Calculus: Early Transcendentals, by Anton, 7th Edition,ch12

Calculus - Early Transcendentals

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492 CHAPTER 12 Three-Dimensional Space; Vectors EXERCISE SET 12.1 1. (a) (0 , 0 , 0) , (3 , 0 , 0) , (3 , 5 , 0) , (0 , 5 , 0) , (0 , 0 , 4) , (3 , 0 , 4) , (3 , 5 , 4) , (0 , 5 , 4) (b) (0 , 1 , 0) , (4 , 1 , 0) , (4 , 6 , 0) , (0 , 6 , 0) , (0 , 1 , 2) , (4 , 1 , 2) , (4 , 6 , 2) , (0 , 6 , 2) 2. corners: (2 , 2 , ± 2), (2 , 2 , ± 2), ( 2 , 2 , ± 2), ( 2 , 2 , ± 2) y x z (–2, –2, 2) (–2, 2, 2) (–2, 2, –2) (–2, –2, –2) (2, 2, –2) (2, –2, –2) (2, –2, 2) (2, 2, 2) 3. corners: (4 , 2 , 2), (4,2,1), (4,1,1), (4 , 1 , 2), ( 6 , 1 , 1), ( 6 , 2 , 1), ( 6 , 2 , 2), ( 6 , 1 , 2) (–6, 2, 1) (–6, 2, –2) y x z (–6, 1, –2) (4, 1, 1) (4, 1, –2) (4, 2, 1) 4. (a) ( x 2 ,y 1 ,z 1 ) , ( x 2 2 1 ) , ( x 1 2 1 )( x 1 1 2 ) , ( x 2 1 2 ) , ( x 1 2 2 ) (b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints of the edges. The midpoint of the edge ( x 1 1 1 ) and ( x 2 1 1 )i s µ 1 2 ( x 1 + x 2 ) 1 1 ; the midpoint of the edge ( x 2 1 1 ) and ( x 2 2 1 s µ x 2 , 1 2 ( y 1 + y 2 ) 1 ; the midpoint of the edge ( x 2 2 1 ) and ( x 2 2 2 )) is µ x 2 2 , 1 2 ( z 1 + z 2 ) . Thus the coordinates of the midpoint of the diagonal are 1 2 ( x 1 + x 2 ) , 1 2 ( y 1 + y 2 ) , 1 2 ( z 1 + z 2 ). 5. The diameter is d = p (1 3) 2 +( 2 4) 2 +(4+12) 2 = 296, so the radius is 296 / 2= 74. The midpoint (2 , 1 , 4) of the endpoints of the diameter is the center of the sphere. 6. Each side has length 14 so the triangle is equilateral. 7. (a) The sides have lengths 7, 14, and 7 5; it is a right triangle because the sides satisfy the Pythagorean theorem, (7 5) 2 =7 2 +14 2 . (b) (2,1,6) is the vertex of the 90 angle because it is opposite the longest side (the hypotenuse). (c) area = (1/2)(altitude)(base) = (1 / 2)(7)(14) = 49 8. (a) 3 (b) 2 (c) 5 (d) p (2) 2 3) 2 = 13 (e) p ( 5) 2 3) 2 = 34 (f) p ( 5) 2 + (2) 2 = 29 9. (a) ( x 1) 2 + y 2 z +1) 2 =16 (b) r = p ( 1 0) 2 +(3 0) 2 +(2 0) 2 = 14, ( x 2 y 3) 2 z 2) 2 =14
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Exercise Set 12.1 493 (c) r = 1 2 p ( 1 0) 2 +(2 2) 2 +(1 3) 2 = 1 2 5, center ( 1 / 2 , 2 , 2), ( x +1 / 2) 2 +( y 2) 2 z 2) 2 =5 / 4 10. r = | [distance between (0,0,0) and (3 , 2 , 4)] ± 1 | = 29 ± 1 , x 2 + y 2 + z 2 = r 2 = ( 29 ± 1 ) 2 =30 ± 2 29 11. ( x 2) 2 y +1) 2 z +3) 2 = r 2 , (a) r 2 =3 2 =9 (b) r 2 =1 2 (c) r 2 =2 2 =4 12. (a) The sides have length 1, so the radius is 1 2 ; hence ( x +2) 2 y 1) 2 z 3) 2 = 1 4 (b) The diagonal has length 1+1+1= 3 and is a diameter, so ( x +2) 2 +( y 1) 2 +( z 3) 2 = 3 4 . 13. ( x +5) 2 y 2 z 2 = 49; sphere, C ( 5 , 2 , 1), r =7 14. x 2 y 1 / 2) 2 + z 2 / 4; sphere, C (0 , 1 / 2 , 0), r / 2 15. ( x 1 / 2) 2 y 3 / 4) 2 z +5 / 4) 2 =54 / 16; sphere, C (1 / 2 , 3 / 4 , 5 / 4), r 6 / 4 16. ( x 2 y 1) 2 z 2 = 0; the point ( 1 , 1 , 1) 17. ( x 3 / 2) 2 y 2 z 4) 2 = 11 / 4; no graph 18. ( x 1) 2 y 3) 2 z 4) 2 = 25; sphere, C (1 , 3 , 4), r 19. (a) y x z (b) y x z (c) y x z 20. (a) y x z x = 1 (b) y x z y (c) y x z z 21. (a) 5 y x z (b) 5 y x z (c) 5 y x z
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494 Chapter 12 22. (a) y x z (b) y x z (c) y x z 23. (a) 2 y + z =0 (b) 2 x + z (c) ( x 1) 2 +( y 1) 2 =1 (d) ( x 1) 2 z 1) 2 24. (a) ( x a ) 2 z a ) 2 = a 2 (b) ( x a ) 2 y a ) 2 = a 2 (c) ( y a ) 2 z a ) 2 = a 2 25. y x z 26. y x z 27. 1 1 x y z 28. y x z 29. 3 3 2 y x z
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Exercise Set 12.1 495 30.
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Calculus: Early Transcendentals, by Anton, 7th Edition,ch12...

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