Question

If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx x=sin^3t/(cos2t)^1/2, y=cos^3t/(cos2t)^1/2

Answer 1

Given, $x=\frac{sin^3t}{\sqrt{cos2t}},y=\frac{cos^3t}{\sqrt{sin2t}}$

$x=\frac{sin^3t}{\sqrt{cos2t}}$
now, differentiate x with respect to t,
$\frac{dx}{dt}=\frac{\sqrt{cos2t}\frac{d(sin^3t)}{dt}-sin^3t\frac{d(\sqrt{cos2t})}{dt}}{\sqrt{cos2t}^2}\\\\=\frac{\sqrt{cos2t}3sin^2t.cost-sin^3t\frac{(-2sin2t)}{2\sqrt{cos2t}}}{cos2t}\\\\=\frac{3sin^2t.cost.cos2t + sin^3t.sin2t}{cos2t\sqrt{cos2t}}------(1)$

similarly $y =\frac{cos^3t}{\sqrt{cos2t}}$
now differentiate y with respect to t,
then ,we get ,
$\frac{dy}{dt}=\frac{-3cos2t.cos^2t.sint+cos^3t.sin2t}{cos2t\sqrt{cos2t}}-----(2)$

dividing equations (2) by (1),
$\frac{dy}{dx}=\frac{-3cos2t.cos^2t.sint+cos^3t.sin2t}{3sin^2t.cost.cos2t+sin^3t.sin2t}\\\\=\frac{cost.sint(-3cos2t.cost+cos^2.2cost)}{sint.cost(3sint.cos2t + sin^2t(2sint)}\\\\=\frac{-3(2cos^2t-1)cost + 2cos^3t}{3sint.(1-2sin^2t)+2sin^3t}\\\\=\frac{3cost-4cos^3t}{3sint - 4sin^3t}\\\\=\frac{-(4cos^3t-3cost)}{3sint-4sin^3t}\\\\=\frac{-cos3t}{sin3t}\\\\=-cot3t$

hence, dy/dx = -cos3t