Question

Convert the polar equation to rectangular form and identify the graph.
r= 7/ (2cosΘ+5sinΘ)

Answer 1

y = r sin(theta)
x = r cos(theta)
x^2 + y^2 = r^2

plug and chug

Answer 2

yea so how would i plug it in

Answer 3

cuz i did this last week and i kinda forgot sorry

Answer 4

Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)

Answer 5

okay. i did that

Answer 6

So you have:
\[2rcos(\theta)+5rsin(\theta)=7 ?\]

Answer 7

why would you put r w/ each one of the terms isnt there just 1?

Answer 8

\[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\]
Distributive property!

Answer 9

oh ok i gotchu

Answer 10

you should have it from here. Use the equations sour gave you.

Answer 11

thannk you(:

Answer 12

what if the question's like r= -3sinΘ ?

Answer 13

Try multiplying r on both sides
recall rsin(theta)=y
and r^2=x^2+y^2

Answer 14

hmm ok...

Answer 15

so r^2= -3rsinΘ?

Answer 16

right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2

Answer 17

okay

Answer 18

then youd have?

Answer 19

the square root of that right?

Answer 20

just do exactly what I said replace r sin(theta) with y
replace r^2 with x^2+y^2

Answer 21

i did tht

Answer 22

you are done
what else do you want?

Answer 23

is tht my final answer

Answer 24

oh ok

Answer 25

putting an equation in terms of just x and y ( or just x or just y) is called a cartesian equation
if that was the directions to just right as a cartesian equation then you are done

Answer 26

write*

Answer 27

right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ

Answer 28

multiply both sides by r

Answer 29

do you square both sides 1st

Answer 30

thts what i emant

Answer 31

replace r^2 with x^2+y^2
replace r sin(theta) with y
replace r cos(theta) with x

Answer 32

ok i did that: x^2+y^2= 2y-4x

Answer 33

yep. :)
that is right you wrote the poloar equation as a cartesian equation
if that is all the directions said to do then you are done.

Answer 34

do you stop there

Answer 35

ok

Answer 36

aand i missed r= 3+3cosΘ

Answer 37

so i multiplied both sides by r 1st

Answer 38

and got x^2+y^2=3r +3x but dont i need to get rid of the r?

Answer 39

you could subtract both sides by 3x
giving you \[x^2+y^2-3x=3r \]
then square both sides and replace the r^2 with x^2+y^2

Answer 40

why square both sides?

Answer 41

I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals

Answer 42

so what would be final answer then

Answer 43

what if you with what I was saying
take my equation and then square both sides
and then replace the r^2 with x^2+y^2
you will have your final answer

Answer 44

yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?

Answer 45

\[(x^2+y^2-3x)^2=(3r)^2 \]
the us what you get when you square both sides
your equation is not equivalent to that
you squared each term sorta but not really
it is whatever you do to one side of equation you can do to the other and the equation still holds

Answer 46

\[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]

Answer 47

so..

Answer 48

im not following are you saying keep left side just squared in parentheses?

Answer 49

It is easier than multiplying it out, don't you think?

Answer 50

yes. so only change right side. like substitue

Answer 51

You can multiply it out if you want. Just do it correctly, but the if the directions just say write as a rectangular (aka cartesian equation) than you are done.
If the left hand side was already in terms of x and y. r^2 was the only thing that needed to be converted.