Saturday
April 25, 2015

Homework Help: calculus

Posted by ms on Wednesday, September 18, 2013 at 4:19am.

Find complete length of curve r=a sin^3(theta/3).
I have gone thus- (theta written as t)
r^2= a^2 sin^6 t/3 and
(dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3)
s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt
=a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int Sqrt[sin^4(t/3)dt
=a Int sin^2(t/3)dt
=a[1/2(t/3)-1/2{3sin(2t/3)}] from t=0 to 2pi
Or, s=a.pi/3-3a[sin(4pi/3)]/4
=a.pi/3+3a[sin(pi/3)]/4
=a.pi/3+3/4*rt3/2=a.pi/3+3rt3/8.
The answer in the book is 3a.pi/2. Where have I gone wrong?

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