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Problem in Plotting the polar equation r = cos2θ

Anonymous Mon 20 Nov 2023 14:05:16 No.1420002 View View Reply Original Report

Quoted By: >>1420012 >>1420030

So I turned it into regular rectangular form equation but the graph isn't coming right. I'll be attaching my conversion, the graph i got and the actual graph.

Anonymous

View Same Google ImgOps iqdb SauceNAO photo_2023-11-20_16-38-26_proc.jpg, 60KiB, 1280x294

Anonymous Mon 20 Nov 2023 14:05:36 No.1420003 Report

Quoted By:

Anonymous

View Same Google ImgOps iqdb SauceNAO firefox_pmHflOs8wb.png, 53KiB, 1252x637

Anonymous Mon 20 Nov 2023 14:06:39 No.1420004 Report

Quoted By:

The Graph from the reduction

Anonymous

View Same Google ImgOps iqdb SauceNAO Acrobat_aa20eG7nUz.png, 25KiB, 306x335

Anonymous Mon 20 Nov 2023 14:07:52 No.1420005 Report

Quoted By: >>1420270

Actual Graph

Anonymous

Anonymous Mon 20 Nov 2023 14:34:32 No.1420012 Report

Quoted By: >>1420014 >>1420025

>>1420002
Hard to follow, but I noticed on the first line
> cos 2θ = √((1+cos θ)/2)
which is backwards. The correct formula is
> cos(θ/2) = √((1+cos θ)/2)

Anonymous

Anonymous Mon 20 Nov 2023 14:41:55 No.1420014 Report

Quoted By: >>1420016

>>1420012
(continued)

So the correct formula is
r = cos(2θ) = 2cos^2(θ) - 1
with x = r cos(θ) we get
r = 2(x/r)^2 - 1
multiplying both sides by r^2:
r^3 = 2x^2 - r^2
r^3 + r^2 = 2x^2
x = √((r^3 + r^2)/2) = r√((1+r^2) / 2)

Anonymous

View Same Google ImgOps iqdb SauceNAO desmos.png, 23KiB, 975x357

Anonymous Mon 20 Nov 2023 14:50:03 No.1420016 Report

Quoted By: >>1420017

>>1420014
(continued)

from
x^2 = (r^3 + r^2) / 2
we get
x^2 = ((x^2 + y^2)^(3/2) + (x^2 + y^2)) / 2

I only get half of the expected plot. I think this is because squaring gets rid of some minus signs. I have to think about it.

Anonymous

View Same Google ImgOps iqdb SauceNAO desmos.png, 28KiB, 866x431

Anonymous Mon 20 Nov 2023 14:56:06 No.1420017 Report

Quoted By: >>1420026

>>1420016
(continued)

A minus sign gives the other half. I'm not sure how to get them both in Desmos.

Anonymous

View Same Google ImgOps iqdb SauceNAO desmos.png, 27KiB, 800x349

Anonymous Mon 20 Nov 2023 15:17:06 No.1420020 Report

Quoted By: >>1420262

(continued)

So the plot from your textbook doesn't restrict r to positive values only. The vertical leaves of the rose correspond to parts where r=cos(2θ) is negative.

So we have
r = ±√(x^2 + y^2)
The minus sign disappears in the term r^2 but not in the term r^3.

Anonymous

Anonymous Mon 20 Nov 2023 15:22:28 No.1420025 Report

Quoted By:

>>1420012
WHAT THE FUCK, I need to sleep. Thank you very much brother.

Anonymous

Anonymous Mon 20 Nov 2023 15:24:02 No.1420026 Report

Quoted By:

>>1420017
There are tabs to write below where you wrote the top equation.

Anonymous

Anonymous Mon 20 Nov 2023 15:35:59 No.1420030 Report

Quoted By: >>1420031

>>1420002
you know you can plot polar plots in desmos right? just use r and theta

Anonymous

View Same Google ImgOps iqdb SauceNAO desmos.png, 43KiB, 1090x372

Anonymous Mon 20 Nov 2023 15:54:52 No.1420031 Report

Quoted By:

>>1420030
Didn't know that, but the point is to practice conversion between polar and rectangular coordinates and check his answer.

Anonymous

Anonymous Tue 21 Nov 2023 09:10:18 No.1420262 Report

Quoted By: >>1420263

>>1420020
How can Root x2 + y2 be negative, sum of two squares will always be positive.

Anonymous

Anonymous Tue 21 Nov 2023 09:14:58 No.1420263 Report

Quoted By: >>1420265 >>1420267

>>1420262
it wont be
but the solution to a^2 = b is +-sqrt(a) = b

Anonymous

Anonymous Tue 21 Nov 2023 09:15:59 No.1420265 Report

Quoted By:

>>1420263
other way around
+-sqrt(b) = a

Anonymous

Anonymous Tue 21 Nov 2023 09:16:24 No.1420267 Report

Quoted By:

>>1420263
Yep i got it, i am a dumb fuck. In case of variables there's always plus and minus case.

Anonymous

Anonymous Tue 21 Nov 2023 09:26:39 No.1420270 Report

Quoted By: >>1420643

>>1420005
What book?

Anonymous

View Same Google ImgOps iqdb SauceNAO stewart-calculus-9ed.png, 58KiB, 839x678

Anonymous Wed 22 Nov 2023 17:28:31 No.1420643 Report

Quoted By: >>1420797

>>1420270
looks like Stewart Calculus, but OP seems to have a different edition

Anonymous

Anonymous Thu 23 Nov 2023 05:21:36 No.1420797 Report

Quoted By:

>>1420643
OP here,

It's from Pre-Calculus by Stalwart, Chapter - 8

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