Question

If $x=2cost-cot2t,y=2sint-sin2t$, then $\frac{d^{2}y}{d{x}^2}$ at t = $\frac{\pi }{2}$ is

• A. $\frac{-5}{2}$
• B. $\frac{-3}{2}$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $\frac{-3}{2}$

$\because \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2cost-2cos2t}{-2sint+2sin2t}=\frac{cost-cos2t}{sin2t-sint}=\frac{2sin\left(\frac{3t}{2}\right)sin\left(\frac{t}{2}\right)}{2cos\left(\frac{3t}{2}\right)sin\left(\frac{t}{2}\right)}=tan\left(\frac{3t}{2}\right)\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(tan\frac{3t}{2}\right)=\frac{d}{dt}tan\left(\frac{3t}{2}\right).\frac{dt}{dx}=\frac{3}{2}.se{c}^2\left(\frac{3t}{2}\right).\frac{1}{(-2sint+2sin2t)}\therefore {\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{2}}=\frac{3}{2}se{c}^2\left(\frac{3\pi }{4}\right).\frac{1}{(-2+0)}=-\frac{3}{4}(\sqrt{2}{)}^2=-\frac{3}{2}$