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Question Number 76883 by john santu last updated on 31/Dec/19
what is the perimeter of the loop?  3ay^2  = x(x−3a)^(2 ) ?
whatistheperimeteroftheloop?3ay2=x(x3a)2?
Commented by MJS last updated on 31/Dec/19
length of a curve y=f(x) within [p, q]  ∫_p ^q (√(1+(y′)^2 ))dx  in this case  y=±(x−3a)(√(x/(3a)))  due to symmetry we can take the upper curve ×2  p=0; q=3a  y′=((3(x−a))/(2(√(3ax))))  this leads to  (√(3/a))∫_0 ^(3a) (√((x^2 −((2a)/3)x+a^2 )/x))dx  and I found no way to exactly solve this  but it′s c×a with c=8.14677558193...
lengthofacurvey=f(x)within[p,q]qp1+(y)2dxinthiscasey=±(x3a)x3aduetosymmetrywecantaketheuppercurve×2p=0;q=3ay=3(xa)23axthisleadsto3a3a0x22a3x+a2xdxandIfoundnowaytoexactlysolvethisbutitsc×awithc=8.14677558193

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