Question

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