Question

If \(x\) and \(y\) are connected parametrically by the equation,without eliminating the parameter,find \(\frac{dy}{dx}\).
\(x=\frac{sin^3t}{\sqrt{cos\,2t}},y=\frac{cos^3t}{\sqrt{cos\,2t}}\)

Answer 1

The correct answer is \(=-cot\,3t\)
The given equations are \(x=\frac{sin^3t}{\sqrt{cos\,2t}},y=\frac{cos^3t}{\sqrt{cos\,2t}}\)
Then,\(\frac{dx}{dt}=\frac{d}{dt}\frac{sin^3t}{\sqrt{cos\,2t}}\)
\(=\frac{\sqrt{cos\,2t}.\frac{d}{dt}(sin^3t)-sin^3t.\frac{d}{dt}\sqrt{cos\,2t}}{cos\,2t}\)
\(=\frac{\sqrt{cos\,2t}.3sin^2t\,.\frac{d}{dt}(sin\,t)-sin^3t.\frac{1}{2\sqrt{cos2t}}\frac{d}{dt}(cos\,2t)}{cos\,2t}\)
\(=\frac{3\sqrt{cos2t}.sin^2t\,cost-\frac{sin^3t}{2\sqrt{cos2t}}.(-2sin2t)}{cos2t}\)
\(=\frac{3cos\,2t\,sin^2t\,cost+sin^3t\,sin2t}{cos2t\sqrt{cos2t}}\)
\(\frac{dy}{dt}=\frac{d}{dt}(\frac{cos^3t}{\sqrt{cos2t}})\)
\(=\frac{\sqrt{cos\,2t}.\frac{d}{dt}(cos^3t)-cos^3t.\frac{d}{dt}(\sqrt{cos2t})}{cos2t}\)
\(=\frac{\sqrt{cos\,2t}.3cos^2t.\frac{d}{dt}(cos\,t)-cos^3t.\frac{1}{2\sqrt{cos2t}}.(-2sin2t)}{cos2t}\)
\(=\frac{-3cos2t.cos^2t.sin\,t+cos^3t\,sin\,2t}{cos2t.\sqrt{cos2t}}\)
\(∴\frac{dy}{dx}=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{-3cos2t.cos^2t.sint+cos^3t\,sin\,2t}{3cos\,2t\,sin^2t\,cos\,t+sin^3t\,sin2t}\)
\(\frac{-3cos2t.cos^2t.sint+cos^3t\,(2sint.cost)}{3cos\,2t\,sin^2t\,cos\,t+sin^3t\,(2sint.cost)}\)
\(=\frac{sin\,t\,cos\,t[-3cos2tcost+2cos^3t}{sint\,cost[3cos2t\,sint+2sin^3t]}\)
\(=\frac{[-3(2cos^2t-1)cost+2cos^3t]}{3(1-2sin^2t)sint+2sin^3t}\)    \(\bigg[cos2t=(2cos^2t-1),cos2t=(1=2sin^2t)\bigg]\)
\(=\frac{-4cos^3t+3cost}{3sint-4sin^3t}\)
\(=\frac{-cos3t}{sin3t}\)      \([cos3t=4cos^3t-3cost,sin3t=3sint-4sin^3t]\)
\(=-cot\,3t\)