Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex ...
Transcript of Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex ...
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 1 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 5.1 – The student will explore methods of vector addition and subtraction. 5.2
– The student will develop strategies for computing a vector’s direction angle and magnitude given
its coordinates. 5.4 – The student will resolve vectors into unit vectors. 5.7 – The student will solve
real-world application problems using vectors in two and three dimensions.
Directed Line Segment: a segment with direction and distance
AB A: Initial Point (start); B: Terminal Point (end)
Coordinates of A: ,x yA A Coordinates of B: ,x yB B
Magnitude (length) of a Directed Line Segment AB : 22
x x y yAB B A B A
Note: This is the distance formula!
Vector (v): the set of all directed line segments that are equivalent to a given directed line segment
Note: Equivalent means same magnitude and direction.
Component Form of a Vector: ,x x y yB A B A
Ex: Graph the vector 3, 2AB and find the magnitude.
One possible graph:
Note: 3, 2AB could be placed anywhere on the coordinate grid. We have
placed it in standard position, which is with the initial point at the origin.
Magnitude: 2 2
3 0 2 0 9 4 13AB
Note: If a vector u is written in component form, 1 2,u uu , then the magnitude of u is
2 2
1 2u uu . This is because the initial point is the origin, 0,0 .
Vector Addition: Let 1 2,u uu and 1 2,v vv . Then 1 1 2 2,u v u vu v .
Scalar Multiplication: Let 1 2,u uu and k be any constant. Then 1 2,k ku kuu .
Note: If 0k , then ku is in the opposite direction.
A
B
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 2 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Ex: Use the graph of the vectors to complete each example below.
1. Show that u v .
Show that u v . 2 2
6 1 4 1 25 9 34u
2 2
4 1 1 4 25 9 34v
Show that the direction of u is the same as the direction of v.
Use slope: Direction of u = 4 1 3
6 1 5; direction of v =
1 4 3
4 1 5
The direction and magnitude are the same, so u v .
2. Find the component form and the magnitude of and u w .
Component form of u: 1 6,1 4 5, 3u 34u (see above)
Component form of w: 2 4 ,5 3 2,8w 2 22 8 4 64 68 2 17w
3. Find the component form of 2 3u w .
2 3 2 5, 3 3 2,8 10, 6 6, 24 10 6, 6 24 16, 30u w
Unit Vector: a vector with a magnitude of 1
A unit vector in the direction of a vector v can be found by dividing v by the magnitude of v.
Unit Vector in the Direction of v: v
v
Standard Unit Vectors: unit vectors i and j in standard position along the positive x- and y-axes
1,0 & 0,1i j
Any vector can be written in terms of the standard unit vectors.
Ex: Write the vector 2,5v in terms of the standard unit vectors.
2,5 2 1,0 5 0,1 2 5v i j
u
v
w
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 3 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Ex: Find a unit vector in the direction of the given vector. Verify your answer is a unit vector
and give your answer in component form and standard unit vector form. 2 4i j
Find the magnitude: 22
2 4 2 4 20 2 5i j
Divide the original vector by its magnitude: 2 4 2 4 2 5 2 5
5 52 5 2 5 2 5 5 5
i j i j i ji j (SUV)
Component Form: 5 2 5
,5 5
Verify magnitude of unit vector:
2 2
5 2 5 5 2 5 5 20 25, 1
5 5 5 5 25 25 25☺
Recall: In the unit circle, cos , sinx y . This leads into another way of expressing a vector, in
terms of its direction angle, θ.
Direction Angle: in standard position, the angle the vector makes with the positive x-axis
(counterclockwise)
Resolving a Vector: in terms of its direction angle, θ, a vector can be written as
cos ,sin cos sinu u i u j
Ex: Find the magnitude and direction angle of 2 6v i j .
Magnitude: 2 2
2 6 40 2 10v
Direction angle: cos sinv i v j 2 cosv
1
2 2 10 cos
1cos
10
1cos 108.43
10
OR 6 sinv
1
6 2 10 sin
3sin
10
3sin 71.57
10
but since we know 2 6v i j is in Quadrant II, 180 71.57 108.43
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 4 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Ex: Find the component form of v given its magnitude and its direction angle. 5, 30v
5 3 5cos sin 5cos30 5sin30
2 2v v i v j v i j v i j
Application: Resultant Force
Ex: Two forces act on an object: 3, 45uu and 4, 30uv . Find the direction and
magnitude of the resultant force.
Write each vector in component form:
3 2 3 2
cos sin 3cos45 3sin 452 2
u uu u i u j u i j u i j
cos sin 4cos 30 4sin 30 2 3 2v vv v i v j v i j v i j
The resultant force is the sum u v : 3 2 3 2
2 3, 22 2
Application: Bearing
Ex: A plane flies due east at 500 km/h and there is a 60 km/h with a bearing of 45°. Find the
ground speed and the actual bearing of the plane.
Sketch a diagram:
Find the vectors p and w: 500 cos ,500sinp
60cos 45 ,60sin 45w Note: The 45° is the direction angle, not the bearing.
Vector v is the sum p + w: 60cos45 500cos ,60sin 45 500sinv
The second component of vector v must equal zero, because the plane is headed due east.
160sin 45 60sin 4560sin 45 500sin 0 sin sin 4.868
500 500
Bearing of the plane: 90 94.868
Ground speed of the plane:
2
260 cos 45 500 cos 0 60 cos 45 500 cos 4.868 540.623km/hrv
500 km/hr
60 km/hr
45°
θ
w v
p
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 5 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
You Try: Find the component form of v given its magnitude and the angle it makes with the positive x-
axis. 2v , direction: 2 3i j
QOD: In the examples in your notes, we used sine or cosine to find the direction angle of a vector.
Explain how you could use tangent to find the direction angle.
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 6 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objective: 5.3 – The student will explore methods of vector multiplication. 5.5 – The
student will determine if two vectors are parallel or perpendicular (orthogonal). 5.6 – The student
will derive an equation of a line or plane by using vector operations. 5.7 – The student will solve
real-world application problems using vectors in two and three dimensions.
Dot Product: Let 1 2,u uu and 1 2,v vv . The dot product is 1 1 2 2u v u vu v .
Note: The dot product is a scalar.
Ex: Evaluate 5, 2 3,4 .
5, 2 3,4 5 3 2 4 15 8 7
Properties of the Dot Product:
1. u v v u
2. 2
u u = u
3. 0 u 0
4. u v + w = u v u w
5. c c cu v u v u v
Ex: Evaluate the following given 3,6 ; 1,0 ; 5, 2u v w
(a) w w 5 5 2 2 29w w
(b) w 2 2
5 2 25 4 29w
(c) v w u 1 5,0 2 6, 2 3,6 6 3 2 6 30v w u u
(d) v u w u 1 3 0 6 5 3 2 6 3 27 30v u w u
Angle Between Two Vectors: 1cos cos
u v u v
u v u v
Proof: Use the triangle.
Law of Cosines: 2 2 2
2 cosv u u v u v
Property of Dot Product: 2 2
2 cosv - u v - u u v u v
Expand: 2 2
2 cosv v v u u v u u u v u v
Property of Dot Product: 2 2 2 2
2 2 cosv u v u u v u v
Property of Equality: 2 2 cos cosu v
u v u vu v
u
v u − v
θ
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 7 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Ex: Find u v , where θ is the angle between u and v. 5
6, 8,6
u v
5 3cos cos 48 24 3
6 6 8 2
u v u vu v u v
u v
Application
Ex: Find the interior angles of the triangle with vertices .
4 3,2 0 1,2 ; 5 3,1 0 2,1AB AC 11 2 2 1 4cos cos 36.87
55 5A A A
3 4,0 2 1, 2 ; 5 4,1 2 1, 1BA BC
11 1 2 1 1cos cos 71.565
105 2B B B
180 36.87 71.565 71.565C Angles: 36.87 , 71.565 ,71.565
Orthogonal Vectors: two vectors whose dot product is equal to 0
What is the angle between two non-zero orthogonal vectors?
0
cos cos cos 0 90u v
u v u v
Note: If the angle between the vectors is 90°, we may also say they are perpendicular. The word
orthogonal is used instead for vectors because the zero vector is orthogonal to any other vector, but is not
perpendicular.
What is the dot product of two vectors that are parallel? The angle “between” them would have to be
either 180° or 360°.
cos cos180 1u v u v
u vu v u v
or cos cos360 1u v u v
u vu v u v
Parallel Vectors: two vectors whose dot product is equal to −1 or 1
Ex: Are the vectors orthogonal, parallel, or neither? 3 2 , 3 4v i j w i j
Find v w : 3 3 2 4 1v w The vectors are parallel.
A
B
C
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 8 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Vector Projection: the projection of u onto v is denoted by: 2
projv
u vu v
v
Ex: Find the projection of v onto w. Then write v as the sum of two orthogonal vectors, with
one the projwv . 1,3 ; 1,1v w
2 22 2
1 1 3 1proj proj 1,1 proj 2 1,1 proj 2, 2
1 1
w w w w
v wv w v v v
w
proj 1,3 2,2 1,1wv v 2,2 1,1v
Application: Force
Ex: Find the force required to keep a 200-lb cart from rolling down a 30° incline.
Draw a diagram and label The force due to gravity: 200g j (gravity acts vertically downward)
Incline vector: 3 1
cos 30 sin 302 2
v i j i j
Force vector required to keep the cart from rolling: 2
projv
g v
vf g v
2 2
3 10, 100 ,
2 2 3 1 3 1, 100 , 50 3, 50
2 2 2 21
g vv
vf
Magnitude of Force: 2 250 3, 50 (50 3) ( 50) 100 poundsf
u
v
u1 = projvu
u2 = u – u1 Note: The projection of u onto v is
the “shadow” formed by vector u
onto v as light comes straight down.
2 23 1 3 1
12 2 4 4
v
g
v
f
20030
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 9 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Application: Work cos force distanceW
Ex: A person pulls a wagon with a constant force of 15 lbs at a constant angle of 40° for 500 ft.
What is the person’s work?
cos 40 15 lbs 500 ft 5745.33 foot lbsw
You Try: Find the projection of v onto u. Then write v as the sum of two orthogonal vectors, with one
the projuv . 2 3 ;v i j u i j
QOD: If u is a unit vector, what is u u ? Explain why.
40°
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 10 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objective: 1.10 – The student will solve problems using parametric equations.
Parametric Curve: the set of all points ,x y , where x f t and y g t are continuous functions of t
on an interval I (called the parameter interval)
Parameter: the variable t
Parametric Equations: x f t and y g t
Orientation: the directions that results from plotting the points as the values of t increase
Graphing Parametric Equations
Ex: Graph 22 , 2 1, 1 2x t y t t
Note: Choose appropriate values for t first. Then substitute in and find x and y.
Make a table:
Plot points:
Eliminating the Parameter
1. Solve one of the equations for t. (Or if a trig function, isolate the trig function.)
2. Substitute for t in the other equation. (Or use an identity if a trig function.)
Ex: Write the parametric equation as a function of y in terms of x.
a) 22 , 2 1x t y t
Solve for t in the x-equation (easier to solve for): 22
xx t t
Substitute into the y-equation: 2
2 212 1 2 1 1
2 2
xy t y y x
b) 1
,22
tx y
tt
Solve for t in the x-equation (easier to solve for):
2 2
2 2
1 1 1 12 1 2 2
22x x t x t t
tt x x
t x y
−1 2 1
1 −2 1
0 0 −1
2 −4 7
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 11 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Substitute into the y-equation:
2
2 22
2 2
1 1 22
1 21 12
2 2
x
t x xy y y y xt
x x
Ex: Write the parametric equation as a function of y and graph.
tan , 2 tan 1, 02
x y
Note: A parametric equation can be written in terms of θ instead of t.
The x-equation is already solved for the trig function: tanx
Substitute into the y-equation: 2 1y x
Graph:
Using a Trig Identity
Ex: Eliminate the parameter. 6cos , 6sin , 0 2x t y t t
Solving for a trig function won’t help, so we need to use the identity 2 2sin cos 1t t .
Square both equations: 2 2 2 236cos , 36sinx t y t
Add the equations: 2 2 2 2 2 2 2 236cos 36sin 36 cos sinx y t t x y t t
Trig identity: 2 2 2 2 2 236 cos sin 36x y t t x y
Note: The graph is a circle. The parameter interval lets us know that it would go around 1 time.
Writing a Parameterization
Ex: Find the parameterization of the line segment through the points 1, 2A and 2, 3B .
Sketch a graph:
OP OA AP ; AP is a scalar multiple of AB , so OP OA t AB
A
B
O
P
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 12 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
, 1, 2 2 1 , 3 2 , 1, 2 3, 1OP OA t AB x y t x y t
, 1 3 , 2x y t t 1 3 , 2x t y t These equations define the LINE.
Find the parameter interval for the line segment: We want 1 2x .
1 3x t : 1 1 3 0t t 2 1 3 1t t So, 0 1t
Solution: 1 3 , 2 , 0 1x t y t t
Simulating Horizontal Motion
Ex: A dog is running on a horizontal path with the coordinates of his position (in meters) given
by 3 20.2 19 100 70s t t t where 0 15t . Use parametric equations and a graphing
calculator to simulate the dog’s motion.
Choose any horizontal line to simulate the motion: We will choose 3y .
Parametric Equations: 2 20.2 19 100 70 , 3, 0 15x t t t y t
Graph (Calculator must be in Parametric mode):
Note: To see the motion, change the type of line to a “bubble”. If you would like the bubble to move
slower, make the Tstep smaller.
Parametric Equations for Projectile Motion
distance: 0 cosx v t height: 20 0
1sin
2y gt v t h
Note: On Earth, 232 ft/secg or 2
9.8 m/secg . 0v is initial velocity; 0
h is initial height.
Ex: A golf ball is hit at 150 ft/sec at a 30° angle to the horizontal.
a) When does it reach its maximum height?
Height: 2 20 0
1sin 16 150sin 30 0
2y gt v t h y t t
( 0 0h because a golf ball is hit from the ground)
Simplify: 2 216 150sin 30 0 16 75y t t y t t
Maximum height is at vertex: 75 75 11
2 sec2 2 16 32 32
bt t
a
b) How far does it go before it hits the ground?
Hits the ground when 0y : 2 1116 75 0 16 75 0 16 75 0 4 sec
16y t t t t t t
Note: We could have doubled the time it took for the ball to reach its highest point!
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 13 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Distance: 0
11cos 150cos 30 4 608.84 ft
16x v t
c) Does the ball hit a 6 ft tall golfer, standing directly in the path of the ball 580 feet away?
Find the time it takes for the ball to be 580 ft away: 0 cos 580 150cos 30 4.46secx v t t t
Find the height of the ball at this time: 22
16 75 16 4.46 75 4.46 16.23fty t t
No – the ball misses him by about 10 feet.
Application: Ferris Wheel
Ex: Zac is on a Ferris wheel of radius 20 ft that turns counterclockwise at a rate of one
revolution every 24 sec. The lowest point of the Ferris wheel (6 o’clock) is 10 ft above ground
level at the point (0, 10) on a rectangular coordinate system. Find the parametric equations for the
position of Zac as a function of time t (in seconds) if the Ferris wheel starts (t = 0) with Zac at the
point (20, 30).
Time to complete one revolution = 24 sec: 2
24 12
So, in one second, the ferris wheel travels through an angle of 12
.
When 0, 20 & 30t x y : 20cos , 30 20sin , 012 12
x t y t t
You Try: A baseball is hit at 3 ft above the ground with an initial speed of 160 ft/sec at an angle of 17°
with the horizontal. Will the ball clear a 20-ft wall that is 400 ft away?
QOD: How would you write a parametrization for a semicircle?
Remember: cos , sinx r y r
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 14 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 3.3 – The student will differentiate between polar and Cartesian (rectangular)
coordinates. 6.2 – The student will transform functions between Cartesian and polar form. 6.4 –
The student will solve real-world application problems using polar coordinates.
Polar Coordinate: ,r ; r: the directed distance from the pole (origin); θ: the directed angle from the
polar axis (x-axis)
Plotting Points on a Polar Graph
Ex: Plot the points 3 5
3, , 8, 240 , & 2,2 6
A B C .
Point A: Start at the polar axis and go counter-clockwise 3
2 (270°). Place the point 3 units from the
pole (origin). Point B: Start at the polar axis and go clockwise 240°. Place the point 8 units from the
pole. (Note: Each radius drawn in the grid is 15°.) Point C: Start by going counter-clockwise 5
3
(300°) from the polar axis. Place a point 2 units from the pole. Because 2r , you must place the point
on the opposite side of the pole.
Writing the Polar Coordinates of a Point
Ex: Find four different polar coordinates of P.
Starting at the polar axis and going counter-clockwise: 6,30
6
4
2
2
4
6
10 5 5 10
P
6
4
2
2
4
6
10 5 5 10
A
B
C
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 15 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Starting at the polar axis and going clockwise: 6, 330
Going counter-clockwise more than one revolution: 6,390
Using 0r and rotating counter-clockwise: 6,210
Note: There are infinitely many correct answers!
Polar Conversions
Polar to Rectangular: cos
sin
r x
r y
Rectangular to Polar:
1
2 2 2 2 2
tany
x
x y r r x y
Converting from Polar to Rectangular Coordinates
Ex: Convert to rectangular coordinates.
a) 5 2,45
cos
sin
r x
r y
15 2 cos 45 5 2 5
2
15 2 sin 45 5 2 5
2
x
y
5,5
b) 3,3
cos
sin
r x
r y
1 33cos 3
3 2 2
3 3 33sin 3
3 2 2
x
y
3 3 3
,2 2
Converting from Rectangular to Polar Coordinates (Note: Be careful with the quadrant!)
Ex: Convert to polar coordinates.
a) 1, 3
1
2 2 2 2 2
tany
x
x y r r x y
1
tan 3 60
1 3 2r
1, 3 is in Quadrant II, so the polar coordinates are 2,120 .
b) 0, 4
This point is on the negative y-axis, so we know 270 . 4,270
Note: There are other possible answers to these!
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 16 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Converting from Polar to Rectangular Equations
Ex: Convert the equations and sketch the graph.
a) 2r
2 2 22 4 4r r x y Graph is a circle with center at origin & radius 2.
b) 4
2cos cos
4 2
2sin sin
4 2
x r x r x r
y r y r y r
y x
c) secr
1sec cos 1 1
cosr r r x
Graphing in Polar Coordinates on the Calculator
We will check our graphs above. Calculator must be in Polar mode.
a) c)
Note: We cannot check the graph of b) on the calculator, but the line y x represents the angle
454
for all values of r.
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 17 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Converting from Rectangular to Polar Equations
Ex: Convert the equations.
a) 4x
cosx r : cos 4 4 secr r
b) 3 6 2 0x y
2 23 6 2 0 3 cos 6 sin 2 0 3cos 6sin 2 or
3cos 6sin 6sin 3cosx y r r r r r
c) 2 2
2 1 5x y
Expand: 2 2 2 24 4 2 1 5 4 2 0x x y y x y x y
Substitute: 24 cos 2 sin 0 4 cos 2sin 0r r r r r
So 0r or 4cos 2sin 0r . But 0r is a single point. So 4 cos 2 sinr
Application: Finding Distance
Ex: The location of two ships from the shore patrol station, given in polar coordinates, are
2 mi, 150 & 3mi, 80 . Find the distance between the ships.
Sketch a diagram:
Note: The angle between the ships (from the patrol station) is 150 80 70 .
Using the Law of Cosines: 2 2 22 3 2 2 3 cos70 13 12cos70 2.983mid d
You Try: Convert the coordinates. Polar: 2, ; Rectangular: 7,7
QOD: How could you write an expression for all of the possible polar coordinates of a point?
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 18 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 6.3 – The student will sketch the graph of a polar function and analyze it.
Tests for Symmetry of Polar Curves
1. Symmetry about x-axis: ,r is equivalent to ,r
2. Symmetry about y-axis: ,r is equivalent to ,r
3. Symmetry about origin: ,r is equivalent to ,r
Graphing Polar Curves
Ex: Graph and find the domain, range, symmetry, and maximum r-value.
a) 2 4cosr
θ 0
3
2
2
3
r 6 4 2 0 −2
Domain: All real numbers Range: 2 6r Max r-value: 6r
Symmetry: Substitute . 2 4 cos 2 4 cosr Symmetric about x-axis
This curve is called a limaçon.
b) 6sin 3r
θ 0
18
6
5
18
3
7
18
2
r 0 3 6 3 0 −3 −6
Domain: All real numbers Range: 6 6r Max r-value: 6r
Symmetry: Substitute &r . 6sin 3 6sin 3 6sin 3r r r Symmetric about y-axis
This curve is called a rose.
c) 24 cos 2r
θ 0
6
4
r 2 2 0
Domain: All real numbers Range: 2 2r Max r-value: 2r
Symmetry:
Substitute . 2 24 cos 2 4 cos 2r r Symmetric about x-axis
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 19 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Substitute r . 2 2
4cos 2 4cos 2r r Symmetric about origin
Substitute &r . 2 2
4cos 2 4cos 2r r Symmetric about y-axis
This curve is called a lemniscate.
Classifications of Polar Curves
Limaçon Curves: sinr a b and cosr a b
Rose Curves: cosr a n and sinr a n
Petals: odd = n and even = 2n
Lemniscate Curves: 2 2cos 2r a and 2 2
sin 2r a
You Try: Use your graphing calculator to explore variations of sinr a n . Describe the effects of
changing the window, the θ-step, a, n, and changing sin to cos .
QOD: Are all polar curves bounded? Explain.
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 20 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 7.1 – The student will graph a complex number on the complex/Argand plane.
7.2 – The student will represent a complex number in trigonometric (polar) form. 7.3 – The student
will simplify expressions involving complex numbers in trigonometric (polar) form. 7.4 – The
student will compute the powers of complex numbers using DeMoivre’s Theorem and find the nth
roots of a complex number.
Complex Number Plane (Argand Plane): horizontal axis: real axis; vertical axis: imaginary axis
Plotting Points in the Complex Plane
Ex: Plot the points 3 4 , 1 3 , & 2A i B i i in the complex plane.
Absolute Value (Modulus) of a Complex Number: the distance a complex number is from the origin on
the complex plane 2 2a bi a b (This can be shown using the Pythagorean Theorem.)
Ex: Evaluate 3 i . 22
3 3 1 10i
Recall: Trigonometric form of a vector: cos ,sinu
Trigonometric Form of a Complex Number z = a + bi: cos sinz r i
Note: This can also be written as cisz r .
2 2cos , sin ,a r b r r a b tan
b
a
r = modulus; θ = argument
Writing a Complex Number in Trig Form
Ex: Find the trigonometric form of 3 i .
Find r: 2 22 2
3 1 4 2r a b
Find θ: 1 11
tan tan6 63
b
a
11 11 113 2 cos sin or 2cis
6 6 6i i
A
B
C
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 21 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Writing a Complex Number in Standard Form (a + bi)
Ex: Write 9cis in standard form.
Expand: 9cis 9 cos sin 9 1 0 9 0 9i i i
Multiplying and Dividing Complex Numbers
Let 1 1 1 1cos sinz r i and 2 2 2 2cos sinz r i .
Multiplication: 1 2 1 2 1 2 1 2cos sinz z r r i
Division: 1 11 2 1 2
2 2
cos sinz r
iz r
Ex: Express the product of 1z and 2z in standard form.
1 24 cos sin , 2 cos sin4 4 6 6
z i z i
1 2
5 54 2 cos sin 4 2 cos sin 1.464 5.464
4 6 4 6 12 12z z i i i
Powers of a Complex Number: De Moivre’s Theorem cos sin cos sinnn n
z r i r n i n
Ex: Evaluate 5
2 2i .
Rewrite in trig form: 1 2tan 45 135
2
2 2
2 2 4 2r
55
2cis135 2 cis 5 135 32cis 675 32cis 315
nth Roots of a Complex Number:
2 2 2cos sin cisn nn k k k
z r i rn n n n n
, 0,1, 2, ... 1k n
Note: Every complex number has a total of “n” nth roots.
Ex: Find the cube roots of 8i.
Write in trig form: 8, 90r 8cis90
Evaluate the roots: 3 3 90 3608cis90 8cis 2cis 30 120
3
kk
0 : 2cis 30 0 2cis 30 2 cos30 sin 30 3k i i
1: 2cis 30 120 2cis 150 2 cos150 sin150 3k i i
2 : 2cis 30 240 2cis 270 2 cos270 sin 270 2k i i
Roots of Unity: the nth roots of 1
Precalculus Notes: Unit 6 – Vectors, Parametrics, Polars, & Complex Numbers
Page 22 of 22 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 6
Ex: Express the fifth roots of unity in standard form and graph them in the complex plane.
5th Roots of Unity: 5 1 0i 1, 0r 1cis0
55 0 2 21cis0 1cis cis
5 5
k k
0 : cis0 1k
21: cis 0.31 0.95
5k i
42 : cis 0.81 0.59
5k i
63 : cis 0.81 0.59
5k i
84 : cis 0.31 0.95
5k i
You Try:
1. Write each complex number in trigonometric form. Then find the product and the quotient.
1 3 , 2 2 3i i
2. Solve the equation 41 0x . (You should have 4 solutions!)
QOD: Is the trigonometric form of a complex number unique? Explain.