# a5f08s - HINTS TO ASSIGNMENT#5 1 Just evaluate both sides...

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HINTS TO ASSIGNMENT #5 1. Just evaluate both sides and see that they are equal (mixed second partials equal!). For notations see formula 5.33 in your text. 2. In all parts it is useful to apply identity 5.28 of your text and simple result r r r = . (a) r r r r r r r r r 12 ) 3 ( 3 ] ) ( ) ( [ 3 3 ) 3 ( 2 3 = + = + = = = r r r r 12 = k (b) 3 2 2 ) 2 ( 1 2 ) 1 ( 2 ) 1 3 1 ( ) ( r r r r r r r r r r r r r = = = = 2 = k (c) k k k k r k r r k r r = = + 3 3 1 ) ( 1 r r r so 3 = k (d) ) 2 ( ) 1 ( ) 1 ( ) ( 3 2 2 2 2 2 r r r r r r r = = = 4 5 4 2 ] 4 3 [ ) 2 ( r r r r = = r r , k . 2 = 3. The area is , here π 80 )] ( 01 . 0 4 [ 2 2 = + ds y x C 2 0 , sin 10 , cos 10 : = = t t y t x C . For both sides the area is 160 . Hence one needs 2 m 6 . 1 liters of paint. 4. (a) 3 2 6 2 sin 4 2 2 )] 2 )(sin 2 cos 2 )( 2 sin [( 4 0 3 4 0 = = = t dt t t t I (b) t dt t s t 2 2 2 2 ) ( 0 = = , that is 2 2 s = t . For 4 , 0 = t we get 2 2 , 0 = s so the arc length parametrization of C is 2 2 0 ), 2 sin , 2 cos 2 , 2 sin ( ) (

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a5f08s - HINTS TO ASSIGNMENT#5 1 Just evaluate both sides...

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