Let x=a sin ^alpha theta cos ^{alpha+1} theta | vedclass

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(A) Given $x=a \sin ^\alpha 	heta \cos ^{\alpha+1} 	heta$ and $y=a \sin ^{\alpha+1} 	heta \cos ^\alpha 	heta$. Calculate $x^2+y^2$: $x^2+y^2 = a^2

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$2 m \alpha=n(2 \alpha+1)$

Vedclass · Answer

(A) Given $x=a \sin ^\alpha 	heta \cos ^{\alpha+1} 	heta$ and $y=a \sin ^{\alpha+1} 	heta \cos ^\alpha 	heta$. Calculate $x^2+y^2$: $x^2+y^2 = a^2 \sin^{2\alpha} 	heta \cos^{2\alpha+2} 	heta + a^2 \sin^{2\alpha+2} 	heta \cos^{2\alpha} 	heta$ $= a^2 \sin^{2\alpha} 	heta \cos^{2\alpha} 	heta (\cos^2 	heta + \sin^2 	heta) = a^2 (\sin 	heta \cos 	heta)^{2\alpha}$. Calculate $xy$: $xy = (a \sin ^\alpha 	heta \cos ^{\alpha+1} 	heta)(a \sin ^{\alpha+1} 	heta \cos ^\alpha 	heta) = a^2 (\sin 	heta \cos 	heta)^{2\alpha+1}$. Now,consider the expression: $\frac{(x^2+y^2)^m}{(xy)^n} = \frac{(a^2 (\sin 	heta \cos 	heta)^{2\alpha})^m}{(a^2 (\sin 	heta \cos 	heta)^{2\alpha+1})^n} = \frac{a^{2m} (\sin 	heta \cos 	heta)^{2m\alpha}}{a^{2n} (\sin 	heta \cos 	heta)^{n(2\alpha+1)}}$ $= a^{2m-2n} (\sin 	heta \cos 	heta)^{2m\alpha - n(2\alpha+1)}$. For the expression to be independent of $	heta$,the exponent of $(\sin 	heta \cos 	heta)$ must be $0$: $2m\alpha - n(2\alpha+1) = 0$ $\Rightarrow 2m\alpha = n(2\alpha+1)$.

Let $x=a \sin ^\alpha \theta \cos ^{\alpha+1} \theta$ and $y=a \sin ^{\alpha+1} \theta \cos ^\alpha \theta$,where $\theta \neq \frac{n \pi}{2}$. If $\frac{(x^2+y^2)^m}{(xy)^n}$ is independent of $\theta$,then the relation between $\alpha, m$ and $n$ is:

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