12.8 Algebraic Vectors & Parametric Equations In 12-7, we focused on the geometric aspect of vectors.
Download ReportTranscript 12.8 Algebraic Vectors & Parametric Equations In 12-7, we focused on the geometric aspect of vectors.
12.8 Algebraic Vectors & Parametric Equations
In 12-7, we focused on the geometric aspect of vectors. 12-8 focuses on the algebraic properties.
(
x, y
) v
v y v
v Note: the book will use ( , ) for a point and a vector! Be
x
careful! We will use ,
v
v
x
2
y
2 is the norm/ magnitude Unit vector: vector with magnitude 1 1,0
i
v (horizontal) 0,1
j
v Component Form: 3, 4 can be expressed as 3 1,0 (vertical) 4 0,1 3
i
v v
v
v
v
is the unit vector 4 w/ same
j
v direction as
v
v Polar Form:
r
cos 1, 0
r
sin is direction angle of 0,1
r
cos
i
v tan
y x
r
sin
j
v
We can determine a vector if we know its initial (
x
1 ,
y
1 ) and terminal points (
x
2 ,
y
2 ).
x
2
x y
1 2
y
1 or
x
2 1
y
2
y
1
j
v Ex 1) Given
v
with initial point (2, 3) & terminal point (7, 9), determine: a) component form v
v
i
v
j
v 5
i
v 6
j
v b) polar form
r
5, 6 5 2 6 2 61 61 cos 50.2
i
v tan 6 5 61sin 50.2
j
v 50.2
c) unit vector in same direction as v
v
5, 6 v
v
5, 6 61 5 61 6 61 , 61 61
Vector operations: If v
v
and v
w
vector sum: v
v
v vector difference: v
v
v scalar multiplication:
x
x
s
s
Ex 2)
s
v v 3,5 and
u v
v 2 3,5 6,10 v 13, 7 v
v
2
s
v
u
v
In this picture,
Q
(
x
,
y
) is a point and
P
(
a
,
b
) is a point.
P
(
a, b
) v
v
If you wanted to get to
Q
(
x
,
y
) from
P
(
a
,
b
), we could add a vector
Q
(
x, y
) (
x
,
y
) = (
a
,
b
) + vector Since we don’t know the size of the vector, we can multiply by a scalar to get to the point.
v
tv
v
v
, the direction vector will be given to you, or you can find it by subtracting the 2 nd point – 1 st point that they give you. so, (
x
,
y
) = (
a
,
b
) +
t c
,
d
(
x
,
y
) = (
a
,
b
) +
tc
,
td
(
x
,
y
) = (
a + tc
,
b + td
) this leads to
x = a + tc
and
y
=
b + td
these two eqtns are called parametric equations with parameter
t
of the line
Ex 3) Determine a direction vector of the line containing the two points
P
(5, 8) and
Q
(11, 2). Then find the equation of the line & a pair of parametric equations of the line.
direction vector =
Q
–
P
= (11 – 5, 2 – 8) = 6, –6 vector equation of line: (
x
,
y
) = (5, 8) +
t
6, –6 parametric:
x
= 5 + 6
t
and
y
= 8 – 6
t
We can also represent other graphs (not just lines) in parametric.
Ex 4) Graph the curve with parametric equations
x
= 3cos
t
and
y
= 5sin
t.
Find an equation of the curve that contains no other variables but
x x x
2 3cos
t
9 cos 2
t
square
y y
2 5sin
t
25sin 2
t x
9 2
y
2 25 1 & Ellipse
y
.
x
9 2 put cos 2
t
together
x
9 2
y
2 25 sin 2
t
y
2 25 cos 2
t
sin 2
t
1