BloodyDiamond
  • BloodyDiamond
how to derive this equation r = a sin 4 (theta) from polar to rectangular?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
FibonacciChick666
  • FibonacciChick666
do you want to change it from polar to rectangular or take the derivative of the polar and change it to cartesian?
FibonacciChick666
  • FibonacciChick666
also, is that r or y?
BloodyDiamond
  • BloodyDiamond
I just want it to change from polar to rectangular

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

BloodyDiamond
  • BloodyDiamond
r
FibonacciChick666
  • FibonacciChick666
ohk, and well, first check out this site https://www.mathsisfun.com/polar-cartesian-coordinates.html
BloodyDiamond
  • BloodyDiamond
thanks
FibonacciChick666
  • FibonacciChick666
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
tkhunny
  • tkhunny
Is it \(\sin^{4}(\theta)\) or \(\sin(4\theta)\)?
BloodyDiamond
  • BloodyDiamond
sin(4 [\theta \])
BloodyDiamond
  • BloodyDiamond
lol
freckles
  • freckles
\[\sin(4 \theta)=2 \sin(2 \theta) \cos(2 \theta)\] use double angle identity for both \[\sin(2 \theta) \text{ and } \cos(2 \theta)\]
freckles
  • freckles
\[\text{ then use } \sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r}\]
BloodyDiamond
  • BloodyDiamond
ok
BloodyDiamond
  • BloodyDiamond
tnx
freckles
  • freckles
np what did you get ?
BloodyDiamond
  • BloodyDiamond
still doing it
freckles
  • freckles
oh
freckles
  • freckles
the result is going to be pretty ugly looking
freckles
  • freckles
\[r = \sin ( 4 \theta) \\ r=2 \sin(2 \theta) \cos(2 \theta) \\ r= 2 (2 \sin(\theta) \cos(\theta)) (\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta)(\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta) ([\cos(\theta)]^2-[\sin(\theta)]^2)\] replace sin(theta) with y/r and replace cos(theta) with x/r
freckles
  • freckles
have you gotten this far?
freckles
  • freckles
if there really is a constant a in there we can throw in there afterwhile right now it is just going to be annoying
BloodyDiamond
  • BloodyDiamond
I did it the wrong way. i substitute the values first
freckles
  • freckles
how did you do that?
BloodyDiamond
  • BloodyDiamond
r = 2 (2y/r) (2x/r)
BloodyDiamond
  • BloodyDiamond
like that
freckles
  • freckles
\[r = \sin ( 4 \theta) \\ r=2 \sin(2 \theta) \cos(2 \theta) \\ r= 2 (2 \sin(\theta) \cos(\theta)) (\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta)(\cos^2(\theta)-\sin^2(\theta)) \\ r=4 \sin(\theta) \cos(\theta) ([\cos(\theta)]^2-[\sin(\theta)]^2) \\ r=4 \frac{y}{r} \frac{x}{r}([\frac{x}{r}]^2-[\frac{y}{r}]^2)\]
freckles
  • freckles
\[r=\frac{4yx}{r^2}(\frac{x^2-y^2}{r^2}) \\ r=\frac{4yx(x^2-y^2)}{r^4}\]
freckles
  • freckles
\[\text{ recall } r^2=x^2+y^2 \]
BloodyDiamond
  • BloodyDiamond
ok
freckles
  • freckles
\[\text{ so } r^4=(x^2+y^2)^2 \\ \text{ and } r= \pm \sqrt{x^2+y^2}\]
freckles
  • freckles
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
freckles
  • freckles
\[\pm \sqrt{x^2+y^2}=\frac{4yx(x^2-y^2)}{(x^2+y^2)^2} \\ \text{ squared both sides so I didn't have to deal with the } \pm \\ x^2+y^2=(\frac{4y(x^2-y^2)}{(x^2+y^2)^2})^2\]
freckles
  • freckles
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) \[\pm \sqrt{x^2+y^2} =a \cdot \frac{ 4yx(x^2-y^2)}{(x^2+y^2)^2} \\ \text{ then \square both sides } \\ x^2+y^2=(a \frac{4 yx (x^2-y^2)}{(x^2+y^2)^2})^2\]
freckles
  • freckles
now if the question really was \[r=a \sin^4(\theta) \\ \text{ then multiply both sides by } r^4 \\ r^5 =a (r \sin(\theta))^4 \\ \text{ recall } \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)=y \\ r^5=a(y)^4 \\ \text{ or } r^5=ay^4 \\ \text{ square both sides } (r^2)^5=a^2 y^8 \\ (x^2+y^2)^5=a^2 y^8\]
freckles
  • freckles
either way crazy equations there
BloodyDiamond
  • BloodyDiamond
wow thank you very much
freckles
  • freckles
np
freckles
  • freckles
were these for fun? or did your teacher really ask you to do these?
freckles
  • freckles
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
freckles
  • freckles
anyways i have to go peace

Looking for something else?

Not the answer you are looking for? Search for more explanations.