Question

find the slope and equation of normal to the curve y=sin theta at theta= pi by 4​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Curve is}\,\;\mathsf{y=sin\,\theta}$

$\underline{\textsf{To find:}}$

$\textsf{Slope and equation of the normal to the curve}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{y=sin\,\theta}$

$\textsf{Differentiate with respect to}\;\mathsf{'\theta'}$

$\mathsf{\dfrac{dy}{d\theta}=cos\,\theta}$

$\textsf{Slope of tangent}$

$\mathsf{m=(\dfrac{dy}{d\theta})_{\theta=\frac{\pi}{4}}}$

$\mathsf{=cos\dfrac{\pi}{4}}$

$\mathsf{=\dfrac{1}{\sqrt{2}}}$

$\textsf{Slope of normal}$

$\mathsf{=\dfrac{-1}{m}}$

$\mathsf{=-\sqrt{2}}$

$\implies\boxed{\textsf{Slope of normal}\;\mathsf{=\sqrt{2}}}$

$\textsf{The equation of normal is}$

$\mathsf{y-y_1=\dfrac{-1}{m}(x-x_1)}$

$\mathsf{y-sin\dfrac{\pi}{4}=-\sqrt{2}(\theta-\dfrac{\pi}{4})}$

$\mathsf{y-\dfrac{1}{\sqrt{2}}=-\sqrt{2}\theta+\sqrt{2}\dfrac{\pi}{4}}$

$\textsf{Rearranging terms, we get}$

$\implies\boxed{\mathsf{\sqrt{2}\theta-y+(\sqrt{2}\dfrac{\pi}{4}+\dfrac{1}{\sqrt{2}})=0}}$