a)
d y / d x = d y d θ d θ d x dy/dx=\frac{dy}{d\theta}\frac{d\theta}{dx} d y / d x = d θ d y d x d θ
d y / d θ = s i n θ dy/d\theta=sin\theta d y / d θ = s in θ
d x / d θ = 1 − c o s θ dx/d\theta=1-cos\theta d x / d θ = 1 − cos θ
d y / d x = s i n θ 1 − c o s θ dy/dx=\frac{sin\theta}{1-cos\theta} d y / d x = 1 − cos θ s in θ
b)
equation of the tangent:
y − y 0 = f ′ ( x 0 ) ( x − x 0 ) y-y_0=f'(x_0)(x-x_0) y − y 0 = f ′ ( x 0 ) ( x − x 0 )
then, θ=π/3:
1 − c o s θ − ( 1 − c o s ( π / 3 ) ) = s i n ( π / 3 ) 1 − c o s ( π / 3 ) ( θ − s i n θ − ( π / 3 − s i n ( π / 3 ) ) ) 1-cosθ-(1-cos(\pi/3))=\frac{sin(\pi/3)}{1-cos(\pi/3)}(θ-sinθ-(\pi/3-sin(\pi/3))) 1 − cos θ − ( 1 − cos ( π /3 )) = 1 − cos ( π /3 ) s in ( π /3 ) ( θ − s in θ − ( π /3 − s in ( π /3 )))
1 / 2 − c o s θ = 3 ( θ − s i n θ − π / 3 + 3 / 2 ) 1/2-cosθ=\sqrt3(θ-sinθ-\pi/3+\sqrt3/2) 1/2 − cos θ = 3 ( θ − s in θ − π /3 + 3 /2 )
c o s θ + 3 ( θ − s i n θ ) = π / 3 − 1 cosθ+\sqrt3(θ-sinθ)=\pi/\sqrt3-1 cos θ + 3 ( θ − s in θ ) = π / 3 − 1
c)
the tangent horizontal if f ′ ( x ) = 0 f'(x)=0 f ′ ( x ) = 0
so,
s i n θ 1 − c o s θ = 0 \frac{sin\theta}{1-cos\theta}=0 1 − cos θ s in θ = 0
s i n θ = 0 sin\theta=0 s in θ = 0
c o s θ ≠ 1 cos\theta\neq 1 cos θ = 1
θ = π \theta=\pi θ = π
y ( π ) = 2 , x ( π ) = π y(\pi)=2,x(\pi)=\pi y ( π ) = 2 , x ( π ) = π
d)
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