Question

If a curve is given by $x=a\cos t+\frac{b}{2}\cos 2t$ and $y=\sin t+\frac{b}{2}\sin 2t$ , then the points for which $\frac{d^{2}y}{d{x}^2}=0$ , are given by

• A. $\sin t=\frac{2a^2+b^2}{3ab}$
• B. $\cos t=\frac{a^2+2b^2}{3ab}$
• C. $\tan t=a/b$
• D. None of the above

Answer 1

Answer: (B)

We have, $x=a\cos t+\frac{b}{2}\cos 2t$ and
$y=a\sin t+\frac{b}{2}\sin 2t$ On-differentiating both
equations w.r.t.t, we get $\frac{dx}{dt}=-a\sin t-b\sin 2t$ and
$\frac{dy}{dt}=a\cos t+b\cos 2t\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t+b\cos 2t}{-a\sin t-b\sin 2}$
$\Rightarrow$
$\frac{dy}{dx}=-\frac{(a\cos t+b\cos 2t)}{(a\sin t+b\sin 2t)}$ Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)$
$\cdot \frac{dt}{dx}\Rightarrow$
$\frac{d^{2}y}{d{x}^2}=-\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$
$=\frac{\left[\begin{array}{l}(a\sin t+b\sin 2t)(-a\sin t-2b\sin 2t)\\ -(a\cos t+b\cos 2t)(a\cos t+2b\cos 2t)\end{array}\right]}{\left[\begin{array}{l}{a}^2{\sin }^{2}t+3ab\sin t\sin 2t+2b^2{\sin }^{2}2t\\ (a\sin t+b\sin 2t{)}^2\end{array}\right]}$ $\times \frac{1}{-a\sin t-b\sin 2t}=\frac{\frac{dab{\cos }^{2}t+3ab\cos t\cos 2t+2b^2{\cos }^{2}2t}{(a)}}{a^2\left({\sin }^{2}t+{\cos }^{2}t\right)+b^2\left({\sin }^{2}2t+{\cos }^{2}2t\right)}$ $+3ab\cos (2t-t)$ $=\frac{(a\sin t+b\sin 2t{)}^3}{\Rightarrow a^2+2b^2+3ab\cos t=0\Rightarrow \cos t}=$ $-(\frac{d^{2}y}{d{x}^2}=-\left[\frac{a^2+2b^2+3ab\cos t}{(a\sin t+b\sin 2t{)}^3}\right]$ Given, $\frac{d^{2}y}{d{x}^2}=0\Rightarrow a^2)$