If ( x + y ) sin u = x 2 y 2 , then x ∂ x ∂ u + y ∂ y ∂ u =

Given : ( x + y ) sin U = x 2 y 2 ⇒ sin U = x + y x 2 y 2 = v ( let ) Here n = 2 − 1 = 1 Euler's theorem x ⋅ ∂ x ∂ v + y ⋅ ∂ y ∂ v = nv ∴ x ∂ x ∂ s i n U + y ∂ y ∂ s i n U = sin U ⇒ x ⋅ cos U ∂ x ∂ U + y ⋅ cos U ⋅ ∂ y ∂ U = sin U ⇒ x ∂ x ∂ U + y ∂ y ∂ U = c o s U s i n U = tan U ⇒ x ∂ x ∂ U + y ∂ y ∂ U = tan U

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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2007

MathematicsMediumsingle choice

If $(x + y) sin u = x^{2} y^{2}$ , then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} =$

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Given : $(x + y) sin U = x^{2} y^{2}$ $\Rightarrow sin U = \frac{x ^{2} y ^{2}}{x + y} = v (let)$ Here $n = 2 - 1 = 1$ Euler's theorem $x \cdot \frac{\partial v}{\partial x} + y \cdot \frac{\partial v}{\partial y} = nv$ $∴ x \frac{\partial s i n U}{\partial x} + y \frac{\partial s i n U}{\partial y} = sin U$ $\Rightarrow x \cdot cos U \frac{\partial U}{\partial x} + y \cdot cos U \cdot \frac{\partial U}{\partial y} = sin U$ $\Rightarrow x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y} = \frac{s i n U}{c o s U} = tan U$ $\Rightarrow x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y} = tan U$

Stream:VITEEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2007

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