖𝑒𝑠
𝑥7
−
7!.(3)
…
HYPERBOLIC FUNCTIONS.
𝐼𝑛 𝑃ℎ𝑦𝑠𝑖𝑐𝑠 𝑖𝑡 𝑖𝑠 𝑝𝑟𝑜𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑎𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑎𝑏𝑙𝑒 ℎ𝑎𝑛𝑔𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑡𝑤𝑜
𝑠𝑎𝑚𝑒 − ℎ𝑖𝑔ℎ 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝑝𝑜𝑠𝑡𝑠 𝑡ℎ𝑒 𝑠ℎ𝑎𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑐𝑎𝑡𝑒𝑛𝑎𝑟𝑦 𝑎𝑛𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑚𝑜𝑑𝑒𝑙𝑒𝑑 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑌=
𝑎 𝑥
(𝑒
2
𝑎
+𝑒
−𝑥
𝑎)
𝑎
X= 2 (𝑒
𝐼𝑓 𝑤𝑒 𝑚𝑎𝑘𝑒 𝑎 = 1 𝑡ℎ𝑒𝑛 Y=
𝑥
𝑎
−𝑒
−𝑥
𝑎
(𝑒 𝑥 +𝑒 −𝑥 )
2
)
X=
(𝑒 𝑥 −𝑒 −𝑥 )
2
𝑤ℎ𝑖𝑐ℎ 𝑠𝑎𝑡𝑖𝑠𝑓𝑦 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑦 2 − 𝑥 2 = 1
𝐷𝑒𝑚𝑜𝑠𝑡𝑟𝑎𝑡𝑖𝑜𝑛:
(𝑒 𝑥 + 𝑒 −𝑥 )
2
2
(𝑒 𝑥 − 𝑒 −𝑥 )
+
2
2
𝑒 2𝑥 + 2 + 𝑒 −2𝑥 𝑒 2𝑥 + 2 + 𝑒 −2𝑥 4
=
+
= =1
4
4
4
𝑊𝑒 𝑠𝑒𝑒
(𝑒 𝑥 +𝑒 −𝑥 )
𝑡ℎ𝑎𝑡
2
𝐼𝑓 𝑤𝑒 𝑢𝑠𝑒 𝑡ℎ𝑒
(𝑒 𝑥 −𝑒 −𝑥 )
𝑎𝑛𝑑
𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 cos
2
ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑒 𝑦 2 − 𝑥 2 = 1 𝑖𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝑡ℎ𝑒
(𝑒 𝑥 + 𝑒 −𝑥 )
cosh 𝑥 =
2
𝑤ℎ𝑒𝑟𝑒
𝑥 𝑎𝑛𝑑 sin 𝑥 𝑖𝑛 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐𝑠.
𝑐𝑖𝑟𝑐𝑙𝑒 𝑥 2 + 𝑦 2 = 1 𝑡ℎ𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒
(𝑒 𝑥 − 𝑒 −𝑥 )
sinh 𝑥 =
2
cosh 𝑥 − sinh 𝑥 = 1
(𝑒 𝑥 + 𝑒 −𝑥 )
𝑌=
2
sinh(𝑥)
tanh 𝑥 =
cosh(𝑥)
(𝑒 𝑥 −𝑒 −𝑥 )
=
2
𝑊𝑒 𝑑𝑒𝑓𝑖𝑛𝑒
2
2
X
Remark 1) sinh and tanh are odd functions and cosh is even.
2) For every relation or formula in Trigonometry there is an analog one in Hyperbolic and sometimes the one sign
could be changed at most.
Example: Cos2x=2co𝑠 2 x -1
---- Cosh2x=2co𝑠ℎ2 -1
sin𝑥 2 (x)+ cos𝑥 2 (x) = 1 - -- - sin𝑥ℎ2 (x) + cos𝑥ℎ2 (x) = 1
INVERSE OF HYPERBOLIC FUNCTION
Type equation here.
Type equation here.
1) sinh−1 (x) = ln|x|+ 𝑥 2 +1 for all x
2) cosh−1 (x) = ln|x+ 𝑥 2 −1 for all |x |≥1
1
1+𝑥
3) tanh−1 (x) = ln|x|=2 ln | 1−𝑥 | for all x < 1
4) sech−1 (x) =
1
| 𝑥| 𝑥 2 −1
for all |x| >1
1) Proof: Let y = sinh−1 (x)  sinh (sinh−1 (x)) =sinh(y)  x =
So working you get
2x =𝑒 𝑦 + 𝑒 −𝑦  y = ln|x|+ 𝑥 2 +1
Same procedure for 2) ,3) and 4)
Type equation here.
(𝑒 𝑦 +𝑒 −𝑦 )
2
for all x.
Type equation here.
Type equation here.
Type equation here.
Type equation here.
Ex 1.
Let x = sinh (u)
dx = cosh(u)du
cosh 𝑢𝑑𝑢
( sinh 𝑢 )2 +1
𝑑𝑥
𝑥 2 +1
Ex 2.
𝑑
𝑑𝑥
dx = ln|x|+ 𝑥 2 + 1 + C 
−1
Find
( 𝑐𝑜𝑠ℎ (sec(x))
𝑑
1
( 𝑐𝑜𝑠ℎ −1 (secx))=
2
𝑑𝑥
−1
du =𝑑𝑢 = u + C = 𝑠𝑖𝑛ℎ (x) + C
𝑠𝑒𝑐 (𝑥)−1
5
2
𝑑𝑥
𝑥 2 +1
𝜆
sec( x) tan(x) = sec(x)
Remark: This example reinforce the correspondence
Trigonometry  Hyperbolic functions
dx = ln ln |5+ 26|
ln | 2+ 5 |
Ex 1. Solve
5
2
𝑑𝑥
𝑥 2 +1
Let x = sinh (u)
dx = cosh(u)du
dx
cosh 𝑢𝑑𝑢
( sinh 𝑢 )2 +1
𝑑𝑥
𝑥 2 +1
Ex 2.
−1
du =𝑑𝑢 = u + C = 𝑠𝑖𝑛ℎ (x) + C
dx = ln|x|+ 𝑥 2 + 1 + C 
5
2
𝑑𝑥
𝑥 2 +1
𝑑
Find
( 𝑐𝑜𝑠ℎ−1 (sec(x))
𝑑𝑥
𝑑
1
( 𝑐𝑜𝑠ℎ −1 (secx))=
2
𝑑𝑥
𝑠𝑒𝑐 (𝑥)−1
= sec( x) tan(x) = sec(x)
Remark: This example reinforce the correspondence
Trigonometry  Hyperbolic functions
dx = ln ln |5+ 26|
ln | 2+ 5 |
VECTORS ON THE REAL PLANE
Vectors are important and necessary to understand Science because work with oriented magnitudes, like force,
velocity, acceleration , etc. The vague idea of a vector is an oriented arrow ( having a starting point an with a starting
point and a length that represent the power or the magnitude of the vector. To added the vectors select a point O
on the plane and we graph the two vectors starting from the point O, finally we complete a parallelogram adding
the two missing sides . The diagonal from the point O is the adding the two given vectors.
B
B
C
C
R
R
A
C
Type equation here.
D
A =O = C
D
A =O = C
Definition: Let A=(𝑎1 , 𝑏1 ), B=(𝑏1 , 𝑏2 ) two point on the real plane we will call a vector v = AB = AB the
oriented segment AB going from A to B (we will use the AB to avoid the arrow on top). And we will call
magnitude of the vector to the length of the segment length of the segment AB.
Ex 1: Let A=(2,3) and B=(5,4) then the length
||AB ||= ((2 − 5)2 +(3 − 4)2 = 10
and the orientation is going down and to the
right. 10 going down and to the right.
Ex 2: Let AB
Y
B
A
We see that the vector start in A and
end in B. Furthermore that = AB = (𝑎2 − 𝑎1 ) , 𝑏2 − 𝑏1 )
and in general for any two points we can define the
vector for any point P and any point Q we can define
The vector PQ = (𝑎𝑞 − 𝑎𝑝 , 𝑎𝑞 − 𝑎𝑝 ) . We see too that
|| AB|| is given by Pythagoras Theorem
(𝐴𝐶)2+ (BC)2 ||= ||AB||.
𝑎2
O
𝑏2 − 𝑏1
C
𝑎2 − 𝑎1
𝑏2
𝑏1
𝑎1
Ex 3: Find a unite vector EN LA DIRECTION
𝑎2
X
Type equation here.
Type equation here.
Type equation here.
Type equation here.
equation here.Type Type equation here.
Type equation here. Type equation here.
Type equation here.
STANDARD REPRESENTATION OF VECTOR IN THE PLANE.
The unit vectors I =(1,0 )and j = (0,1) Find a direction vector pointing in the direction positive X axis and Yaxis respectively are called standard basis. Any vector v= (𝑣1 , 𝑣2 ) can be expressed as a linear combination
of the i and j because any (𝑣1 , 𝑣2 ) = 𝑣1 (1,0) + 𝑣2 (0,1) = 𝑣1 i + 𝑣2 j
Ex 1: Find a direction v in the direction of vector V= ( 3, - 4 ).
Since ||v||= 32 + 42 = 5
v (𝟑,𝟒)
3 4
Let u =
=
= ( , )
|𝑽|
𝟐𝟓
5
5
Ex 2: Tree forces F1 = i – 2 j , F2 = 3 I – 7 j , F3 = i – 2 j act on an object. What additional force F4 should
be applied keep the body at rest ?
Total Force = ( 1+3-1) i + ( -2-7-3) j = 3i -11 j , so F4 = -3 i – 11 j
Ex 3: Find standard representation of the vector PQ for points P(3,-4) and Q (-2,6).
The component form of PQ= < (-2) -3 , 6 – (- 4] = (<5,10> , which means that PQ has standard
representation PQ = -5 i + 10 j .
VECTORS IN R𝟑
A vector v in space is a trio of real numbers) ( a, b , c) . We’ll define the dot product of a
and b as
||v|| ||w|| cos( ) where is the angle (in space) defined by v and w.
a
h
ß
b
a.b = ||a||.|| b|||cos( ) (Since h = a cos( )
a . b = a b cos(ß)
) = Area of the parallelogram
ß
If beta = Type equation here. a.b = 0 . This allows us to check if two vectors are
perpendicular the angle defined by two vector are perpendicular.
Ex. Are the vectors defined by a=(3,5,1) , b =(2,-3, 9) are perpendicular?
Type equation here.
a.b = 6 – 15 +9 = 0 , so the angle is aright one.
CROSS PRODUCT
Definition: The cross product of two vector a and b in R3 is defined as the determinant
axb=
I j k
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
It is clear
= that a x b = o if the third row is proportional to the second, e.g. vectors a is parallel
to vector b.
.
Type equation here.
Ex 1. Check vectors a = (2 ,-3 ,6) and b = (6,9,18) for parallelism .
I j k
I (3)(18)- j ( 2)(18) + k( 2)(9)-(6)(3) = 0 or quikly 3 times
=
axb=
2 3 6
second row = third row.
6 9 18
So second and third vectors are parallel.
Remark : It is easy to prove that both dot and cross multiplication are distributives e.g.
( a + b ) . c = a . c + b . c and (a + b) X c = a X c + b X c .
If we are using standard notation for vector we are going to need multiplication for
I, j and k unitary vectors .
Ex 3. Find the values of I , j ,k
dot product I.I = j . j = k .k = 1 (Angle 0 )
I.J = J . K = I .K = 0 ( Angle 90 )
(𝒙1 𝒊 + 𝑥2 𝒋 + x 3 k) .( 𝒚1 𝒊 + 𝑦2 𝒋 + y 3 k) = 𝑥1 𝑦1 + 𝑥2 𝑦2 + 𝑥3 𝑦3
This a second way to calculate the dot product.
Ex 4 Calculate (2,1,5) . (2,3,1)
=2.2+1.3+5.1 = 12
Now we can calculate the angle as a.b = ||a|| ||b|| cos ( )  cos ( ) =
<(a,b) = 𝑐𝑜𝑠 −1 [
6
105
]=
12
(30) (14)
Type equation here.
Ex 5. Find the proyection of v= 2i-3j+5k onto w=2i-2j+k.
( v).(w) = 4+6+5=15 , so the proyection over v onto w is
(𝑣.𝑤)
W=
𝑤.𝑤
15
9
(2 I -2 J+K).
Ex 6. Prove that a x b =||a||.||b|| sin ß
Type equation here.
𝑎.𝑏
𝑎 𝑏
sin ß= 1 − 𝑐𝑜𝑠 2 ß  1 −
a x b = ||a|| . || b||sin .
2
𝑎| 𝑏

2
− (𝑎.𝑏)2
(||a|| . || b||)
2
=
𝑎𝑥𝑏
||a|| . || b||
𝜃
Remark: This probe that a x b = area of the parallelogram defined by a and b. Besides the triple product a.(b x c)
measured the volumen of the parallepiped defined by a , b , c. the determinant:
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝑐1 𝑐2 𝑐3
𝜃
= a.(b x c)
SURFACES IN R𝟑
I ) PLANES: A plane is the most simple in space , similar to the line in two dimension. The equation of a plane is
a x + b y + c z =d where a , b , c , where a , b , c , d are real numbers
Since a x + b y + c z = (a , b , c) . (x , y , z) we have that the vector (a , b , c ) is perpendicular to with every vector (x
, y , z) in the plane. We will call this vector the normal vector to the plane.
Ex1: Find the plane containing vectors v = (1,3, -2 ) , w = (2,5,1) and containing the point P= (1,2,4)
We need to find a vector perpendicular to u and v and v x w do that.
Type equation here.
I j k
1 3 -2 = 13 I - 5 j – 8 k , this vector is normal to the plane , and the equation of the parallel plane
2 5 1
plane passing thought (0,0,0) is 13x-5y-8z = 0 . The plane parallel containing
P = (1,2,4) is 13x-5y-8z = d, meaning 13(1) - 5(2) – 8(4)= - 29 , so d=-29
Finally the plane equation is is 13x-5y-8z = = -29.
𝑛1
Ex 2. Find the acute angle define by the planes 3x + 2y –z = 3 , 5x + 4y + z = 5
𝑛2
We have to see that the angle define by the planes is equal to the
corresponding angle defines by the normal to the planes meaning
7
7
Cos (𝜃) =(3,2,-1)). (5,4,1)/( 14 42 =
= 𝑐𝑜𝑠 −1 (
).
2 3
2 3.
Ex 3. Find the equation of the common line to the planes 2x+5y+3z = 4 and x+4y+z = 5
By eliminating z we get -3y+z =- 6. Now shifting y try to left 0 in the second member and
the second member 3(y+2) – 6 +z = - 6  3(y+2)+ z=0  calling y+2 = 𝜆 we get z=-3𝜆
and since y = 𝜆 − 2 , we can get x as function of 2 +5(𝜆 − 2) +3z = 0  2+ 5𝜆 – 4 = -3z so
x= 5𝜆 -2. So we have.
𝑥+2
𝑦+2
𝑧
Isolating 𝜆 from each equations we get 𝜆 =
=
=
the equation of the common
5
1
3
line that is v = (5,1,3) is vector on the direction of the line that pass through P=(-2,-2,0)
To get the parametric expression could be done in a different way .
For example from -3y + z =-6  z = 3y + 6 if we make y= 𝜆 and then z=3 𝜆+6 , and we
calculate x from x+4y+z = 5  x+4 𝜆 +3 𝜆 = 5  x=5 -7 𝜆 . Now we need to write the
equation of the line.
𝑥−5
𝑦−0
𝑧−6
In summary x=5-7𝜆 , y = 𝜆 , z = 3𝜆 finally the equation is
=
=
−7
1
3
Type equation here.
EX Find the equation of the line that passes through the point Q=(2-1,3) and is orthogonal to the plane
3 x- 7y +5z+ 55 = 0. Where does the line intersect the plane?
We know that the orthogonal to the plane is n = (2,-7,5), the line has to be parallel to n and contains the
the point Q(2,-1,3) , so we replace on the equation of the line we get:
x = 2+ 3 t
𝑥−2
3
==
𝑦+1
−7
=
𝑧−3
5

y= - 1-7t
and substituting on the equation of the plane
z= 3 +5t
n
Q*
3(3t +2) - 7(-7t-1 ) +5 (5t +3) = -55 
9t +6 +49t +7 +25t +15 = -55  83t =-83  t=-1
P(-1,6,-2)
Now using t= - 1 , we get
x= -1, y-6 , z=-2  so (-1,6,-2)
Q=( 2,-1,3)
MORE APPLICATION OF DOT AND CROSS MULTIPLICATION OF VECTORS .
1) TRIPLE SCALAR PRODUCT.
2) u x v =||u|| ||v||sin∝
||h|||cosß= 𝐴
So | u x v .w| = Volume of parallelepide.
3) Exercise. Prove that
axb.c =
uxv
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝑐1 𝑐2 𝑐3
2
Type equation here.
2
w ∝
h
*
v
u
Y
2
5) Find the plane tangent to the sphere 𝑥 + 𝑦 +𝑧 = 10 at P=(1,0,3).
p
*
The plane is perpendicular to the radius OP , and the equation of the radius is
QP = (1, 0, 3 ) -- (0, 0, 1 ) = (1, 0 ,2 ) and
OQ = (0 ,0 ,1 )
So, equation of the plane x + 2z = t , since P (1 , 0, 3 ) we get t=7
Finally the equation of the plane is 2z=7
O
Z
Type equation here.
Q
X
2
i
3
2
-30
2
j
3
k
2 0
0
0
2
2
Z
T = PQ x F
y
P
F
X
In Physics when a force F is applied to a point Q of an “arm” vector
PQ , then the Torque of F around P is defined as the cross product
of the arm PQ with the force F. The magnitude of the torque ||T||
provide a measure of the tendency of the arm PQ to rotate
counter clock about the an axis perpendicular to the plane
determinate by PQ and F.
EXAMPLE. The half-open door open that is 3 ft. The horizontal
force of 30 lb. is applied at the edge of the door. Find the cos
torque of the force about the hinge on the door.
3
3
= 3(cos(45) i+sin45j) = 2 𝑖 +
2j+k
Type
equation
here.
Q
3 FT
= 45 2 k
y
The magnitude of the torque is 45 2 ft-lb.
F
Force =-30i
X