qr code vb.net open source Illustration for the solution to Problem 4 in Chap 18 in .NET framework

Drawing USS Code 39 in .NET framework Illustration for the solution to Problem 4 in Chap 18

Illustration for the solution to Problem 4 in Chap 18
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5 Stated again for convenience, the parametric equations are x = t2 + 2t y=t z=0 According to the last equation, the whole object lies in the plane z = 0, which coincides with the xy plane In that system, the object is a parabola defined by x = t2 + 2t and y=t Substituting y for t in the first equation, we obtain x = y 2 + 2y Figure B-33 is a graph of this curve as it looks when we observe the xy plane broadside from a point on the +z axis at a considerable distance from the origin
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Figure B-33
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Illustration for the solution to Problem 5 in Chap 18
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568 Worked-Out Solutions to Exercises: 11-19
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6 We have been given the parametric equations x=t y = 7 z = t 2 /2 5 According to the second equation, our object lies entirely in the plane y = 7 This plane is perpendicular to the y axis, parallel to the xz plane, and 7 units distant from the xz plane on the y side When we draw projections of the three-space x and z axes onto the plane y = 7, we create a coordinate grid for a parabola defined by x=t and z = t2 /2 5 Let s substitute x directly into the second of these equations to obtain z = x 2 /2 5 Figure B-34 is a graph of this equation as it appears when seen from a point of view broadside to the plane y = 7 We re looking in the +y direction from somewhere along the y axis, but quite a lot farther away from the origin than the point where y = 7
z 6 4 2 x 6 4 2 2 4 6 2 4 6
Figure B-34
Illustration for the solution to Problem 6 in Chap 18
18
7 Repeated for convenience, the parametric equations are x = 4 cos t y = 4 sin t z=1 According to the last equation, the object is contained entirely in the plane z = 1, which is parallel to the xy plane and 1 unit away from it on the +z side Within this plane, the parametric equations of the object reduce to x = 4 cos t and y = 4 sin t The graph is a circle in the plane z = 1, centered on the point (0,0,1) and having a radius of 4 units Figure B-35 is a graph of this object as we would see it looking broadside at the plane z = 1, from a point fairly far from the origin on the +z axis
2 x 6 2 2 2 6
Figure B-35
Illustration for the solution to Problem 7 in Chap 18
570 Worked-Out Solutions to Exercises: 11-19
8 Repeated for convenience, the parametric equations are x = 5 cos t y=0 z = 5 sin t According to the middle equation, the entire object lies in the plane y = 0, which is the xz plane Within the Cartesian xz system, the equations describing the object are x = 5 cos t and z = 5 sin t The graph is a circle in the xz plane, centered at the origin and having a radius of 5 units Figure B-36 illustrates this circle as seen from somewhere along the y axis We re fairly far from the origin, and we re looking in the +y direction
z 6 4 2 x 6 4 2 2 4 6 2 4 6
Figure B-36
in Chap 18
Illustration for the solution to Problem 8
18
9 Repeated for convenience, the parametric equations are x = 5 cos t y = 3 sin t z=p According to the last equation, the object is contained entirely in the plane z = p, which is parallel to the xy plane and p units away from it on the +z side Within that plane, the parametric equations are x = 5 cos t and y = 3 sin t The graph is an ellipse in the plane z = p and centered on (0,0,p ) The major semi-axis is parallel to the x axis, and measures 5 units wide The minor semi-axis is parallel to the y axis, and measures 3 units high Figure B-37 is a graph of this ellipse as we gaze broadside at the plane z = p, from some location on the +z axis that s considerably farther from the origin than (0,0,p ) 10 Stated again for reference, the parametric equations are x = 2 cos t y = t /(2p ) z = 2 sin t
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