Chapter 9, Part 2 Precalc

Graph r= -1 / (2+4sinθ) ON TEST

Divide by 2 (-1/2) / (1+2sinθ) e>1, hyperbola e=2, ep=-1/2, so p=-1/4 REMEMBER- since this is a sin function and there's a horizontal directrix, this means directrix is at y=-1/4 Since this is a sideways hyperbola, the vertices will be on the y-axis. So plug in π/2 and 3π/2 to find them (they're 0, -1/6 and 0, -1/2, respectively) Then find the 2 x-intercepts by plugging in 0 and π (You get 1/2, 0 and -1/2, 0) Then you're done! Double check with desmos

Find polar equation for conic w/ e=3/4 and directrix at y=-4

E less than 1, elipse Directrix is horizontal and negative. p=4 r= (3/4×4) / (1- 3/4sinθ) r=3/ [1- (3/4)sinθ]

If you were going to graph r=3+2cosθ how would you use symmetry

Has x-axis symmetry. Everything below x axis is a reflection. So make a table from 0 to π See example 2, 9-6 notes

Graph θ=5π/3 and put it into rectangular form

It's just a line (put it into desmos) at the 5π/3 place tanθ=y/x tan(-5π/3)=√3 y=√3x or just recognize that this is y=mx+b form and find the slope. Slope is √3

What would r= 7/( 7+ sinθ) look like

Need to get it into ep/ (1+sinθ) form. Divide by 7 r= 1/ ( 1+ (1/7)sinθ) so e<1, elipse Sin is in denominator, so skinny elipse You know p must be 7 (because e=1/7 and ep=1) Plug in π/2 and π to find the points on the major axis Then, once you have the length of the major axis a, use e=c/a and then b²=a²-c² to find the length of the minor (in this case, horizontal) axis

Identify the type of conic represented by r= 15/(3-2cosθ)

Needs to be in form r=ep/(1±ecosθ) so divide everything by 3 r= 5/(1-(2/3)cosθ) then you can see e<1, because 2/3<1 elipse

STUDY THIS- was on review and might be on test Write equation of elipse with vertices (5, 0) and (1, π)

Put into rectangular coordinates (5, 0) and (-1, 0) Find center at (2, 0) need eccentricity a/c a=distance from vertex to center a=3 c=distance from center to focus (0, 0) c=2 e=c/a=2/3 now plug into equation r= (2/3)(p) / 1+ 2/3cosθ Plug in a point like (1,π) to solve for p p=5/3 r= (5/3) / 1+ 2/3cosθ

All polar equations of conics are of the form

Reminder: |p| is the distance between the center and the directrix

What does it mean if the conic equation is in the form ep/1+esinθ

The directrix will be horizontal, and above the center (pole)

What does it mean if the conic equation is in the form ep/1-esinθ

The directrix will be horizontal, and below the center (pole)

What does it mean if the conic equation is in the form ep/1-ecosθ

The directrix will be vertical, and it will be to the left of the center (pole)

What does it mean if the conic equation is in the form ep/1+ecosθ

The directrix will be vertical, and it will be to the right of the center (pole)

If e=1, then the conic is a(n)

parabola remember eccentricity is c/a

Find a polar equation for a parabola with directrix x=-1

parabola so e=1 Directrix is vertical and negative, so -cos equation p=1 since directrix is -1 (and p is the distance from center to focus) ep / (1-ecosθ) 1/ (1-cosθ)

what is the x-axis of with polar coordinate system

polar axis

put θ=11π/6 into rectangular form

remember tanθ=y/x tan11π/6= -1/√3 then y= -x/√3 (look at the slope of the line 11π/6 creates)

Convert (5, -7π/6) to rectangular coordinates

remember x=rcosθ and y=rcosθ you should get (-5√3)/2, 5/2

Put x²+y²-6x into polar form ON TEST

remember x²+y²=r², and that x=rcosθ r²-6rcosθ=0 then just solve for r r²=6rcosθ r=6cosθ

Convert y=x² into polar form

remember y=rsinθ (rule) and x=rcosθ (rule) rsinθ= (rcosθ)² rsinθ=r²cos²θ sinθ=rcos²θ solve for r r=(sinθ)/(cos²θ) r=tanθ * (1/cosθ) r=tanθsecθ

Coordinate conversion- y=

rsinθ

Convert (3, -√3) in rectangular form to polar form

r²=x²+y² r²=3²+(-√3)² r=√12 tanθ= y/x = -√3/3 θ=π/6 PAY ATTENTION TO THE QUADRANT YOURE IN

If youre converting from rectangular to coordinate (from x,y to r,theta), and you're given some coordinates, use

r²=x²+y² then to find the directed angle, θ, use tanθ=y/x

1+tan²θ

sec²θ

analyze r= 32 / (3+5sinθ) ON TEST

see 9-7 notes, example 3

out of sin(x), csc(x), cos(x), sec(x), tan(x), and cot(x) when you do the sin(-x) thing, which ones become negative?

sin(-x)=-sinx csc(-x)=-cscx cos(-x)=cosx sec(-x)=secx tan(-x)=-tanx cot(-x)=-cotx

tanθ=

sinθ/cosθ

elipse w sin in denominator

skinny elipse

Convert θ=π/3 to rectangular

tanθ=y/x tanπ/3=√3 y=√3x If this doesn't make sense, think about π/3 in the polar coordinate system (see paper he gave us) It's just a line, and it's y-intercept is 0, with a slope of √3

what is the directed angle with polar coordinate system

the distance you rotate, counterclockwise from polar axis is positive clockwise from polar axis is negative

What is the pole with polar coordinate system

the origin

Write equation of parabola with vertex at (5, π)

this is a sideways parabola with a vertical directrix- cosine equation It's to left of origin so -cos plug into equation r= 1p / (1-cosθ) p/2 = 5 so p=10 r= 10 / 1-cosθ

figure out how to sketch stupid petal things

ugh

r=(ep)/(1-ecosθ)

vertical directrix p units left of origin

r=(ep)/(1+ecosθ)

vertical directrix p units right of origin

If something has cos, it has _________symmetry

x-axis

sin²θ+cos²θ=

1

secθ=

1/cosθ

cscθ=

1/sinθ

sin2θ=

2sinθcosθ

eccentricity of mars is .09, closest distance to sun is 128000000, find eq of mars

Example 4, 9-7 notes

Limacon r=a±bcosθ a/b<1

Limacon w inner loop

Find equation of elipse with vertices (2, π/2) and (4, 3π/2)

This is going to be a skinny elipse because vertices on x-axis Means sin in the denominator First convert those points to rectangular form. It's (0,2) and (0, -4) The center must be (0, -1) Now find a, the distance from the center to the vertex. It's 3 c, distance from center to focus (which is 0,0 here), is 1 e=a/c=1/3 Now that you know e, plug in (2, π/2) to solve for p r= 1/3p / (1+1/3sinθ) p=8 So final equation r= 8/ 3+sinθ

What is p again?

abs value of distance between pole and directrix

With these conic equations, one focus is always __________

at the pole

Limacon r=a±bcosθ a/b=1

cardioid

cotθ=

cosθ/sinθ

IMPORTANT: how do you know an equation is a rose curve

form of r=acosnθ basically theres a number in front of the θ

r=(ep)/(1+esinθ)

horizontal directrix p units above origin

r=(ep)/(1-esinθ)

horizontal directrix p units below origin

If e>1, then the conic is a(n)

hyperbola remember eccentricity is c/a

Graph r=4/ (1-2cosθ)

hyperbola bc e>1 p=-2 so that's where directrix is. It's vertical directrix bc cos is is denominator Plug in 0 and π to find vertices They're -4 and -4/3 Plug in π/2 and 3π/2 You get 4 but somehow know there's a -4 idk

If you see an r² its probably a

lemniscate-looks like an infinity sign

An equation of the form r=a±bcosθ or r=a±bcosθ will be a

limacon basically a weird circle

Put r²=sin2θ into rectangular form

multiply both sides by r² r⁴=r²(2sinθcosθ) next step idk? r⁴=2rsinθrcosθ then remember rcosθ=x and rsinθ=y and also r²=x²+y² 2xy=(x²+y²)²

Put r=2cosθ into rectangular form

multiply by r on both sides r²=2rcosθ remember r²=x²+y² x²+y²=2rcosθ then remember that rcosθ=x x²+y²=2x

How do you know how many petals a rose curve equation has

n petals if n is odd 2n petals if n is even

What would r=4/(1-cosθ) look like

of form ep/(1-ecosθ) e=1 so parabola -ecosθ so because it's cosine that's a vertical directrix (sideways parabola) the - means it'll open to the right (directrix is left of pole) Check with desmos

If something has r², has __________ symmetry

origin

one focus is always at the

origin

Put r=6/(2cosθ-3sinθ) into rectangular form

r(2cosθ-3sinθ)=6 2rcosθ-3rsinθ=6 remember rcosθ=x and rsinθ=y 2x-3y=6

find eq for parabola w/ focus at pole, directrix x=-4

r= 4 / (1-cosθ)

Rose curve standard equation form

r=acosnθ if it has x-axis symmetry r=asinnθ if it has y-axis symmetry

How would you sketch r=4sin2θ

r=asinnθ since it's sin, y-axis symmetry n is even, and there's 2n petals if n is even (2)(2)= 4 petals since a=4, it will go out to 4 Can plot some points using a θ and r table Go to desmos to double check

Coordinate conversion- x=

rcosθ

Convert xy=5 to polar form

rcosθrsinθ=5 r²cosθsinθ=5 r²= 5/(cosθsinθ) r²=5 × (1/cosθ) ×(1/sinθ) r²=5secθcscθ

put r=10 into rectangular form

remember r²=x²+y² 10²=x²+y² 100=x²+y²

Convert r=2 into rectangular form

remember r²=x²+y² r²=2² r²=4 x²+y²=4

Convert (4, π/4) to rectangular form

x=rcosθ (rule) x=4cos(π/4) x=2√2 y=rsinθ (rule) y=4sin(π/4) y=2√2

Coordinate conversion- r²=

x²+y²

Convert (5, -5) to polar coordinates

x²+y²=r² r=√50=5√2 tanθ=y/x= -5/5 =-1 θ=7π/4 (5√2, 7π/4)

If something has a sin, it has __________symmetry

y-axis

Coordinate conversion- tanθ=

y/x

If youre converting from coordinate to rectangular (from r, theta to x,y), and youre given some coordinates, use

y=rsinθ x=rcosθ

Put x²+y²=9 into polar form

y=rsinθ and x=rcosθ (rsinθ)²+(rcosθ)²=9 r²(sin²θ)+r²(cos²θ)=9 r²(sin²θ+cos²θ)=9 remember that cos²θ+sin²θ=1 r²=9 r=3

An equation of the form r=acosθ or r=asinθ will be a

circle

Limacon r=a±bcosθ a/b≥2

convex limacon

How do you sketch the graph of r=4sinθ by hand

create a table with values 0 through π, and find the corresponding r, plot the points To check you can put it in rectangular form (ON TEST) r=4sinθ r²=4rsinθ x²+y²=4y x²+y²-4y=0 complete the square x²+(y-2)²=4 Circle with center at (0,2), radius of 2

1+cot²θ

csc²θ

Limacon r=a±bcosθ 1<a/b<2

dimpled limacon

What is a

distance from center to vertex

what is c

distance from focus (which will be the origin) to vertex

Find polar eq for conic w/ e=2 and directrix at x=1

e=2 and p=1 Directrix vertical and pos, so +cos / (1+2cosθ)

What would r=4/1-3sinθ look like

e>1 so hyperbola -3sinθ. The sin means horizontal directrix - means directrix below center, e check w/ desmos

If e<1, then the conic is a(n)

elipse remember eccentricity is c/a