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प्रश्न
Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4 ?
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उत्तर
\[x = \theta + \sin \theta \text { and } y = 1 + \cos \theta\]
\[\frac{dx}{d\theta} = 1 + \cos \theta \text { and } \frac{dy}{d\theta} = - \sin \theta\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- \sin \theta}{1 + \cos \theta}\]
\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{4}} =\frac{- \sin \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}=\frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}=\frac{-1}{\sqrt{2} + 1}=\frac{-1}{\sqrt{2} + 1}\times\frac{\sqrt{2} - 1}{\sqrt{2} - 1}=1 - \sqrt{2}\]
\[\left( x_1 , y_1 \right) = \left( \frac{\pi}{4} + \sin\frac{\pi}{4}, 1 + \cos \frac{\pi}{4} \right) = \left( \frac{\pi}{4} + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}} \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - \left( 1 + \frac{1}{\sqrt{2}} \right) = \left( 1 - \sqrt{2} \right)\left[ x - \left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right) \right]\]
\[ \Rightarrow y - 1 - \frac{1}{\sqrt{2}} = \left( 1 - \sqrt{2} \right)\left[ x - \frac{\pi}{4} - \frac{1}{\sqrt{2}} \right]\]
