Question

write each polar equation in rectangular form:
r=-3sec(Theta)
r=costheta+sintheta
r=5/3costheta+8sin

Answer 1

^theta after the last sin

Answer 2

You need to use these equations:
\[r^2=x^2+y^2 ; rcos(\theta)=x ; rsin(\theta)=y \]

Answer 3

i did, but like it ended up like x^2+y^2= 25+9x^2r^2-30xr/64y^2 for the last one

Answer 4

We want the equations just in terms of x and y instead of r and theta.

Answer 5

i don't know what i did wrong

Answer 6

so the last one is:
\[r=\frac{5}{3}\cos(\theta)+8\sin(\theta) \]?

Answer 7

no uh (5)/(3costheta+8sintheta)

Answer 8

oh so you meant 5/(3costheta)+8sin(theta)) and not 5/3cos(theta)+8sin(theta)

Answer 9

mhm
5 / (3cos+8sin)

Answer 10

\[r= \frac{5}{3\cos(\theta)+8\sin(\theta)}\]
Multiply both sides by that one fraction's denominator

Answer 11

\[r(3\cos(\theta)+8\sin(\theta))=5 \]
Distribute then substitute the equations I gave you

Answer 12

3xr+8ry=5 but then you'll need to put r on one side cause you would need to square it. and the answer comes out all weird

Answer 13

well rcos(theta) is x not just cos(theta
same thing with the y deal
rsin(theta) is y not just sin(theta)

Answer 14

whoops how about the secant one

Answer 15

put it in terms of cos by using an identity

Answer 16

then put it in some form where you can use the equations I gave you

Answer 17

sosoososososo it would be x^3+y^2x=-3 AND uh the other one x^2+y^2=x+y?

Answer 18

for the first one? no...

Answer 19

I haven't looked at the second one...
You are making this way too hard it looks like.

Answer 20

first one
you did this as first step right?:
\[r=\frac{-3}{\cos(\theta)}\]

Answer 21

yep

Answer 22

Notice this only has a cos in it...
how do you get cos with the r over there so you can use the equation rcos(theta)=x

Answer 23

multiply both sides with r

Answer 24

no

Answer 25

try again

Answer 26

How do you undo multiplication?

Answer 27

Or division?

Answer 28

oh omg

Answer 29

we have division by cos
to undo that you ?

Answer 30

x=-3

Answer 31

yes

Answer 32

thanks

Answer 33

so for that second one...

Answer 34

you actually got it right

Answer 35

but how did you go about it
you didn't choose a complicated way right?

Answer 36

yep uh
multiplied both side with r and then substituted the stuff in

Answer 37

oh good

Answer 38

do you have any other questions? do you think you got this better?

Answer 39

uh yea i have a few more

Answer 40

for standard forms of polar equation uh how would graph r=2cos3theta

Answer 41

Plug in values of theta and see what the r output is

Answer 42

then remember how you graph using polar coordinates

r is the distance from the origin and direction is given by r to from the angle
theta is the angle

Answer 43

like how would you do this with a graphing calculator

Answer 44

So for example if I plug in theta=0
\[r=2 \cos (3 \cdot 0)=2\cos(0)=2(1)=2 => \text{ graph the point } (2,0^o)\]

Answer 45

find 0 degrees and then go to the circle with radius 2

Answer 46

cause wksht says not to use a table

Answer 47

then plug in more points

Answer 48

oh i know how to graph but not without plugging in points

Answer 49

I haven't looked at a calculator in ages

Answer 50

did you try looking up how to input polar equations into whatever calculator type you have

Answer 51

some calculators require a difference procedure in order to do certain things

Answer 52

http://mathbits.com/MathBits/TISection/PreCalculus/polargraphs.htm

Answer 53

find mode
and choose polar or pol or polargc

Answer 54

then just type your equation in

Answer 55

u what's the website

Answer 56

*uh

Answer 57

What ?