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प्रश्न
Find the slope of tangent to the curve x = sin θ and y = cos 2θ at θ = `pi/6`
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उत्तर
Given, x = sin θ and y = cos 2θ
Differentiating w.r.t. θ, we get
`("d"x)/("d"theta)` = cos θ and `("d"y)/("d"theta)` = – 2sin 2θ
∴ `("d"y)/("d"x) = ((("d"y)/("d"theta)))/((("d"x)/("d"theta))) = (-2sin2theta)/(cos theta)`
Slope of the tangent at θ = `pi/6` is
`(("d"y)/("d"x))_(theta = pi/6) = (-2sin2(pi/6))/(cos(pi/6))`
= `(-2sin(pi/3))/(cos(pi/6))`
= `(-2 xx sqrt(3)/2)/(sqrt(3)/2)`
= – 2
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