Question

convert the polar equation to rectangular form
r=4sin(theta)

Answer 1

how do you do this?

Answer 2

please explain

Answer 3

do you know what a "perfect square trinomial" is?

Answer 4

no....

Answer 5

yeah i do

Answer 6

hmmm ok... what section is this exercise covering anyway?

Answer 7

is taking the square

Answer 8

polar coordinates

Answer 9

perfect square trinomial
\(\large { \begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad

% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array} }\)

looks familiar?

Answer 10

its like a^2+2ab+b^22

Answer 11

well.. more or less.. yes
one sec

Answer 12

okay

Answer 13

\(\bf r=4sin(\theta)\qquad
\begin{cases}
x=rcos(\theta)\\
y={\color{brown}{ rsin(\theta) }} \\
tan(\theta)=\cfrac{y}{x}\\
x^2+y^2={\color{blue}{ r^2}}
\end{cases}\qquad thus
\\ \quad \\
r\cdot r=r\cdot 4sin(\theta)\implies {\color{blue}{ r^2 }} =4{\color{brown}{ rsin(\theta)}} \implies x^2+y^2=4y\)

so... now it's just a matter of "completing the square"
to make it look like a conic equation

Answer 14

okay so the 4y would go to the left of the equation right?

Answer 15

yes
\(\bf x^2+y^2=4y\implies x^2+y^2-4y=0
\\ \quad \\
x^2+(y^2-4y+{\color{red}{ \square ?}}^2)=0\)

so.. any ideas on what we're missing there to "complete the square"?

Answer 16

(-4/2)^2

Answer 17

hmm

well.. the middle term is 2 * "other terms"

so let's see the middle term here is "4y" leaving the negative out
...hmm

Answer 18

well.... -4/2 will be -2

but yes

the 3rd term in a "perfect square trinomial" is positive though
so we'll use +2 :) one sec

Answer 19

okay

Answer 20

keep in mind that
if we ADD \(2^2\)
we also have to SUBTRACT \(2^2\)

since all we're really doing, is borrowing from our good fellow Mr Zero, "0"

thus \(\bf x^2+y^2=4y\implies x^2+y^2-4y=0
\\ \quad \\
x^2+(y^2-4y+{\color{red}{ 2}}^2)-{\color{red}{ 2}}^2=0\implies x^2+(y-2)^2-4=0
\\ \quad \\
x^2+(y-2)^2=4\implies (x-0)^2+(y-2)^2=2^2\)

Answer 21

so... as you can see
is just a circle with a center at (0,2) with a radius of 2

Answer 22

oh okay

Answer 23

so the rectangular for is (x-0)^2+(y-2)^2=4

Answer 24

yeap

Answer 25

thank you

Answer 26

yw