If x2+y22=xy, find dydx.ORIf x=a(2θ−sin⁡2θ) and y=a(1−cos⁡2θ), find dydx, when θ=π3.

Given: x2+y22=xy......(1)Differentiate the given equation w.r.t x, it follows2x2+y22x+2y⋅dydx =xdydx+y4xx2+y2+4ydydxx2+y2 =xdydx+y⇒ dydx4yx2+y2−x =y−4xx2+y2⇒ dydx =y−4xx2+y24yx2+y2−x⇒ dydx =y−4x3−4xy24x2y+4y3−xTherefore, dydx=y−4x3−4xy24x2y+4y3−x . Which is the required answer.ORGiven: x=a(2θ−sin⁡2θ) and y=a(1−cos⁡2θ)Now, differentiate the x=a(2θ−sin⁡2θ) and y=a(1−cos⁡2θ) w.r.t. , it followsdxdθ=a(2−cos⁡2θ⋅2)=2a(1−cos⁡2θ)and dydθ=a(sin⁡2θ⋅2)=2asin⁡2θ∴ dydx=dydθdxdθ=2asin⁡2θ2a(1−cos⁡2θ)=sin⁡2θ1−cos⁡2θ=2sin⁡θcos⁡θ2sin2⁡θ=cot⁡θ∴ dydxθ=π3=cot⁡π3=13Therefore, dydx=13 when θ=π3.

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If ${(x^{2} + y^{2})}^{2} = xy, find \frac{dy}{dx}$ .
OR
If $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ), find \frac{dy}{dx}, when θ = \frac{π}{3}$ .

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Detailed Solution

Given: ${(x^{2} + y^{2})}^{2} = xy$ ......(1)

Differentiate the given equation w.r.t x, it follows
$\begin{array}{l} 2 (x^{2} + y^{2}) (2 x + 2 y \cdot \frac{dy}{dx}) = x \frac{dy}{dx} + y \\ 4 x (x^{2} + y^{2}) + 4 y \frac{dy}{dx} (x^{2} + y^{2}) = x \frac{dy}{dx} + y \\ \Rightarrow \frac{dy}{dx} [4 y (x^{2} + y^{2}) - x] = y - 4 x (x^{2} + y^{2}) \\ \Rightarrow \frac{dy}{dx} = \frac{y - 4 x (x^{2} + y^{2})}{4 y (x^{2} + y^{2}) - x} \\ \Rightarrow \frac{dy}{dx} = \frac{y - 4 x^{3} - 4 {xy}^{2}}{4 x^{2} y + 4 y^{3} - x} \end{array}$
Therefore, $\frac{dy}{dx} = \frac{y - 4 x^{3} - 4 {xy}^{2}}{4 x^{2} y + 4 y^{3} - x}$ . Which is the required answer.

Given: $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ)$
Now, differentiate the $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ)$ w.r.t. 𝜃, it follows
$\begin{array}{l} \frac{dx}{dθ} = a (2 - \cos 2 θ \cdot 2) \\ = 2 a (1 - \cos 2 θ) \end{array}$
and $\begin{array}{l} \frac{dy}{dθ} = a (\sin 2 θ \cdot 2) \\ = 2 asin 2 θ \end{array}$
$\begin{array}{l} ∴ \frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}} = \frac{2 asin 2 θ}{2 a (1 - \cos 2 θ)} \\ = \frac{\sin 2 θ}{1 - \cos 2 θ} \\ = \frac{2 \sin θcos θ}{2 \sin^{2} θ} \\ = \cot θ \\ ∴ {[\frac{dy}{dx}]}_{θ = \frac{π}{3}} = \cot \frac{π}{3} = \frac{1}{\sqrt{3}} \end{array}$
Therefore, $\frac{dy}{dx} = \frac{1}{\sqrt{3}}$ when $θ = \frac{π}{3}$ .