JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
Let $x(t)=2\sqrt2\cos t\sqrt{\sin2t}$ and $y(t)=2\sqrt2\sin t\sqrt{\sin2t}$, $t\in\left(0,\frac\pi2\right)$. Then $\frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}$ at $t=\frac\pi4$ is equal to :
Differentiation and its Application
Solution
Concept. Write $x=r\cos t$, $y=r\sin t$ with $r=2\sqrt2\sqrt{\sin2t}$. Find $r$, $r'$, $r''$ at $t=\frac\pi4$, then use the parametric formulas. Formulas used. (1) $\dfrac{dy}{dx}=\dfrac{\dot y}{\dot x}$, $\ \dfrac{d^2y}{dx^2}=\dfrac{\dot x\ddot y-\dot y\ddot x}{\dot x^3}$(2) $\dot x=\dot r\cos t-r\sin t$, $\ \ddot x=\ddot r\cos t-2\dot r\sin t-r\cos t$(3) $\dot y=\dot r\sin t+r\cos t$, $\ \ddot y=\ddot r\sin t+2\dot r\cos t-r\sin t$Step 1 - $r$ at $\frac\pi4$. $r=2\sqrt2$, $\dot r=\frac{2\sqrt2\cos2t}{\sqrt{\sin2t}}=0$, and $$\ddot r=2\sqrt2\left(-2\sqrt{\sin2t}-\frac{\cos^22t}{(\sin2t)^{3/2}}\right)=-4\sqrt2.$$ Step 2 - Derivatives. With $\sin t=\cos t=\frac{1}{\sqrt2}$, by (2) and (3): $$\dot x=-2,\qquad\dot y=2,$$ $$\ddot x=-4-2=-6,\qquad\ddot y=-4-2=-6.$$ Step 3 - Use (1). $\frac{dy}{dx}=-1$ and $$\frac{d^2y}{dx^2}=\frac{(-2)(-6)-(2)(-6)}{(-2)^3}=\frac{24}{-8}=-3.$$ Step 4 - Value. $\dfrac{1+1}{-3}=-\dfrac23$. Answer: (D)