Question

..

Answer 1

for every point you are given the polar coordinates (r, theta)

to find the rectangular (x,y) coordinates simply find

x = r*cos(theta)
y = r*sin(theta)

Answer 2

still there? ^^

Answer 3

Yes I was doing the math lol

Answer 4

would this be right

Answer 5

almost

on the first one the y value would be positive sqrt(2) not negative

everything else is good

Answer 6

wait the sin one right

Answer 7

yes, instead of 2sin(3pi/4) = -sqrt(2) it's sqrt(2)

Answer 8

this is a bit more complicated but for every value (x,y), to find r and theta:

r = sqrt(x^2 + y^2)
and
theta = arctan(y/x)

Answer 9

sqrt(4^2 + 0^2)

Answer 10

4

Answer 11

arctan (0/4)

Answer 12

0

Answer 13

good, so putting everything together you get r= 4 and theta = 0

so (4,0) = your solution

Answer 14

sqrt(3^2 + 4^2) = 5
arctan (4/3) = 53.13 degrees

Answer 15

(5,53.13 degrees)

Answer 16

well done

Answer 17

sqrt(2^2+(-2)^2) = 2 sqrt 2
Arctan (-2/2) = -57.3 degrees

(2sqrt2,-57.3)

Answer 18

arctan(-2/2) = arctan(-1) which has the value -45 degrees

Answer 19

sqrt(2^2+(-2)^2) = 2 sqrt 2
Arctan (-2/2) = -45 degrees
(2sqrt2,-45degrees)

Answer 20

actually since theta is usually given as an angle between 0 and 2pi it would be a good idea to add 360 to that to get 315 degrees

Answer 21

sqrt(2^2+(-2)^2) = 2 sqrt 2
Arctan (-2/2) = -45 degrees + 360 = 315
(2sqrt2, 315 degrees)

Answer 22

good, perfect

Answer 23

uh let's take it one at a time

for the first one start by identifying the angle and radius

Answer 24

5pi/3

Answer 25

good, what about the radius? notice how the red line stops at about halfway between 2 and 3

Answer 26

2.5

Answer 27

good so that gives us one set of polar coordinates

to get another set of polar coordinates we can multiply the r by -1 and subtract pi from the angle value

Answer 28

2.5*-1

Answer 29

good, that gives us -2.5 as the second possible r-value

Answer 30

so 2.5 and -2.5

Answer 31

you must also consider the theta values

if we subtract pi from 5pi/3 we get 2pi/3

so two possible coordinates are (2.5,5pi/3) and (-2.5, 2pi/3)

Answer 32

if you'd like you can try the same process with b)

Answer 33

oh okay i think I get it

Answer 34

i will try it and then you can check it

Answer 35

(4,3pi/2) and (-4,pi/2)

Answer 36

perfect

Answer 37

from what I remember r = 9cos(5theta) should be a five-petalled rose with a radius of 9 (so the farthest parts of the petals should reach 9)

I don't remember the rules for how to orient the rose but you could try plugging in various theta values into the equation to see where the points should be

Answer 38

oh, right, since it's a cos rose there should be a petal aligned exactly with the x-axis and then the other petals should be separated by 2pi/5 radians (you will have to do some estimation)

so something like:

Answer 39

almost, gotta make sure that first petal is symmetric about the x-axis

Answer 40

otherwise good

Answer 41

that petal needs to be right along the x-axis

you can use this as a reference

https://www.desmos.com/calculator/ms3eghkkgz

Answer 42

not above or below it

Answer 43

wait is this file supposed to look like a rose?

Answer 44

i dont see it :/

Answer 45

on desmos you need to plug in 9cos(5theta) as the reference function

it's just called a rose for naming purposes, it's more like a daisy tbh

Answer 46

that petal could be a bit lower but otherwise good

Answer 47

okay so now we have to name out the symmetries for the polar equation. WOULd it just be pi/2

Answer 48

and have to tell what scale I used for the polar axis

Answer 49

oh

well, there's a couple forms of symmetry here:

notice how the rose can be divided in half along each of the five petals

so it's not just the polar axis, it's also theta = 2pi/5, 4pi/5, 6pi/5, 8pi/5

there's also symmetry with respect to the pole (notice how each corresponding point on each petal is the same distance from the origin)

Answer 50

as for the scale, since we made each tick mark equal to 1 unit, simply state that each dotted radial mark represents 1 unit

Answer 51

Okay so we are on the second part to this question

Answer 52

Right

Is there anything about part a) that's still unclear?

Answer 53

I used desmos as a reference

Answer 54

What's the function

Answer 55

NO I think I got part A in the bag. I think I understand it

Answer 56

that I graphed? r=2 cos theta

Answer 57

yeah I think that's alright

need some help with the symmetry and scale?

Answer 58

hold on let me look at the previous question and see if I can figure it out from what you said ^~^

Answer 59

well the scale used is that each dotted radical mark represents 1 unit.

Answer 60

*radial but yes

Answer 61

anyway for symmetry there's only one line, the polar axis

Answer 62

The picture this function makes on the graph is a circle. This is also theta = 0.2pi, 5pi/3, pi/3, The scale used is that each dotted radial mark represents 1 unit.

Answer 63

OH okay

Answer 64

So the theta part is incorrect?

Answer 65

yeah don't include that part

Answer 66

everything else is fine

Answer 67

9a: since we don't have r we only need to consider

arctan(y/x) = theta

taking tan of both sides gives us

y/x = tan(theta)

so y = xtan(theta) = xtan(1.34 radians) so this is just a straight line with slope tan(1.34)

Answer 68

uh

not sure where g is coming from

Answer 69

Lol ,

Answer 70

I think we shall move on , no ? :)

Answer 71

OKay so for the second one, I got all real numbers . I put it into Wolfran Alpha

Answer 72

0 = pi n

Answer 73

sure

so if r = sec(theta)tan(theta) we can re-write the function in terms of sin and cos

r = [1/cos(theta)] * [sin(theta)/cos(theta)]

multiplying both sides by r*cos^2(theta) gives

r^2 * cos^2(theta) = r*sin(theta)

since x = rcos(theta) and y = r(sintheta) we can simply re-write this as

x^2 = y

Answer 74

so it's a parabola