ns am suppose come uncover the polar curve the ns inside loop. Below ins itns graph, courtesy the Wolfram|Alpha,
i am having actually trouble functioning out this polar feature ~ above a cartesian graph device so my manga comes from findinns ns limits the integration because that ns inside loop. I thsquid if $r = 2cos heta -1$ that means ns have actually $r( heta)$ therefore $ heta$ is mine x and r ins my y. For this reason i understand that at $ heta = 0$ i have actually two y values, 0 and 1. Just how does thins work-related though? clearly $2cons (0) -1$ can never before it is in 0. Therefore what is going on here? just how doens ins obtain two values.
You are watching: The graph of the polar curve r=1-2cos
Ins seems prefer no one understands my question, i have 2 worths in ~ theta 0. Exactly how do i get one arc size if ns have actually 2 arcs?
calculus integration polar-works with
edited Auns 13 "13 at 10:03
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inquiry Aug 12 "13 at 20:12
Paul the PiratePaul ns Pirate
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replacing $(r, heta) o(r,- heta)$, us have actually a very same equatitop top therefore ns graph ins symmetric with respecns to the polar axe.
us deserve to have actually the adhering to polar point. Since ns graph is symmetric so we need to draw a half of the graph:
$$ heta=0 o r=1$$$$ heta=pi/6come r=sqrt3-1$$$$ heta=pi/3 o r=0$$$$ heta=pi/2come r=-1$$$$ heta=2pi/3come r=-2$$$$ heta=5pi/6come r=-sqrt3-1$$$$ heta=pi o r=-3$$
If $r=0$, climate $cos( heta)=1/2$ and also so $ heta=pi/3$ offered $0le heta
answer Auns 12 "13 in ~ 20:47
MikasaMikas it is
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friend don"t have actually 2 values. In ~ $ heta = 0$, $r = 1$. As $ heta$ increases, as much as ns allude wright here $ heta = fracpi3$, $r$ decreases, until in ~ that point, $r = 0$.
Going backwardns from $ heta = 0$, $r = 1$, as $ heta$ decreases, $r$ decreases, approximately the point where $ heta = -fracpi3$, in ~ i m sorry point $r = 0$ again.
Thus, ns bounds the integratitop top to be $-fracpi3$ and $fracpi3$, together tracing ns curve via this limits gives the inner loop, native $(-fracpi3, 0)$ roughly through $(0, 1)$, every the method come $(fracpi3, 0)$.
(Or, in cartesian coordinates, $(0,0)$ through $(0,1)$, back come $(0,0)$ again.)
answer Aug 12 "13 in ~ 20:18
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Keep in mind that 1) ns function $r( heta)=2cos heta-1$ is an $2pi$-routine function. 2) Cartesian works with $(x, y)$ deserve to it is in calculate native polar by $$left lbgyeongju egingather x(r, heta)=r( heta) cos heta, \y(r, heta)=r( heta) sin heta.endgather ight.$$
answered Aug 12 "13 at 20:42
M. StrochykM. Strochyk
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