Question

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Answer 1

$$\begin{gathered} x=\theta +\sin \theta \mathrm{and}y=1+\cos \theta \\ \frac{dx}{d\theta }=1+\cos \theta \mathrm{and}\frac{dy}{d\theta }=-\sin \theta \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-\sin \theta }{1+\cos \theta } \\ \text{Slope of tangent=}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}\text{=}\frac{-\sin \frac{\mathrm{\pi }}{4}}{1+\cos \frac{\mathrm{\pi }}{4}}\text{=}\frac{{\displaystyle \frac{-1}{\sqrt{2}}}}{1+{\displaystyle \frac{1}{\sqrt{2}}}}\text{=}\frac{\text{-1}}{\sqrt{2}+1}\text{=}\frac{\text{-1}}{\sqrt{2}+1}\text{×}\frac{\sqrt{2}-1}{\sqrt{2}-1}\text{=}1-\sqrt{2} \\ \left(x_1,y_1\right)=\left(\frac{\mathrm{\pi }}{4}+\sin \frac{\mathrm{\pi }}{4},1+\cos \frac{\mathrm{\pi }}{4}\right)=\left(\frac{\mathrm{\pi }}{4}+\frac{1}{\sqrt{2}},1+\frac{1}{\sqrt{2}}\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\left(1+\frac{1}{\sqrt{2}}\right)=\left(1-\sqrt{2}\right)\left[x-\left(\frac{\mathrm{\pi }}{4}+\frac{1}{\sqrt{2}}\right)\right] \\ \Rightarrow y-1-\frac{1}{\sqrt{2}}=\left(1-\sqrt{2}\right)\left[x-\frac{\mathrm{\pi }}{4}-\frac{1}{\sqrt{2}}\right] \end{gathered}$$