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If $x$ = $cos\theta \; - \; r sin\theta$ , y = $sin\theta \; + \; r cos\theta \\ \; \\ Prove \ that \ \frac{\partial r}{\partial x} \; = \; \frac{x}{ r}$
partial differentiation • 574  views
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Now, $x = cos\theta \; - \; r sin\theta \\ Squaring \; both \; sides \\ \therefore x^2 \; = \; cos^2\theta \; + \; r^2sin^2\theta \; - \; 2rsin\theta cos\theta \;\;\; \ldots (i) \\ y = sin\theta \; + \; r cos\theta \\ Squaring \ both \ sides \\ \therefore y^2 \; = \; sin^2\theta \; + \; r^2cos^2\theta \; + \; 2rsin\theta cos\theta \;\;\; \ldots (ii) \\ Adding \ equation \ (i) \ and \ (ii) \\ \therefore x^2 \; + \; y^2 \; = \; \Big(cos^2\theta \; + \; sin^2\theta\Big) \; + \; r^2 \Big( cos^2\theta \; + \; sin^2\theta \Big) \\ \therefore x^2 \; + \; y^2 \; = \; 1 \; + \; r^2 \; \; \; or \; \; \; r^2 \; = \; x^2 \; + \; y^2 \; - \; 1 \;\;\;\;\; \ldots (iii) \\ \therefore r \; = \; \sqrt{x^2 \; + \; y^2 \; - \; 1} \\ Diff. \; r \; partially \; w.r.t.x \\ \therefore \frac{\partial r}{\partial x} \; = \; \frac{1}{2\sqrt{x^2 \; + \; y^2 \; - \; 1}} \; \frac{\partial }{\partial x} \Big( x^2 \; + \; y^2 \; - \; 1 \Big) \; = \; \frac{2x}{2r} \; \; \; \; \ldots From \; (iii) \\ \; \\ \therefore \frac{\partial r}{\partial x} \; = \; \frac{x}{ r} \ \ \ \ \ \ \ \ \ \ \ \ Hence \; Proved$

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