Question

If $(x^2+y^2{)}^2=xy$ Find $\frac{dy}{dx}$.

OR If x = $a(2\theta -sin2\theta )$ and y = a $(1-cos2\theta )$, find $\frac{dy}{dx}$ when $\theta =\frac{\pi }{3}$.

Answer 1

Here $(x^2+y^2{)}^2=xy\Rightarrow 2(x^2+y^2)\{2x+2y{y}^{\prime }\}=x{y}^{\prime }+y.1\Rightarrow 4(x^2+y^2)\{x+y{y}^{\prime }\}=x{y}^{\prime }+y\Rightarrow 4x(x^2+y^2)+4y(x^2+y^2)y^{\prime }=x{y}^{\prime }+y\Rightarrow \left\{4y\right(x^2+y^2)-x\}{y}^{\prime }=y-4x(x^2+y^2)\therefore \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4x^{2}y+4y^3-x}$

OR We have x = a $(2\theta -sin2\theta )$ and y = a $(1-cos2\theta )$
$\Rightarrow \frac{dy}{d\theta }=a(2-2cos2\theta )=2a(1-cos2\theta )\text{and}\frac{dy}{d\theta }=a(0+2sin2\theta )=2asin2\theta \therefore \frac{dy}{dx}=\frac{dy}{d\theta }÷\frac{dx}{d\theta }=\frac{2asin2\theta }{2a(1-cos2\theta )}=\frac{2sin\theta cos\theta }{2si{n}^2\theta }=cot\theta \text{At}\theta =\frac{\pi }{3},\frac{dy}{dx}=cot\frac{\pi }{3}=\frac{1}{\sqrt{3}}$