a)
dy/dx=dθdydxdθ
dy/dθ=sinθ
dx/dθ=1−cosθ
dy/dx=1−cosθsinθ
b)
equation of the tangent:
y−y0=f′(x0)(x−x0)
then, θ=π/3:
1−cosθ−(1−cos(π/3))=1−cos(π/3)sin(π/3)(θ−sinθ−(π/3−sin(π/3)))
1/2−cosθ=3(θ−sinθ−π/3+3/2)
cosθ+3(θ−sinθ)=π/3−1
c)
the tangent horizontal if f′(x)=0
so,
1−cosθsinθ=0
sinθ=0
cosθ=1
θ=π
y(π)=2,x(π)=π
d)
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