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.
The Graph from the reduction
>>1420002Hard 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)
>>1420012(continued)So the correct formula isr = cos(2θ) = 2cos^2(θ) - 1with x = r cos(θ) we getr = 2(x/r)^2 - 1multiplying both sides by r^2:r^3 = 2x^2 - r^2r^3 + r^2 = 2x^2x = √((r^3 + r^2)/2) = r√((1+r^2) / 2)
>>1420014(continued)fromx^2 = (r^3 + r^2) / 2we getx^2 = ((x^2 + y^2)^(3/2) + (x^2 + y^2)) / 2I 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.
>>1420016(continued)A minus sign gives the other half. I'm not sure how to get them both in Desmos.
(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 haver = ±√(x^2 + y^2)The minus sign disappears in the term r^2 but not in the term r^3.
>>1420012WHAT THE FUCK, I need to sleep. Thank you very much brother.
>>1420017There are tabs to write below where you wrote the top equation.
>>1420002you know you can plot polar plots in desmos right? just use r and theta
>>1420030Didn't know that, but the point is to practice conversion between polar and rectangular coordinates and check his answer.
>>1420020How can Root x2 + y2 be negative, sum of two squares will always be positive.
>>1420262it wont bebut the solution to a^2 = b is +-sqrt(a) = b
>>1420263other way around+-sqrt(b) = a
>>1420263Yep i got it, i am a dumb fuck. In case of variables there's always plus and minus case.
>>1420270looks like Stewart Calculus, but OP seems to have a different edition
>>1420643OP here, It's from Pre-Calculus by Stalwart, Chapter - 8