Question

Find the equation of the curve passing through the point $\left(1,\frac{\mathrm{\pi }}{4}\right)$ and tangent at any point of which makes an angle tan⁻¹ $\left(\frac{y}{x}-{\cos }^2\frac{y}{x}\right)$ with x-axis.

Answer 1

The slope of the curve is given as $\frac{dy}{dx}=\tan \theta$.
Here,
$\theta ={\tan }^{-1}\left(\frac{y}{x}-{\cos }^2\frac{y}{x}\right)$

$$\begin{gathered} \therefore \frac{dy}{dx}=\tan \left\{{\tan }^{-1}\left(\frac{y}{x}-{\cos }^2\frac{y}{x}\right)\right\} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-{\cos }^2\frac{y}{x} \end{gathered}$$

$$\begin{gathered} \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=v-{\cos }^{2}v \\ \Rightarrow x\frac{dv}{dx}=-{\cos }^{2}v \\ \Rightarrow {\sec }^{2}vdv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \int {\sec }^{2}vdv=-\int \frac{1}{x}dx \\ \Rightarrow \tan v=-\log \left|x\right|+C \\ \Rightarrow \tan \frac{y}{x}=-\log \left|x\right|+C \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(1,\frac{\mathrm{\pi }}{4}\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ \therefore \tan \frac{\mathrm{\pi }}{4}=-\log \left|1\right|+C \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get} \\ \tan \frac{y}{x}=-\log \left|x\right|+1 \\ \Rightarrow \tan \frac{y}{x}=-\log \left|x\right|+\log e \\ \Rightarrow \tan \frac{y}{x}=\log \left|\frac{e}{x}\right| \end{gathered}$$