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,$\frac{dx}{dt}=\frac{d}{dt}\frac{sin^3t}{\sqrt{cos\,2t}}$
$=\frac{\sqrt{cos\,2t}.\frac{d}{dt}(sin^3t)-sin^3t.\frac{d}{dt}\sqrt{cos\,2t}}{cos\,2t}$
$=\frac{\sqrt{cos\,2t}.3sin^2t\,.\frac{d}{dt}(sin\,t)-sin^3t.\frac{1}{2\sqrt{cos2t}}\frac{d}{dt}(cos\,2t)}{cos\,2t}$
$=\frac{3\sqrt{cos2t}.sin^2t\,cost-\frac{sin^3t}{2\sqrt{cos2t}}.(-2sin2t)}{cos2t}$
$=\frac{3cos\,2t\,sin^2t\,cost+sin^3t\,sin2t}{cos2t\sqrt{cos2t}}$
$\frac{dy}{dt}=\frac{d}{dt}(\frac{cos^3t}{\sqrt{cos2t}})$
$=\frac{\sqrt{cos\,2t}.\frac{d}{dt}(cos^3t)-cos^3t.\frac{d}{dt}(\sqrt{cos2t})}{cos2t}$
$=\frac{\sqrt{cos\,2t}.3cos^2t.\frac{d}{dt}(cos\,t)-cos^3t.\frac{1}{2\sqrt{cos2t}}.(-2sin2t)}{cos2t}$
$=\frac{-3cos2t.cos^2t.sin\,t+cos^3t\,sin\,2t}{cos2t.\sqrt{cos2t}}$
$∴\frac{dy}{dx}=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{-3cos2t.cos^2t.sint+cos^3t\,sin\,2t}{3cos\,2t\,sin^2t\,cos\,t+sin^3t\,sin2t}$
$\frac{-3cos2t.cos^2t.sint+cos^3t\,(2sint.cost)}{3cos\,2t\,sin^2t\,cos\,t+sin^3t\,(2sint.cost)}$
$=\frac{sin\,t\,cos\,t[-3cos2tcost+2cos^3t}{sint\,cost[3cos2t\,sint+2sin^3t]}$
$=\frac{[-3(2cos^2t-1)cost+2cos^3t]}{3(1-2sin^2t)sint+2sin^3t}$ $\bigg[cos2t=(2cos^2t-1),cos2t=(1=2sin^2t)\bigg]$
$=\frac{-4cos^3t+3cost}{3sint-4sin^3t}$
$=\frac{-cos3t}{sin3t}$ $[cos3t=4cos^3t-3cost,sin3t=3sint-4sin^3t]$
$=-cot\,3t$