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Problem1.5.3 - Problem 1.5.3(a For this part we need to...

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Problem 1.5.3 (a) For this part we need to make use of the following relations: (1) r 2 = x 2 + y 2 , y ÅÅÅÅ x = Tan @ q D Now differentiating the first equation in (1) with respect to x and then again with respect to y gives (2) 2 r r ÅÅÅÅÅÅÅ x = 2 x ï r ÅÅÅÅÅÅÅ x = x ÅÅÅÅ r = Cos @ q D (3) 2 r r ÅÅÅÅÅÅÅ y = 2 y ï r ÅÅÅÅÅÅÅ y = y ÅÅÅÅ r = Sin @ q D Similary, Differentiating the second equation in (1) (4) Sec @ q D 2 q ÅÅÅÅÅÅÅ x = - y ÅÅÅÅÅÅ x 2 = - rSin @ q D ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ r 2 Cos @ q D 2 Thus (5) q ÅÅÅÅÅÅÅ x = - 1 ÅÅÅÅ r Sin @ q D Similary we get (6) Sec @ q D 2 q ÅÅÅÅÅÅÅ y = 1 ÅÅÅÅ x = 1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ rCos @ q D Thus (7) q ÅÅÅÅÅÅÅ y = 1 ÅÅÅÅ r Cos @ q D (b) Consider a point P in the x-y plane defined by the position vector (8) r = x i + y j Substituting for x and y gives (9) r = rCos @ q D i + rSin @ q D j

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Problem1.5.3 - Problem 1.5.3(a For this part we need to...

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