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Question
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
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Solution
Given, x = cos θ – cos 2θ, y = sin θ – sin 2θ
⇒ x = cos θ – cos 2θ
Differentiating both sides with respect to θ,
`dx/(dθ) = - sin θ - (-sin 2θ) d/(dθ) (2θ)`
= − sin θ + 2 sin 2θ
⇒ y = sin θ – sin 2θ
`dy/(dθ) = cos θ - cos 2θ d/(dθ) (2θ)`
= cos θ − 2 cos 2θ
Hence, `dy/dx = (dy/(dθ))/(dx/(dθ))`
= `(cos θ - 2 cos 2θ)/(- sin θ + 2 sin 2θ)`
= `(cos θ - 2 cos 2 θ)/(- (sin θ - 2 sin 2 θ))`
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