Mathematics › Applications of Derivatives › Tangent Normal
A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point…
A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$, then, the equation of the curve is
$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$ $\operatorname{cosec}\left(\frac{y}{x}\right)=\log (x)+2$ $\sec \left(\frac{2 y}{x}\right)=\log (x)+2$ $\cos \left(\frac{2 y}{x}\right)=\log (x)+\frac{1}{2}$ Solution
$\begin{aligned}
& \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \\
& \text { Put } y=\mathrm{v} x \\
& \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x} \\
& \therefore \quad \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x}=\mathrm{v}+\sec \mathrm{v} \\
& \quad \Rightarrow \cos \mathrm{v} \mathrm{dv}=\frac{1}{x} \mathrm{~d} x
\end{aligned}$
Integrating both sides, we get
$\begin{aligned}
& \sin \mathrm{V}=\log (x)+\mathrm{c} \\
& \Rightarrow \sin \left(\frac{y}{x}\right)=\log (x)+\mathrm{c}
\end{aligned}$
The curve passes through $\left(1, \frac{\pi}{6}\right)$.
$\sin \left(\frac{\pi}{6}\right)=\log (1)+c \Rightarrow c=\frac{1}{2}$
$\therefore \quad$ Equation (ii) becomes
$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$
Asked in: MHT CET 2023 (11 May Shift 1)
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