Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-13 JWDD027-Salas-v1 November 30, 2006 13:43 SECTION 13.1 687 CHAPTER 13 SECTION 13.1 1. length AB :2 5 midpoint: (1 , 0 , 2) 2. length AB 10 midpoint: (0 , 1 , 3) 3. length AB :5 2 midpoint: ( 2 , 1 2 , 5 2 ) 4. length AB :9 midpoint: (1 , 3 2 , 3) 5. z = 2 6. y =1 7. y 8. z = 2 9. x =3 10. x 11. x 2 +( y 2) 2 z +1) 2 =9 12. ( x 1) 2 + y 2 z +2) 2 =16 13. ( x 2) 2 y 4) 2 z +4) 2 =36 14. x 2 + y 2 + z 2 15. ( x 3) 2 y 2) 2 z 2) 2 =13 16. ( x 2) 2 y 3) 2 z 2
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-13 JWDD027-Salas-v1 November 30, 2006 13:43 688 SECTION 13.1 17. x 2 + y 2 + z 2 +4 x 8 y 2 z +5=0 x 2 x +4+ y 2 8 y +16+ z 2 2 z +1= 5+4+16+1 ( x +2) 2 +( y 4) 2 z 1) 2 =16 center: ( 2 , 4 , 1) , radius: 4 18. Rewrite as x 2 4 x y 2 + z 2 2 z 1+4+1=4 = ( x 2) 2 + y 2 z 1) 2 = 4 center (2 , 0 , 1); radius 2 19. (2 , 3 , 5) 20. (2 , 3 , 5) 21. ( 2 , 3 , 5) 22. (2 , 3 , 5) 23. ( 2 , 3 , 5) 24. ( 2 , 3 , 5) 25. ( 2 , 3 , 5) 26. (0 , 3 , 5) 27. (2 , 5 , 5) 28. (2 , 3 , 3) 29. ( 2 , 1 , 3) 30. (6 , 3 , 3) 31. d ( PR )= 14 ,d ( QR 45 ( PQ 59; [ d ( )] 2 +[ d ( QR )] 2 =[ d ( )] 2 32. Let the vertices be ( x i ,y i ,z i ) ,i =1 , 2 , 3 . Then ± x 1 + x 2 2 , y 1 + y 2 2 , z 1 + z 2 2 ² =(5 , 1 , 3); ± x 2 + x 3 2 , y 2 + y 3 2 , z 2 + z 3 2 ² =(4 , 2 , 1); ± x 1 + x 3 2 , y 1 + y 3 2 , z 1 + z 3 2 ² =(2 , 1 , 0) Solving simultaneously gives vertices (3 , 2 , 2) , (7 , 0 , 4) , (1 , 4 , 2). 33. The sphere of radius 2 centered at the origin, together with its interior. 34. The exterior of the sphere of radius 3 centered at the origin. 35. A rectangular box in the Frst octant with sides on the coordinate planes and dimensions 1 × 2 × 3, together with its interior. 36. A cube of side length 4, together with its interior; the origin in at the center of the cube. 37. A circular cylinder with base the circle x 2 + y 2 = 4 and height 4, together with its interior. 38. x 2 + y 2 + z 2 = 4 and x 2 + y 2 + z 2 = 9 are concentric spheres; Ω is the region between the two spheres.
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-13 JWDD027-Salas-v1 November 30, 2006 13:43 SECTION 13.1 689 39. Let B =( x,y,z ). Then x +2 2 =1= x =0 , y +3 2 =2= y =1 , z +4 2 =3= z =2 . Therefore B =(0 , 1 , 2). 41. Let P 1 ) be the trisection point closest to A . Then −→ AP 1 = 1 3 AB = ( x a 1 ,y a 2 ,z a 3 )= 1 3 ( b 1 a 1 ,b 2 a 2 3 a 3 ) . Solving for gives ( ± 2 a 1 + b 1 3 , 2 a 2 + b 2 3 , 2 a 3 + b 3 3 ² . Similarly, if P 2 ) is the trisection point closest to B , then ( ± a 1 b 1 3 , a 2 b 2 3 , a 3 b 3 3 ² . 42. The points on the line segment AB are given by x =1+ t, y = 2+3 t, z = 2 2 t, 0 t 1. The line segment AP has length 3 if ³ t 2 +(3 t ) 2 +( 2 t ) 2 = 12 t 2 t 3=3= t = 1 2 3 . Thus, the point P on the line segment AB that is 3 units from A has coordinates: 1+ 1 2 3 , 2+ 3 2 3 , 2 2 2 3 . 43. Substituting the coordinates of the points into the equation Ax + By + Cz + D = 0, we get the equations Ax 0 + D ,By 0 + D ,Cz 0 + D = 0 which implies Ax 0 = 0 = 0 .
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