Question

how to derive this equation r = a sin 4 (theta) from polar to rectangular?

Answer 1

do you want to change it from polar to rectangular or take the derivative of the polar and change it to cartesian?

Answer 2

also, is that r or y?

Answer 3

I just want it to change from polar to rectangular

Answer 4

r

Answer 5

ohk, and well, first check out this site https://www.mathsisfun.com/polar-cartesian-coordinates.html

Answer 6

thanks

Answer 7

and this, because the way your problem is given. You actually have a rose http://jwilson.coe.uga.edu/EMAT6680Fa09/Samples/Assignment%2011/Polar%20Equations.html

Answer 8

Is it \(\sin^{4}(\theta)\) or \(\sin(4\theta)\)?

Answer 9

sin(4 [\theta \])

Answer 10

lol

Answer 11

\[\sin(4 \theta)=2 \sin(2 \theta) \cos(2 \theta)\]
use double angle identity for both \[\sin(2 \theta) \text{ and } \cos(2 \theta)\]

Answer 12

\[\text{ then use } \sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r}\]

Answer 13

ok

Answer 14

tnx

Answer 15

np what did you get ?

Answer 16

still doing it

Answer 17

oh

Answer 18

the result is going to be pretty ugly looking

Answer 19

\[r = \sin ( 4 \theta) \\ r=2 \sin(2 \theta) \cos(2 \theta) \\ r= 2 (2 \sin(\theta) \cos(\theta)) (\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta)(\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta) ([\cos(\theta)]^2-[\sin(\theta)]^2)\]
replace sin(theta) with y/r
and replace cos(theta) with x/r

Answer 20

have you gotten this far?

Answer 21

if there really is a constant a in there we can throw in there afterwhile
right now it is just going to be annoying

Answer 22

I did it the wrong way. i substitute the values first

Answer 23

how did you do that?

Answer 24

r = 2 (2y/r) (2x/r)

Answer 25

like that

Answer 26

\[r = \sin ( 4 \theta) \\ r=2 \sin(2 \theta) \cos(2 \theta) \\ r= 2 (2 \sin(\theta) \cos(\theta)) (\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta)(\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta) ([\cos(\theta)]^2-[\sin(\theta)]^2) \\ r=4 \frac{y}{r} \frac{x}{r}([\frac{x}{r}]^2-[\frac{y}{r}]^2)\]

Answer 27

\[r=\frac{4yx}{r^2}(\frac{x^2-y^2}{r^2}) \\ r=\frac{4yx(x^2-y^2)}{r^4}\]

Answer 28

\[\text{ recall } r^2=x^2+y^2 \]

Answer 29

ok

Answer 30

\[\text{ so } r^4=(x^2+y^2)^2 \\ \text{ and } r= \pm \sqrt{x^2+y^2}\]

Answer 31

http://www.wolframalpha.com/input/?i=+r%3D+sin%284+*theta%29+to+rectangular+

http://www.wolframalpha.com/input/?i=graph++x%5E2%2By%5E2%3D%284yx%28x%5E2-y%5E2%29%2F%28%28x%5E2%2By%5E2%29%5E2%29%29%5E2

as you see both have the same number of petals

Answer 32

\[\pm \sqrt{x^2+y^2}=\frac{4yx(x^2-y^2)}{(x^2+y^2)^2} \\ \text{ squared both sides so I didn't have to deal with the } \pm \\ x^2+y^2=(\frac{4y(x^2-y^2)}{(x^2+y^2)^2})^2\]

Answer 33

this is the rectangular equation for the polar equation r=sin(4 theta)
just multiply that first one by in just the previous post to this one by a
since you had r=a sin(4 theta)
\[\pm \sqrt{x^2+y^2} =a \cdot \frac{ 4yx(x^2-y^2)}{(x^2+y^2)^2} \\ \text{ then \square both sides } \\ x^2+y^2=(a \frac{4 yx (x^2-y^2)}{(x^2+y^2)^2})^2\]

Answer 34

now if the question really was
\[r=a \sin^4(\theta) \\ \text{ then multiply both sides by } r^4 \\ r^5 =a (r \sin(\theta))^4 \\ \text{ recall } \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)=y \\ r^5=a(y)^4 \\ \text{ or } r^5=ay^4 \\ \text{ square both sides } (r^2)^5=a^2 y^8 \\ (x^2+y^2)^5=a^2 y^8\]

Answer 35

either way crazy equations there

Answer 36

wow thank you very much

Answer 37

np

Answer 38

were these for fun?
or did your teacher really ask you to do these?

Answer 39

that last equation I did
I don't know if wolfram has problems with the rectanglular form

but here is the graph when I enter in polar form:
http://www.wolframalpha.com/input/?i=r%3Dsin%5E4%28theta%29

here is the graph when I enter in rectangular form:
http://www.wolframalpha.com/input/?i=%28x%5E2%2By%5E2%29%5E5%3Dy%5E8+to+polar+form

notice they look very much similar but the second one seems a little jagged in some areas

Answer 40

anyways i have to go
peace