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Find the Slope of the Tangent and the Normal to the Following Curve at the Indicted Point X = a (θ − Sin θ), Y = A(1 − Cos θ) at θ = π/2 ? - Mathematics

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प्रश्न

Find the slope of the tangent and the normal to the following curve at the indicted point  x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2 ?

बेरीज
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उत्तर

\[ x = a\left( \theta - \sin \theta \right)\]

\[ \Rightarrow \frac{dx}{d\theta} = a\left( 1 - \cos \theta \right)\]

\[ y = a\left( 1 - \cos \theta \right)\]

\[ \Rightarrow \frac{dy}{d\theta} = a\left( \sin \theta \right)\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\left( \sin \theta \right)}{a\left( 1 - \cos \theta \right)} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = \cot \frac{\theta}{2}\]

\[\text { Now,} \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}} =cot\left( \frac{\frac{\pi}{2}}{2} \right)=cot\left( \frac{\pi}{4} \right)=1\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}}}=\frac{- 1}{1}=-1\]

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पाठ 16: Tangents and Normals - Exercise 16.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 16 Tangents and Normals
Exercise 16.1 | Q 1.07 | पृष्ठ १०

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