Posted by john on Thursday, October 14, 2010 at 8:53pm.
take y' = dy / dx
y' = [x cos x -sin x]/x^2
at x = pi/3
y' = [ pi/3 cos pi/3 - sin pi/3 ] / (pi^2/9)
=[ pi/3 *1/2 - sqrt3 /2 ]/(pi^2/9)
= [ pi/6 - 3 sqrt 3 /6 ] 9/pi^2
= 1.5(pi-3 sqrt 3)/pi^2
that is m in y = mx+b
to get b once you calculate m
at x = pi/3
y = sin (pi/3) / (pi/3)
y = (3/pi) sqrt3 / 2
so
1.5 sqrt3/ pi = m (pi/3) + b
solve for b
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