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Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Word Count: 4780

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Calculus Stewart ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes 1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each scalar multiples of the third direction vector. Then the first two direction vectors are also scalar multiples of each other, so these vectors, and hence the two...

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Calculus Stewart ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes 1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each scalar multiples of the third direction vector. Then the first two direction vectors are also scalar multiples of each other, so these vectors, and hence the two lines, are parallel. (b) False; for example, the x and y axes are both perpendicular to the z axis, yet the x and y axes are not parallel. (c) True; each of the first two planes has a normal vector parallel to the normal vector of the third plane, so these two normal vectors are parallel to each other and the planes are parallel. (d) False; for example, the xy and yz planes are not parallel, yet they are both perpendicular to the xz plane. (e) False; the x and y axes are not parallel, yet they are both parallel to the plane z=1 . (f) True; if each line is perpendicular to a plane, then the lines' direction vectors are both parallel to a normal vector for the plane. Thus, the direction vectors are parallel to each other and the lines are parallel. (g) False; the planes y=1 and z=1 are not parallel, yet they are both parallel to the x axis. (h) True; if each plane is perpendicular to a line, then any normal vector for each plane is parallel to a direction vector for the line. Thus, the normal vectors are parallel to each other and the planes are parallel. (i) True; see Figure 9 and the accompanying discussion. (j) False; they can be skew, as in Example 3. (k) True. Consider any normal vector for the plane and any direction vector for the line. If the normal vector is perpendicular to the direction vector, the line and plane are parallel. Otherwise, the vectors meet at an angle , 0 <90 , and the line will intersect the plane at an angle 90 . 2. For this line, we have r =i 3 k and v=2 i 4 j+5 k , so a vector equation is 0 r=r +t v=(i 3 k)+t(2 i 4 j+5 k)=(1+2t) i 4t j+( 3+5t) k and parametric equations are x=1+2t , y= 4t , 0 z= 3+5t . 3. For this line, we have r = 2 i+4 j+10 k and v=3 i+ j 8 k , so a vector equation is 0 r=r +t v= ( 2 i+4 j+10 k ) +t(3 i+ j 8 k)=( 2+3t) i+(4+t) j+(10 8t) k and parametric equations are 0 x= 2+3t , y=4+t , z=10 8t . 4. This line has the same direction as the given line, v=2 i j+3 k . Here r =0 i+0 j+0 k , so a vector 0 equation is r=(0 i+0 j+0 k)+t(2 i j+3 k)=2t i t j+3t k and parametric equations are x=2t , y= t , z=3t . 5. A line perpendicular to the given plane has the same direction as a normal vector to the plane, such as n= 1,3,1 . So r =i+6 k , and we can take v=i+3 j+k . Then a vector equation is 0 r=(i+6 k)+t(i+3 j+k)=(1+t) i+3t j+(6+t) k , and parametric equations are x=1+t , y=3t , z=6+t . 6. The vector v= 1 0,2 0,3 0 = 1,2,3 is parallel to the line. Letting P = ( 0,0,0 ) , parametric 0 1 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes equations are x=0+1 t=t , y=0+2 t=2t , z=0+3 t=3t , while symmetric equations are x= y z = . 2 3 7. The vector v= 4 1,3 3,0 2 = 5,0, 2 is parallel to the line. Letting P = ( 1,3,2 ) , parametric 0 equations are x=1 5t , y=3+0t=3 , z=2 2t , while symmetric equations are here that the direction number b=0 , so rather than writing write the equation y=3 separately. x 1 z 2 = ,y=3 . Notice 5 2 y 3 in the symmetric equation we must 0 8. v= 2 6,4 1,5 ( 3) = 4,3,8 , and letting P = ( 6,1, 3) , parametric equations are x=6 4t , y=1+3t 0 , z= 3+8t , while symmetric equations are 1 , 3 1 2 1 , 4 2 x 6 y 1 z+3 = = . 4 3 8 9. v= y=1+ 2 0,1 = 2, , and letting P = ( 2,1, 3) , parametric equations are x=2+2t , 0 1 x 2 y 1 z+3 x 2 z+3 t , z= 3 4t , while symmetric equations are = = or =2y 2= . 2 2 1/2 4 2 4 10. v=(i+ j) ( j+k)= 0 i j k 1 1 0 =i j+k is the direction of the line perpendicular to both i+ j 0 1 1 and j+k . With P = ( 2,1,0 ) , parametric equations are x=2+t , y=1 t , z=t and symmetric equations are y 1 x 2= =z or x 2=1 y=z . 1 11. The line has direction v= 1,2,1 . Letting P = ( 1, 1,1 ) , parametric equations are x=1+t , y= 1+2t 0 , z=1+t and symmetric equations are x 1= y+1 =z 1 . 2 12. Setting x=0 , we see that ( 0,1,0 ) satisfies the equations of both planes, so they do in fact have a line of intersection. v=n n = 1,1,1 1,0,1 = 1,0, 1 is the direction of this line. Taking the point 1 2 ( 0,1,0 ) as P0 , parametric equations are x=t , y=1 , z= t , and symmetric equations are x= z , y=1 . 13. Direction vectors of the lines are v = 2 ( 4),0 ( 6), 3 1 = 2,6, 4 and 1 v = 5 10,3 18,14 4 = 5, 15,10 , and since v = 2 2 5 v , the direction vectors and thus the lines are 2 1 2 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes parallel. 14. Direction vectors of the lines are v = 2,4,4 and v = 8, 1,4 . Since v v = 16 4+16 0 , the 1 2 1 2 vectors and thus the lines are not perpendicular. 15. (a) A direction vector of the line with parametric equations x=1+2t , y=3t , z=5 7t is v= 2,3, 7 and the desired parallel line must also have v as a direction vector. Here P = ( 0,2, 1 ) , so symmetric 0 equations for the line are x y 2 z+1 = = . 2 3 7 x y 2 1 2 11 = = or x= , y= . Thus 2 3 7 7 7 . Similarly for the yz plane, we need (b) The line intersects the xy plane when z=0 , so we need the point of intersection with the xy plane is x=0 0= y 2 z+1 = 3 7 2 11 , ,0 7 7 y=2 , z= 1 . Thus the line intersects the yz plane at ( 0,2, 1 ) . For the xz x 2 z+1 = = 2 3 7 x= 4 11 , z= . So the line intersects the xz plane at 3 3 plane, we need y=0 4 11 ,0, 3 3 . 16. (a) A vector normal to the plane 2x y+z=1 is n= 2, 1,1 , and since the line is to be perpendicular to the plane, n is also a direction vector for the line. Thus parametric equations of the line are x=5+2t , y=1 t , z=t . (b) On the xy plane, z=0 . So z=t=0 in the parametric equations of the line, and therefore x=5 and 5 7 y=1 , giving the point of intersection ( 5,1,0 ) . For the yz plane, x=0 which implies t= , so y= 2 2 5 7 5 and z= and the point is 0, , . For the xz plane, y=0 which implies t=1 , so x=7 and z=1 2 2 2 and the point of intersection is ( 7,0,1 ) . 17. From Equation 4, the line segment from r =2 i j+4 k to r =4 i+6 j+k is 0 1 r(t)=(1 t) r +t r =(1 t)(2 i j+4 k)+t(4 i+6 j+k)=(2 i j+4 k)+t(2 i+7 j 3 k) , 0 t 0 1 1. 18. From Equation 4, the line segment from r =10 i+3 j+k to r =5 i+6 j 3 k is 0 1 r(t) =(1 t) r0+t r1=(1 t)(10 i+3 j+k)+t(5 i+6 j 3 k) =(10 i+3 j+k)+t( 5 i+3 j 4 k) , 0 t 1 3 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes The corresponding parametric equations are x=10 5t , y=3+3t , z=1 4t , 0 t 1 2 1 1. 2 19. Since the direction vectors are v = 6,9, 3 and v = 2, 3,1 , we have v = 3v so the lines are parallel. 20. The lines aren't parallel since the direction vectors 2,3, 1 and 1,1,3 aren't parallel. For the lines to intersect we must be able to find one value of t and one value of s that produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations: 1+2t= 1+s , 3t=4+s , 2 t=1+3s . Solving the first two equations we get t=6 , s=14 and checking, we see that these values don't satisfy the third equation. Thus L and L aren't parallel and don't 1 2 intersect, so they must be skew lines. 21. Since the direction vectors 1,2,3 and 4, 3,2 are not scalar multiples of each other, the lines are not parallel, so we check to see if the lines intersect. The parametric equations of the lines are L : 1 x=t , y=1+2t , z=2+3t and L : x=3 4s , y=2 3s , z=1+2s . For the lines to intersect, we must be able 2 to find one value of t and one value of s that produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations: t=3 4s , 1+2t=2 3s , 2+3t=1+2s . Solving the first two equations we get t= 1 , s=1 and checking, we see that these values don't satisfy the third equation. Thus the lines aren't parallel and don't intersect, so they must be skew lines. 22. Since the direction vectors 2,2, 1 and 1, 1,3 aren't parallel, the lines aren't parallel. Here the parametric equations are L : x=1+2t , y=3+2t , z=2 t and L : x=2+s , y=6 s , z= 2+3s . Thus, for the 1 2 lines to intersect, the three equations 1+2t=2+s , 3+2t=6 s , and 2 t= 2+3s must be satisfied simultaneously. Solving the first two equations gives t=1 , s=1 and, checking, we see that these values do satisfy the third equation, so the lines intersect when t=1 and s=1 , that is, at the point ( 3,5,1 ) . 23. Since the plane is perpendicular to the vector 2,1,5 , we can take 2,1,5 as a normal vector to the plane. ( 6,3,2 ) is a point on the plane, so setting a= 2 , b=1 , c=5 and x =6 , y =3 , z =2 in 0 0 0 Equation 7 gives 2(x 6)+1(y 3)+5(z 2)=0 or 2x+y+5z=1 to be an equation of the plane. 24. j+2 k= 0,1,2 is a normal vector to the plane and ( 4,0, 3) is a point on the plane, so setting a=0 , b=1 , c=2,x =4 , y =0 , z = 3 in Equation 7 gives 0(x 4)+1(y 0)+2[z ( 3)]=0 or y+2z= 6 to be an 0 0 0 equation of the plane. 25. i+ j k= 1,1, 1 is a normal vector to the plane and ( 1, 1,1 ) is a point on the plane, so setting a=1 , b=1 , c= 1 , x =1 , y = 1 , z =1 in Equation 7 gives 1( x 1 ) +1[y ( 1)] 1(z 1)=0 or x+y z= 1 to be 0 0 0 an equation of the plane. 4 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes 26. Since the line is perpendicular to the plane, its direction vector 1,2, 3 is a normal vector to the plane. An equation of the plane, then, is 1[x ( 2)]+2(y 8) 3(z 10)=0 or x+2y 3z= 16 . 27. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 2, 1,3 , and an equation of the plane is 2(x 0) 1(y 0)+3(z 0)=0 or 2x y+3z=0 . 28. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 1,1,1 , and an equation of the plane is 1[x ( 1)]+1(y 6)+1[z ( 5)]=0 or x+y+z=0 . 29. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 3,0, 7 , and an equation of the plane is 3(x 4)+0[y ( 2)] 7(z 3)=0 or 3x 7z= 9 . 30. First, a normal vector for the plane 2x+4y+8z=17 is n= 2,4,8 . A direction vector for the line is v= 2,1, 1 , and since n v=0 we know the line is perpendicular to n and hence parallel to the plane. Thus, there is a parallel plane which contains the line. By putting t=0 , we know the point ( 3,0,8 ) is on the line and hence the new plane. We can use the same normal vector n= 2,4,8 , so an equation of the plane is 2(x 3)+4(y 0)+8(z 8)=0 or x+2y+4z=35 . 31. Here the vectors a= 1 0,0 1,1 1 = 1, 1,0 and b= 1 0,1 1,0 1 = 1,0, 1 lie in the plane, so a b is a normal vector to the plane. Thus, we can take n=a b= 1 0,0+1,0+1 = 1,1,1 . If P is the 0 point ( 0,1,1 ) , an equation of the plane is 1(x 0)+1(y 1)+1(z 1)=0 or x+y+z=2 . 32. Here the vectors a= 2, 4,6 and b= 5,1,3 lie in the plane, so n=a b= 12 6,30 6,2+20 = 18,24,22 is a normal vector to the plane and an equation of the plane is 18(x 0)+24(y 0)+22(z 0)=0 or 18x+24y+22z=0 . 33. Here the vectors a= 8 3,2 ( 1),4 2 = 5,3,2 and b= 1 3, 2 ( 1), 3 2 = 4, 1, 5 lie in the plane, so a normal vector to the plane is n=a b= 15+2, 8+25, 5+12 = 13,17,7 and an equation of the plane is 13(x 3)+17[y ( 1)]+7(z 2)=0 or 13x+17y+7z= 42 . 34. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane. Since the given line lies in the plane, its direction vector a= 3,1, 1 is one vector in the plane. We can verify that the given point ( 1,2,3) does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and find a vector connecting the points. If we put t=0 , we see that ( 0,1,2 ) is on the line, so b= 1 0,2 1,3 2 = 1,1,1 and n=a b= 1+1, 1 3,3 1 = 2, 4,2 . Thus, an equation of the plane is 2(x 1) 4(y 2)+2(z 3)=0 or 2x 4y+2z=0 . (Equivalently, we can write x 2y+z=0 .) 35. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to 5 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes the plane. Since the given line lies in the plane, its direction vector a= 2,5,4 is one vector in the plane. We can verify that the given point ( 6,0, 2 ) does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and find a vector connecting the points. If we put t=0 , we see that ( 4,3,7) is on the line, so b= 6 4,0 3, 2 7 = 2, 3, 9 and n=a b= 45+12,8 18,6 10 = 33, 10, 4 . Thus, an equation of the plane is 33(x 6) 10(y 0) 4[z ( 2)]=0 or 33x+10y+4z=190 . y z 1 1 = , lies in the plane, its direction vector a= 1, , is 1/2 1/3 2 3 parallel to the plane. The point ( 0,0,0 ) is on the line (put t=0 ), and we can verify that the given point ( 1, 1,1 ) in the plane is not on the line. The vector connecting these two points, b= 1, 1,1 , is therefore parallel to the plane, but not parallel to 1,2,3 . Then 36. Since the line x=2y=3z , or x= 1 1 1 1 5 2 3 + , 1, 1 = , , is a normal vector to the plane, and an equation of 2 3 3 2 6 3 2 5 2 3 the plane is (x 0) (y 0) (z 0)=0 or 5x 4y 9z=0 . 6 3 2 a b= 37. A direction vector for the line of intersection is a=n n = 1,1, 1 1 2 2, 1,3 = 2, 5, 3 , and a is parallel to the desired plane. Another vector parallel to the plane is the vector connecting any point on the line of intersection to the given point ( 1,2,1 ) in the plane. Setting x=0 , the equations of the 7 3 planes reduce to y z=2 and y+3z=1 with simultaneous solution y= and z= . So a point on the 2 2 3 1 7 3 line is 0, , 1, , and another vector parallel to the plane is . Then a normal vector 2 2 2 2 to the plane is n= 2, 5, 3 3 1 , = 2,4, 8 and an equation of the plane is 2 2 2(x+1)+4(y 2) 8(z 1)=0 or x 2y+4z= 1 . 1, 38. n = 1,0, 1 and n = 0,1,2 . Setting z=0 , it is easy to see that ( 1,3,0 ) is a point on the line of 1 2 intersection of x z=1 and y+2z=3 . The direction of this line is v =n n = 1, 2,1 . A second vector 1 1 2 parallel to the desired plane is v = 1,1, 2 , since it is perpendicular to x+y 2z=1 . Therefore, a 2 normal of the plane in question is n=v v = 4 1,1+2,1+2 = 3,3,3 , or we can use 1,1,1 . Taking 1 2 ( x ,y ,z ) = ( 1,3,0) , the equation we are looking for is ( x 1 ) + ( y 3) +z=0 0 0 0 x+y+z=4 . 39. Substitute the parametric equations of the line into the equation of the plane: (3 t) (2+t)+2(5t)=9 6 Stewart Calculus ET 5e 0534393217;12. Vectors and the of Geometry Space; 12.5 Equations of Lines and Planes 8t=8 t=1 . Therefore, the point of intersection of the line and the plane is given by x=3 1=2 , y=2+1=3 , and z=5(1)=5, that is, the point ( 2,3,5) . 40. Substitute the parametric equations of the line into the equation of the plane: (1+2t)+2(4t) (2 3t)+1=0 13t=0 t=0 . Therefore, the point of intersection of the line and the plane is given by x=1+2(0)=1 , y=4(0)=0 , and z=2 3(0)=2, that is, the point ( 1,0,2 ) . 41. Parametric equations for the line are x=t , y=1+t , z= plane gives 4(t) (1+t)+3 intersection is ( 2,3,1 ) . 42. A direction vector for the line through ( 1,0,1 ) and ( 4, 2,2 ) is v= 3, 2,1 and, taking P = ( 1,0,1 ) , 0 parametric equations for the line are x=1+3t , y= 2t , z=1+t . Substitution of the parametric equations into the equation of the plane gives 1+3t 2t+1+t=6 t=2 . Then x=1+3(2)=7 , y= 2(2)= 4 , and z=1+2=3 so the point of intersection is ( 7, 4,3) . 43. Setting x=0 , we see that ( 0,1,0 ) satisfies the equations of both planes, so that they do in fact have a line of intersection. v=n n = 1,1,1 1,0,1 = 1,0, 1 is the direction of this line. Therefore, 1 2 1 t =8 2 9 t=9 2 1 t and substituting into the equation of the 2 1 t=2 . Thus x=2 , y=1+2=3 , z= (2)=1 and the point of 2 direction numbers of the intersecting line are 1 , 0 , 1 . 44. The angle between the two planes is the same as the angle between their normal vectors. The normal vectors of the two planes are 1,1,1 and 1,2,3 . The cosine of the angle between these 1,1,1 1,2,3 1+2+3 6 6 = = = two planes is cos = . 7 1,1,1 1,2,3 1+1+1 1+4+9 42 45. Normal vectors for the planes are n = 1,4, 3 and n = 3,6,7 , so the normals (and thus the 1 2 planes) aren't parallel. But n n = 3+24 21=0 , so the normals (and thus the planes) are 1 2 perpendicular. 46. Normal vectors for the planes are n = 1,4, 2 and n = 3, 12,6 . Since n = 3n , the normals 1 2 2 1 (and thus the planes) are parallel. 47. Normal vectors for the planes are n = 1,1,1 and n = 1, 1,1 . The normals are not parallel, so 1 2 neither are the planes. Furthermore, n n =1 1+1=1 0 , so the planes aren't perpendicular. The 1 2 angle between them is given by 7 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes n n cos = 1 2 2 n 1 n = 1 1 = 3 3 3 =cos 1 1 3 70.5 . 48. The normals are n = 2, 3,4 and n = 1,6,4 so the planes aren't parallel. Since 1 2 n n =2 18+16=0 , the normals (and thus the planes) are perpendicular. 1 2 49. The normals are n = 1, 4,2 and n = 2, 8,4 . Since n =2n , the normals (and thus the planes) 1 2 2 1 are parallel. 50. The normal vectors are n = 1,2,2 and n = 2, 1,2 . The normals are not parallel, so neither are 1 2 the planes. Furthermore, n n =2 2+4=4 0 , so the planes aren't perpendicular. The angle between 1 2 n n them is given by cos = 1 2 2 n 1 n = 4 4 = 9 9 9 =cos 1 4 9 63.6 . 51. (a) To find a point on the line of intersection, set one of the variables equal to a constant, say z=0 . (This will only work if the line of intersection crosses the xy plane; otherwise, try setting x or y equal to 0 .) Then the equations of the planes reduce to x+y=2 and 3x 4y=6 . Solving these two equations gives x=2 , y=0 . So a point on the line of intersection is ( 2,0,0 ) . The direction of the line y z is v=n n = 5 4, 3 5, 4 3 = 1, 8, 7 , and symmetric equations for the line are x 2= = . 1 2 8 7 n n 6 3 4 5 1 2 . Therefore (b) The angle between the planes satisfies cos = = = 5 3 50 n n 1 2 =cos 1 6 5 119 (or 61 ). n = 1, 2,1 and 2x+y+z=1 1 1 52. (a) x 2y+z=1 n = 2,1,1 . The vector that gives the direction of 2 2 the line of intersection of these two planes is v=n n = 2 1,2 1,1+4 = 3,1,5 . Setting x=y=0 , we see that both planes contain ( 0,0,1 ) so that this point must lie on their line of intersection. Then x z 1 =y= . symmetric equations for this line are 3 5 n n 2 2+1 1 1 1 1 2 (b) cos = = = =cos 80 . 6 1+4+1 4+1+1 6 n n 1 2 8 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes 53. Setting x=0 , the equations of the two planes become z=y and 5y+z= 1 , which intersect at y= and z= 1 6 1 1 1 . Thus we can choose x ,y ,z = 0, , . The vector giving the direction of this 0 0 0 6 6 6 intersecting line, v , is perpendicular to the normal vectors of both planes. So v=n n = 2, 5, 1 1,1, 1 = 6,1,7 . Therefore, by Equations 2, parametric equations for this line ( ) 1 2 are x=6t , y= 1 1 +t , z= +7t . 6 6 54. Setting y=0 , the equations of the two planes become 2x+5z= 3 and x+z= 2 , which intersect 7 1 7 1 and z= . Thus we can choose x ,y ,z = ,0, . The vector giving the direction of at x= 0 0 0 3 3 3 3 this intersecting line, v , is perpendicular to the normal vectors of both planes. So v=n n = 2,0,5 1, 3,1 = 15,5 2, 6 =3 5,1, 2 . Therefore, by Equations 2, parametric equations ( ) 1 2 of the line of intersection of the two planes are x= 7 1 +5t , y=t , z= 2t . 3 3 55. The plane contains all perpendicular bisectors of the line segment joining ( 1,1,0 ) and ( 0,1,1 ) . All 1 1+1 1 1 1 of these bisectors pass through the midpoint of this segment , , = ,1, . The 2 2 2 2 2 direction of this line segment 1 0,1 1,0 1 = 1,0, 1 is perpendicular to the plane so that we can 1 1 choose this to be n . Therefore the equation of the plane is 1 x +0(y 1) 1 z =0 x=z . 2 2 56. The plane will contain all perpendicular bisectors of the line segment joining the two points. Thus, a point in the plane is P = ( 1, 1,2 ) , the midpoint of the line segment joining the two given points, 0 and a normal to the plane is n= 6, 6,2 , the vector connecting the two points. So an equation of the plane is 6(x+1) 6(y+1)+2(z 2)=0 or 3x 3y+z=2 . 57. The plane contains the points ( a,0,0 ) , ( 0,b,0 ) and ( 0,0,c ) . Thus the vectors a= a,b,0 and b= a,0,c lie in the plane, and n=a b= bc 0,0+ac,0+ab = bc,ac,ab is a normal vector to the plane. The equation of the plane is therefore bcx+acy+abz=abc+0+0 or bcx+acy+abz=abc . Notice x y z that if a 0 , b 0 and c 0 then we can rewrite the equation as + + =1 . This is a good a b c equation to remember! 58. (a) For the lines to intersect, we must be able to find one value of t and one value of s satisfying the three equations 1+t=2 s , 1 t=s and 2t=2 . From the third we get t=1 , and putting this in the second gives s=0 . These values of s and t do satisfy the first equation, so the lines intersect at the 9 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes point P = ( 1+1,1 1,2(1) ) = ( 2,0,2 ) . 0 (b) The direction vectors of the lines are 1, 1,2 and 1,1,0 , so a normal vector for the plane is 1,1,0 1, 1,2 = 2,2,0 and it contains the point ( 2,0,2 ) . Then the equation of the plane is 2(x 2)+2(y 0)+0(z 2)=0 x+y=2 . 59. Two vectors which are perpendicular to the required line are the normal of the given plane, 1,1,1 , and a direction vector for the given line, 1, 1,2 . So a direction vector for the required line is 1,1,1 1, 1,2 = 3, 1, 2 . Thus L is given by x,y,z = 0,1,2 +t 3, 1, 2 , or in parametric form, x=3t , y=1 t , z=2 2t . 60. Let L be the given line. Then ( 1,1,0 ) is the point on L corresponding to t=0 . L is in the direction of a= 1, 1,2 and b= 1,0,2 is the vector joining ( 1,1,0 ) and ( 0,1,2 ) . Then 1, 1,2 1,0,2 1 3 1 b proj b= 1,0,2 , ,1 is a direction 1, 1,2 = 1,0,2 1, 1,2 = a 2 2 2 2 2 2 1 +( 1) +2 3 1 , ,1 = 3,1,2 is also a direction vector, and the line has vector for the required line. Thus 2 2 2 parametric equations x= 3t , y=1+t , z=2+2t . (Notice that this is the same line as in Exercise 59.) 61. Let P have normal vector n . Then n = 4, 2,6 , n = 4, 2, 2 , n = 6,3, 9 , n = 2, 1, 1 . i i 1 2 3 4 2 n , so n and n are parallel, and hence P and P are parallel; similarly P and P are 1 1 3 1 3 2 4 3 3 1 parallel because n =2n . However, n and n are not parallel. 0,0, lies on P , but not on P , 2 4 1 2 1 3 2 so they are not the same plane, but both P and P contain the point ( 0,0, 3) , so these two planes are Now n = 2 4 identical. 62. Let L have direction vector v . Then v = 1,1, 5 , v = 1,1, 1 , v = 1,1, 1 , v = 2,2, 10 . v i i 1 2 3 4 2 and v are equal so they're parallel. v =2v , so L and L are parallel. L contains the point ( 1,4,1 ) , 3 4 1 4 1 3 but this point does not lie on L , so they're not equal. ( 2,1, 3) lies on L , and on L , with t=1 . So 2 4 1 L and L are identical. 1 4 63. Let Q= ( 2,2,0 ) and R= ( 3, 1,5) , points on the line corresponding to t=0 and t=1 . Let P= ( 1,2,3) . Then a=QR= 1, 3,5 , b=QP= 1,0,3 . The distance is 10 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes a b d= = a 1, 3,5 1,0,3 1, 3,5 = 9, 8, 3 1, 3,5 = 9 +8 +3 1 +3 +5 2 2 2 2 2 = 154 35 = 2 22 . 5 64. Let Q= ( 5,0,1 ) and R= ( 4,3,3) , points on the line corresponding to t=0 and t=1 . Let P= ( 1,0, 1 ) . Then a=QR= 1,3,2 and b=QP= 4,0, 2 . The distance is a b d= = a 1,3,2 4,0, 2 1,3,2 = 6, 10,12 1,3,2 = 2 3 +5 +6 1 +3 +2 2 2 2 2 2 = 2 70 14 =2 5 . 2 65. By Equation 9, the distance is D= 1 25 (1)(2)+( 2)(8)+( 2)(5) 1 = . 3 1+4+4 1 26 4(3)+( 6)( 2)+1(7) 5 = . 16+36+1 53 66. By Equation 9, the distance is D= 67. Put y=z=0 in the equation of the first plane to get the point ( 1,0,0 ) on the plane. Because the planes are parallel, the distance D between them is the distance from ( 1,0,0 ) to the second plane. By 7 6 7 3( 1)+6(0) 3(0) 4 Equation 9, D= = or . 18 3 6 2 2 2 3 +6 +( 3) 68. Put y=z=0 in the equation of the first plane to get the point 4 ,0,0 3 on the plane. Because the 4 ,0,0 3 to the second plane. planes are parallel the distance D between them is the distance from 1 By Equation 9, D= 4 3 2 +2(0) 3(0) 1 = 1 +2 +( 3) 2 2 1 . 3 14 69. The distance between two parallel planes is the same as the distance between a point on one of the planes and the other plane. Let P = x ,y ,z be a point on the plane given by ax+by+cz+d =0 . Then 0 ( 0 0 0 ) 1 ax +by +cz +d =0 and the distance between P and the plane given by ax+by+cz+d =0 is, from 0 0 0 1 0 2 ax +by +cz +d Equation 9, D= 0 0 0 2 d +d = 2 1 2 2 2 d d = 1 2 2 2 . a +b +c 2 2 2 a +b +c a +b +c 2 70. The planes must have parallel normal vectors, so if ax+by+cz+d=0 is such a plane, then for some 11 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes t 0 , a,b,c =t 1,2, 2 = t,2t, 2t . So this plane is given by the equation x+2y 2z+e=0 , where 1 e e=d/t . By Exercise 69, the distance between the planes is 2= 6= 1 e e=7 or 5 . 2 2 2 1 +2 +( 2) So the desired planes have equations x+2y 2z=7 and x+2y 2z= 5 . 71. L : x=y=z 1 x=y (1). L : x+1=y/2=z/3 2 x+1=y/2 (2). The solution of (1) and (2) is x=y= 2 . However, when x= 2 , x=z z= 2 , but x+1=z/3 z= 3 , a contradiction. Hence the lines do not intersect. For L , v = 1,1,1 , and for L , v = 1,2,3 , so the lines are not parallel. Thus the lines are 1 1 2 2 skew lines. If two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be perpendicular to both 1,1,1 and 1,2,3 , the direction vectors of the two lines. So set n= 1,1,1 1,2,3 = 3 2, 3+1,2 1 = 1, 2,1 . From above, we know that ( 2, 2, 2 ) and ( 2, 2, 3) are points of L and L respectively. So in the notation of Equation 8, 1 2 1( 2) 2( 2)+1( 2)+d =0 1 d =0 and 1( 2) 2( 2)+1( 3)+d =0 1 2 d =1 . 2 0 1 1 = . 1+4+1 6 Alternate solution (without reference to planes): A vector which is perpendicular to both of the lines is n= 1,1,1 1,2,3 = 1, 2,1 . Pick any point on each of the lines, say ( 2, 2, 2 ) and ( 2, 2, 3) , and form the vector b= 0,0,1 connecting the two points. The distance between the two skew lines is n b 1 0 2 0+1 1 1 the absolute value of the scalar projection of b along n , that is, D= = = . n 1+4+1 6 By Exercise 69, the distance between these two skew lines is D= 72. First notice that if two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be perpendicular to both v = 1,6,2 and v = 2,15,6 , 1 2 the direction vectors of the two lines respectively. Thus set n=v v = 36 30,4 6,15 12 = 6, 2,3 . 1 2 Setting t=0 and s=0 gives the points ( 1,1,0 ) and ( 1,5, 2 ) . So in the notation of Equation 8, 6 2+0+d =0 d = 4 and 6 10 6+d =0 d =10 . Then by Exercise 69 , the distance between the 1 1 2 2 4 10 14 = =2 . 7 36+4+9 Alternate solution (without reference to planes): We already know that the direction vectors of the two lines are v = 1,6,2 and v = 2,15,6 . Then n=v v = 6, 2,3 is perpendicular to both lines. two skew lines is given by D= 1 2 1 2 Pick any point on each of the lines, say ( 1,1,0 ) and ( 1,5, 2 ) , and form the vector b= 0,4, 2 connecting the two points. Then the distance between the two skew lines is the absolute value of the scalar projection of b along n , that is, 12 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes D= n b 1 14 = 0 8 6 = =2 . n 7 36+4+9 73. If a 0 , then ax+by+cz+d=0 a ( x+d/a ) +b(y 0)+c(z 0)=0 which by (7) is the scalar equation of the plane through the point ( d/a,0,0 ) with normal vector a,b,c . Similarly, if b 0 (or if c 0 ) the equation of the plane can be rewritten as a(x 0)+b ( y+d/b ) +c(z 0)=0 which by (7) is the scalar equation of a plane through the point ( 0, d/b,0 ) with normal vector a,b,c . 74. (a) The planes x+y+z=c have normal vector 1,1,1 , so they are all parallel. Their x , y , and z intercepts are all c . When c>0 their intersection with the first octant is an equilateral triangle and when c<0 their intersection with the octant diagonally opposite the first is an equilateral triangle. (b) The planes x+y+cz=1 have x intercept 1 , y intercept 1 , and z intercept 1/c . The plane with c=0 is parallel to the z axis. As c gets larger, the planes get closer to the xy plane. , which are perpendicular to the (c) The planes ycos +zcos =1 have normal vectors 0,cos ,sin x axis, and so the planes are parallel to the x axis. We look at their intersection with the yz plane. These are lines that are perpendicular to cos ,sin and pass through ( cos ,sin ) , since cos +sin =1 . So these are the tangent lines to the unit circle. Thus the family consists of all planes tangent to the circular cylinder with radius 1 and axis the x axis. 2 2 13
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TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.6 Cylinders and Quadric Surfaces1. (a) In R , the equation y=x represents a parabola.22(b)In R , the equation y=x doesn't involve z , so any horizontal plane with e
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.7 Cylindrical and Spherical Coordinates1. See Figure 1 and the accompanying discussion; see the paragraph accompanying Figure 3. 2. See Figure 5 and the accompanying discus
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves1. The component functions t , t 1 , and 5 t are all defined when t 1 0 t 5 , so the domain of r(t) is 1,5 . 2. The component functions2t1 and 5 t
FSU - HIST - 08758
Prcis 1-28-08The Chinese Exclusion Act of 1882 denied Chinese immigrants the right to obtain citizenship in America. It states that any captain of a ship facilitating Chinese immigrants into the country would be subject to criminal charges. Another
FSU - HIST - 08758
Prcis 2-11-08This passage of Harry Emerson Fosdick's &quot;Shall the Fundamentalists Win?&quot;(1922) shows Fosdick's distaste in the people who call themselves fundamentalists who feel there is no place for liberal Christians in the church. Evidence of this
FSU - HIST - 08758
Prcis 2-20-08The passage from Meridel Le Sueur's &quot;Women on the Breadlines&quot; (1932) displays how The Great Depression affected an endless number of people of any race, gender or religion. In this particular passage, Sueur describes a typical day of a
FSU - HIST - 08758
Prcis 2-22-08Franklin D Roosevelt's &quot;First Inaugural Address (1933)&quot;, gives a glimpse on how he planned to revive the nation through his presidency by using any tactics he felt that was necessary to restore American prosperity. He shows the desire
Kansas State - PHYS - 213
1. With speed v = 11200 m/s, we findK= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 22. (a) The change in kinetic energy for the meteorite would be1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2()()2= -5 1014 J
FSU - HIST - 08758
Prcis 3-26-08In George F. Kennan's &quot;The Sources of Soviet Conduct&quot; (1947), he states, &quot;it is easy to regard to the Soviet Union as a rival not a partner, in the political arena. It must continue to expect that Soviet policies will reflect no abstra
Kansas State - PHYS - 214
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
Kansas State - PHYS - 214
1. The amplitude of the induced emf in the loop is m = A 0 ni0 = (6.8 10-6 m 2 )(4 10 -7 T m A)(85400 / m)(1.28 A)(212 rad/s)= 1.98 10-4 V.2. (a) =d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dtchb g(b) Appealing to
Kansas State - PHYS - 214
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
FSU - HIST - 08758
Prcis 3-28-08The Supreme Court's decision on Brown v. Board of Education of Topeka (1954) served as a turning point and as a milestone in the start of the civil rights movement. It reviewed the &quot;separate but equal&quot; principle decided by the case of
FSU - HIST - 08758
Prcis 4-2-08George C. Wallace displayed his prominence among white conservatives as Governor of Alabama in a speech on&quot; The Civil Rights Movement: Fraud, Sham, and a Hoax&quot; (1964). In June of 1963, Wallace defiantly prevented the first black student
FSU - HIST - 08758
Prcis 4-4-08Martin Luther King's Letter from Birmingham Jail (1963)is a letter Dr. King wrote after being jailed in Birmingham Alabama after organizing a economic boycott of white businesses, using a smuggled pen and piece of paper. King's statemen
FSU - GLY - 01636
Origin of the Universe: The Big Bang: Supernova: Life elements: Leon Sinks: Karst topography: Caves, caverns: Ground water: Porosity: Permeability Water table: Springs: Sink holes: o Wet Sinks: o Dry sinks: o Sink vs. Swamp Natural Bridge: Evolution
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 1 ~ Eras of Roman History Reading: BHR 2-4.Spring 2008Roman history is traditionally divided into three phases: monarchy, republic and empire. The dates are as follows (keep in mind that the very earliest date
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 2: Roman Society and Roman Values Readings: BHR 129-131; Shelton 128, 163-4, 171, 194-5, 197Spring 2008Roman Values and Virtues Roman virtues public, not private: virtuous man acts on behalf of the state Virtu
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books I-IVSpring 2008Virgil's Aeneid, strongly modeled on Homer's Iliad and Odyssey, was considered even in antiquity the great epic of Rome. It tells the story of the Trojan Aeneas who, afte
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books V-VIIISpring 2008Book V provides a transition between the high emotion of Book IV and the sombre majesty of the descent to the underworld in Book VI. Most of the book is taken up with g
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books IX-XIISpring 2008Book IX. War finally breaks out, the full-scale battles spoken of in Book VII. The book divides into three sections: (1) Turnus and the Rutulians attack the Trojan ship
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 5. Roman Imperialism and Expansion Readings: BHR 44-67; Shelton 291-293 I. Roman Imperialism - definition of `imperialism' - older notion of `defensive imperialism': how realistic? - was Rome more warlike/aggressi
FSU - CLA - 2123
Classics 2123: The Roman Way Roman Names A Roman had three names: praenomen (first name) nomen (name of the gens or clan) cognomen (family branch) Thus for: Publius praenomen Cornelius nomen Scipio cognomenSpring 2008his given name was &quot;Publius,&quot;
Stevens - PEP - 111
AC CIRCUITS35.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.1, the phase angle isU? = 180'n rad - 30&quot; = 150 x -= 2.618 rad180&quot;w=2*618ra
Stevens - PEP - 111
15.1. Solve: The density of the liquid is=m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 10 -3 10 -3 m 3Assess: The liquid's density is more than that of water (1000 kg/m3) and is a reasonable number.15.2. Solve: The volume of the helium
Stevens - PEP - 111
16.1. Solve: The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have thesame mass its volume must beVwater =mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 316.2. Solve: The volume of the uranium nucleus isV
Stevens - PEP - 111
17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =Kmicro. Solve: The number of atoms isN=M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kgBecause helium atoms have an atomic mass number A
Stevens - PEP - 111
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):1.013 10 5 Pa N p = 2.69 10 25 m -3 = = V kB T (1.38 10 -23 J K )(273 K )()18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-1
Stevens - PEP - 111
19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycleconsists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the a
Stevens - PEP - 111
20.1. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is h
Stevens - PEP - 111
21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by
Stevens - PEP - 111
22.1. Visualize: Please refer to Figure Ex22.1.Solve: (a)(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates
Stevens - PEP - 111
23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist=x 1.0 m = = 3.33 10 -9 s = 3.33 ns c 3 10 8 m / s(b) The refractive indices for water, glass, and zircon are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, l
Stevens - PEP - 111
24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,=91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6where n = 3, 4, 5, 6, . and wher
Stevens - PEP - 111
ELECTROMAGNETIC AND WAVES FIELDSw.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a &quot;net pole&quot; within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
FSU - CLA - 2123
CLA 2123: The Roman Way First Exam. February 8, 2007Name _Please read all directions carefully; no credit will be given for doing more than is required in each section. Part I. Identifications (35 points). Choose FIVE of the following and identif
Stevens - PEP - 111
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, henceT= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
Stevens - PEP - 111
1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Stevens - PEP - 111
2.1.Solve:Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0(b)2.2. Solve:Diagram (a) (b) (c)Position Negative Negative PositiveVelocity
Stevens - PEP - 111
3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the
Stevens - PEP - 111
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
Stevens - PEP - 111
5.1.Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:Solve:Written in component form, Newton's first law is( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 NT1 x = - T1T1y = 0 N Using Newton's first l
Stevens - PEP - 111
6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.Visualize:Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
Stevens - PEP - 111
7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
Stevens - PEP - 111
8.1. Visualize:Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
Stevens - PEP - 111
Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .9.1. Model: Model the car and the baseball as particles.9.2. Model: Model the bicycle and its rider as a particl
Stevens - PEP - 111
10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:Solve:For the bullet,KB =For the bowling ball,1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
Stevens - PEP - 111
11.1. Visualize:r Please refer to Figure Ex11.1. rSolve: (b) (c)(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.11.2. Visualize:r Please refer to Figure Ex11.2. rSolve
Stevens - PEP - 111
12.1.Solve: (b)Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)Fs on e =Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -Fm on e =GMm Me (6.67 10 -1
Stevens - PEP - 111
13.1. Model: The crankshaft is a rotating rigid body.Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 3. The Early History of Rome and Roman Republican Government Readings: BHR 15-41; Shelton 2-4, 7-8, 251-3, 255-259, 262, 264-5Spring 2008Early History of Rome Regal period: dimly known, mostly through legends;
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 6. Internal Disorders; Roman Slavery Readings: BHR 82-92; Shelton 207-209, 219, 227-229, 317-318Spring 2008I. Establishment of Roman provincial government Provincia: sphere of action of a magistrate with imper
UMiami - ACC - 212
Chapter 10Standard CostingAccounting 21210 - 1Learning Objective 1 Describe standard costing and indicate why standard costing is important.Accounting 212 10 - 2Why is Standard Costing Used?A standard is a preestablished benchmark for des
FSU - CLA - 2123
Classics 2123: The Roman Way Outline for Lecture 7. Roman Religion Readings: BHR 41-44; Shelton nos. 402-419, 423-428Spring 2008Problems Studying Ancient Religious Systems - time and culture (modern politics separated from religion) - vocabulary
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 8. The Strains of Empire (I) Readings: BHR 72-77, 92-110; Shelton 187-189, 266, 317-318Spring 2008I. Continuation of Roman Imperialism Macedonian Wars - Rome fights four wars in Greece, first against the Maced
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 9. The Strains of Empire (II) Readings: BHR 111-128, 132-140Spring 2008I. New Developments in the Roman State in the Late Republic Breakdown of concordia in the late Republic the result of many things, includi
UMiami - ACC - 212
Chapter 9The Operating Budget2004 Prentice Hall Business Publishing Introduction to Management Accounting , 2/e Werner/Jones9-1Learning Objective 1Describe some of the benefits of the operating budget.2004 Prentice Hall Business Publishing
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 10. The Fall of the Roman Republic Readings: BHR 124-179.Fall 2008I. Career of Pompey (Gn. Pompeius Magnus) (106-48 BCE) - begins as supporter of Sulla: raises legions from his father's troops (client army); a
UMiami - ACC - 212
Chapter 11Evaluating PerformanceAccounting 21211 - 1Learning Objective 1Describe centralized and decentralized management styles.Accounting 212 11 - 2Centralized ManagementTop management makes most of the decisions.The most experience