math192_hw3sp06 - 4 lim x,y →(2-3 ± 1 x 1 y ² 2 = 1 36...

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HW3 Solutions 13.3 Arc Length and the Unit Tangent Vector T 8. T = v | v | = (cos t ) i - (sin t ) j and Length = R 2 2 tdt = h t 2 2 i 2 2 10. Let P( t 0 ) denote the point. Then v = (12 cos t ) i + (12 sin t ) j + 5 k and - 13 π = R t 0 0 p 144 cos 2 t + 144 sin 2 t + 25 t dt = R t 0 0 13 dt = 13 t 0 t 0 = - π , and the point is P ( - π ) = (12 sin( - π ) , - 12 cos( - π ) , - 5 π ) = (0 , 12 , - 5 π ) 14. r (1 + 2 t ) i + (1 + 3 t ) j + (6 - 6 t ) k v = 2 i + 3 j - 6 k ⇒ | v | = p 2 2 + 3 3 + ( - 6) 2 = 7 s ( t ) = R 6 0 7 = 7 t Length = s (0) - s ( - 1) = 0 - ( - 7) = 7 14.1 Functions of Several Variables 8. (a) Domain: all ( x, y ) satisfying x 2 + y 2 9 (b) Range: 0 z 3 (c) level curves are circles centered at the origin with radii r 3 (d) boundary is the cirle x 2 + y 2 = 9 (e) closed (f) bounded 13. f 14. e 15. a 16. c 17. d 18. b
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20. -6 -4 -2 0 2 4 6 -5 0 5 -40 -30 -20 -10 0 10 x f(x,y) = 4-y 2 y Figure 1: (a) x y z = 4-y 2 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 Figure 2: (b) 42. f ( x, y, z ) = ln( x 2 + y + z 2 ) at ( - 1 , 2 , 1) w = ln( x 2 + y + z 2 ); at ( - 1 , 2 , 1) w = ln(1+2+1) = ln 4 ln 4 = ln( x 2 + y + z 2 ) 4 = x 2 + y + z 2 46. Minimum of f along the line is f (1 , 0 , 9) = (1)(0)
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Unformatted text preview: 4. lim ( x,y ) → (2 ,-3) ± 1 x + 1 y ² 2 = 1 36 12. lim ( x,y ) → ( π 2 , ) =-2 18. lim ( x,y ) → (2 , 2) x + y-4 √ x + y-2 = lim ( x,y ) → (2 , 2) ( √ x + y-2)( √ x + y +2) √ x + y-2 = lim ( x,y ) → (2 , 2) ( √ x + y + 2) = 4 24. lim P → (-1 4 , π 2 , 2 ) tan-1 ( xyz ) = tan-1 ³-π 4 ´ 30. (a) All ( x, y ) so that x 2-3 x + 2 6 = 0 ⇒ ( x-2)( x-1) 6 = 0 ⇒ x 6 = 2 and x 6 = 1 (b) All ( x, y ) so that y 6 = x 2 40. Along y = kx, k 6 = 0, lim ( x,y ) → (0 , 0) x + y x-y = lim x → x + kx x-kx = 1 + k 1-k ⇒ diferent limits For diferent values oF k , k 6 = 1 ⇒ Limit DNE...
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