bearing complex polar coordinates binomial theorem (terms) trig equations law of cosines binomial...
TRANSCRIPT
Bearing
Complex Polar Coordinates
Binomial Theorem (Terms)
Trig Equations
Law of Cosines
Binomial Theorem (Basic)
Vectors
SSA Triangles
Comments
Please report any errors ASAP by email to [email protected] or IM at kimtroymath.
Problems may be more difficult on test. Consult homework assignment. Not all topics covered.
Ones in read are the ones that have been completed. Remember, some material is on other powerpoints.
Green are always changing.
1) Figure out angles
2) Make vectors
3) Add vectors
4) Find magnitude
5) Find Bearing
A plane is traveling 400 miles per hour west. A wind from a direction of N 60o W is 10 mph. Find the ground speed and bearing of the plane. (I will round to hundredths)
)jsin180i180(cos400: Plane
180o
i400
Wind is coming FROM this direction, which is different from where it is heading.
60o
So it’s really heading E 30o S.
330o
)jsin330i330(cos10: Wind j5i67.8
Distribute the magnitude.
j5i33.391
Remember, magnitude is the same as ground speed.
22 baMagnitude
mph36.391
)5()33.391( 22
Remember inverse tangent is either quadrant I or IV. Make a sketch of the vector to see what quadrant the angle is supposed to be in.
QI Nothing QII Add 180 QIII Add 180 QIV Add 360
Then find the bearing afterwards. componentx
componenty
1tan
73.
33.391
5tan 1 QIII 73.180180
The angle for the bearing is .73. You can either subtract 180, or logically deduce it, or whatever you may need. You don’t always subtract 180. It depends on what quadrant it’s in.
SW 73
Read problems carefully, whether the wind is coming FROM a direction or is HEADING IN A direction. Heading in a direction is straight forward, coming from a direction is trickier.
22 )()( bar
iz 3 iw 2222 Putting into complex polar coordinate form.
1) Find radius
2) Find argument (angle)
a) Inverse Tangent
b) Figure out angle
i) QII, QIII add 180
ii) QIV, add 360
3 1
24 r
1tan
3 1
30
IIIQuadrant
210180
)240sin240(cos2 i
Convert the other complex number into complex polar form. Next click will give answer.
13518045
451tan22
22tan
42222
11
22
IIQuadrant
r
)135sin135(cos4 i
212
1
2
1
212121
222
111
)(
)(
if Remember,
cisr
r
z
z
cisrrzz
cisrz
cisrz
w
z zw
Find
360. subtractedthen
360, and 0between t isn'
argumentyour if Remember,
)345(8
135210)4)(2(
cis
ciszw
)75(2
1
))135210((4
2
w
z
cis
cis
Binomial Theorem
n
j
jjnn axj
nax
0
)()()(Common errors:
1) Parenthesis, you need them. Otherwise your powers will be messed up. (Math kryptonite)
2) Set up the bottom factorial carefully.
3) Keep sign.
)!(!
!
jnj
n
j
nC jn
3)32( yx
3
0
3 )3()2(3
j
jj yxj
)3)(2(3
3)3)(2(
2
3)3)(2(
1
3)3)(2(
0
3yxyxyxyx
3 0 2 1 21 30
Notice:
1) First term starts with exponent, goes down by 1.
2) Second term starts with 0, goes up by 1.
3) Bottom number matches up with second term exponent.
4) Exponents add up to n
5) Term number is ONE MORE than bottom number.
3223
3223
2754368
)27()9)(2)(3()3)(4)(3(8
yxyyxx
yyxyxx
1st term 2nd term 3rd term 4th term
62
62
62
112
)()2(6
8
6
862
yx
yx
isnumberBottom
becauseymeansx
62
26
2
112
)2()(2
8
2,
,)(
yx
xy
jx
xajn
nxwithtermax jjnjn
Binomial Theorem (Terms) Methods
1) Be safe, list them all, pick the one you need
2) Logic
3) Formula
List
Clear
Logic
Clear
Formula
Clear
6th term y with Term with x termFind
)2(42
8yx
80716253
4435261708
)()2(8
8)()2(
7
8)()2(
6
8)()2(
5
8
)()2(4
8)()2(
3
8)()2(
2
8)()2(
1
8)()2(
0
8
yxyxyxyx
yxyxyxyxyx
62
62
112
)()2)(28(
yx
yx
44
44
1120
)()2)(70(
yx
yx
53
53
448
)()2)(56(
yx
yx
1) Bottom number matches up with second term exponent.
2) Exponents add up to n
3) Term number is one more than bottom number.
44
44
1120
)()2(4
8
4
4
yx
yx
ispowerOther
isnumberBottom
53
53
448
)()2(5
8
5
yx
yx
isnumberBottom
The rest, you use logic to set it up so that you can use the formula. Refer to other slide or logic button for logical rules. You use logic to set up the x term.
||||||||cos
,, 2211
vu
vu
bavbauGiven
1212
22
11
yy,xx
:is vector The
)y,Q(xpoint terminaland
)y,P(xpoint initialGiven
jrsinircos
:by form
jbiain written becan vector the
,argument andr magnitudeGiven
)1,1(
)4,3(
Q
P
)7,2(
)5,3(
Q
P
54
41)3(1
,
,
125
57)3(2
,
,
left. on the vectors
twoebetween th angle theFind
||13||||41||
12,55,4cos
4113
80
4113
6020cos
04.16
2121
2211 ,,
bbaavu
bavbauGiven
21
21
11
||||
,
bau
bauGiven
formjbiacomponentinwrite
rwithvectorGiven
)ˆˆ(
200,4
ji ˆ200sin4ˆ200cos4 ji ˆ37.1ˆ76.3
Check the General Trig Powerpoint Ch 6 for good equation examples. If you want a specific hw problem done, e-mail me. I’ll try to fit 1 or 2 in here.
Law of Cosines – To be used with SSS or SAS triangle. There are no ambiguous cases for these triangles.
654 cba 7068 Cba
SSS – 3 sides given SAS – 2 sides given, name of angle is not the letter of the other sides. (or make a sketch)
Baccab
Abccba
Cabbac
cos2
cos2
cos2
222
222
222
Quick Check, small angle small side, middle angle middle side, big angle big side.
CHECK MODE OF CALCULATOR!
Ccos)5)(4(2546 222 Ccos404136
Ccos405
Ccos8
1
Cm81.82
Just showing LOC step. Use LOS to finish the problem.
A
B
Cb
ac
A
B
Cb
ac
70cos)6)(8(268 222 c8339.321002 c
1661.672 c20.8c
SSA triangle. Use law of sines. Rules
h = bsinA
a is the side opposite the angle.
b is side adjacent to angle
a < h, no triangle
a = h, one right triangle
a < b and h < a, 2 triangles
a ≥ b, one triangle
A
b a
A
A
A
a
a a
a
b
b
b
You can use common sense. Set up a law of sines. And then try to find the supplement. See if 0, 1, or both triangles work. Common sense is nice because even if you use the rules, if you notice there are 2 triangles, you will need to use the supplement anyways.m A = 40o; a = 4, b = 5
cCm
bBm
aAm
5
40o
40o 4
4
5
464.534
40sin5sin
5
sin
4
40sin
B
B
B
53.46o
cCm
bBm
aAm
cCm
bBm
aAm
cCm
bBm
aAm
Check supplement and the third angle.
180-53.46o = 126.54o
126.54o
86.54o
13.46o
Both triangles are ok, you can finish using law of sines.
triangles2
421.340sin5
54 :Rules Using
m C = 30o; c = 6, b = 4
630o
4
630o
4
47.196
30sin4sin
4
sin
6
30sin
B
B
B
19.47o
Check supplement and the third angle.
180-19.47o = 160.53o
160.53o
130.53o
-10.53o
2nd triangle is impossible, only need to solve for the top one.
triangle1
46 :Rules Using
9
10
11
12
13
Comments
Make sure you really understand graphing and solving equations.
Graphing, know the formulas, and how to find period, amplitude and shift.