Under derivatives in calculus trigonometric questions in finding calculus
asked Nov 25, 2016 in Calculus Answers by Martin Kishao

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1 Answer

x/y=(a/b)cotø.

x'=-asinø, y'=bcosø, dy/dx=y'/x'=-(b/a)cotø=-(b^2/a^2)x/y.

But x^2=a^2cos^2(ø) and y^2=b^2sin^2(ø)=b^2(1-x^2/a^2).

d2y/dx2 = -(b/a)^2(y-xdy/dx)/y^2 = -(b/a)^2(bsinø-acosø(-b/a)cotø)/(bsinø)^2 =

-(b/a^2)(sin^2(ø)+cos^2(ø))/sin^3(ø)=-(b/a^2)cosec^3(ø) QED

answered Nov 25, 2016 by Rod Top Rated User (486,780 points)
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