polar to cartesian equation, plz help me

Test Preparation AP Test Preparation
1

<p>replace the polar equation by an equivalent cartesian equation</p>

<p>O= symbol for data</p>

<p>r= 2cosO + 2sinO</p>

<p>here's what i did:</p>

<p>square both sides so i get: r^2 = 4cos^2O + 4sin^2O
factor out 4: r^2 = 4(cos^2O + sin^2O)
use trig identity to simplify: r^2= 4(1) (since cos^20 + sin^2O = 1)
replace r^2 with x^2 + y^2 : x^2 + y^2 =4</p>

<p>FINAL ANSWER IN CARTESIAN FORM IS: x^2 + y^2 =4</p>

<p>however this answer is wrong according to the answer in the back of the book, it should be: x^2 + y^2 = 2x + 2y</p>

<p>what have i done wrong???</p>

2

<p>you squared both sides wrong</p>

3

<p>It is spelled "theta." By the way, when you square "r= 2cosO + 2sinO", shouldn't you use foil on the right side since the two terms are separated by an operation?</p>

4

<p>i forget to square correctly
so i have now:
r^2 = 4cos^2O + 4sin^2O + 8sinOcosO
which simplifies to
x^2 + y^2 = 4 + 8sinOcosO</p>

<p>Now how do i get rid of that 8sinOcosO??</p>

5

<p>x = r*cosO
= 2(cosO + sinO)cosO
= 2cos^2O + 2sinOcosO</p>

<p>y = r*sinO
= 2(cosO + sinO)sinO
= 2sin^2O + 2sinOcosO</p>

<p>2x + 2y = 4cos^2O + 4sinOcosO + 4 sin^O + 4sinOcosO
= 4 + 8sinOcosO</p>

<p>so x^2 + y^2 = 2x + 2y</p>

6

<p>BassDad you proved that: x^2 + y^2 = 2x + 2y
but in the problem i was ONLY given "r= 2cosO + 2sinO"
i need to convert that to cartesian
so i was doing it as follows:
r= 2cosO + 2sinO
r^2 = 4cos^2O + 4sin^2O + 8sinOcosO
x^2 + y^2 = 4 + 8sinOcosO</p>

<p>NOW i am left with "4 + 8sinOcosO" on the right, i need to convert that to terms of x and y. (u should get 2x + 2y according to the answer in the back of the book BUT how do you convert 4 + 8sinOcosO into 2x + 2y ? you solved the problem backwards by using the answer, how do you solve the problem forwards?</p>

7

<p>Run my solution backwards then,</p>

<p>x^2 + y^2 = 4 + 8sinOcosO
= 4(sin^2O + cos^2O) + 8sinOcosO
= 2(2cosO + 2sinO)(cosO) + 2(2cosO + 2sinO)(sinO)
= 2r<em>cosO + 2r</em>sinO
= 2x + 2y</p>

<p>On all these polar-cartesian conversion problems, you have three equations that will not change:</p>

<p>x^2 + y^2 = r^2
x = r<em>cosO
y = r</em>sinO</p>

<p>and you are given r in terms of O. You also almost always have to use
sin^2O + cos^2O = 1 as well. After that, it is just a matter of doing a few of them to get used to manipulating the equations to get rid of all the r and O terms.</p>

8

<p>why did you make
4 + 8sinOcosO
INTO
4(sin^2O + cos^2O) + 8sinOcosO
i thought i simplified the problem by making
4(sin^2O + cos^2O) + 8sinOcosO
INTO
4 + 8sinOcosO
is there any way to solve this problem without using the trig identity
sin^2O + cos^2O = 1 ???</p>

9

<p>Multiply everything by r.
r^2 = 2rcosO + 2rsinO</p>

<p>Using the identities stated above,</p>

<p>x^2+y^2=2x+2y. </p>

<p>Simple as that. You just have to know that x=rcosO, y=rsinO, and r^2=x^2+y^2</p>

10

<p>An elegant solution indeed, evilmonkey32.</p>

11

<p>woa that works evilmonkey ur a genius!!!</p>