course Mth 272 7/31 2:30 pm Query 27*********************************************
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Given Solution: `a The level curves are the curves for which e^(xy) = c. Solving e^(xy) = c for y in terms of x: first take the natural log of both sides of the equation to get xy = ln(c). Then solve for y, obtaining y = ln(c) / x. Think of this as y = ln(c) * 1/x. The graph of 1/x has vertical asymptotes at the y axis, approaching +infinity as we approach x = 0 from the right and -infinity as we approach x = 0 from the left, approaching the x axis as x -> infinity or x -> -infinity and passing through the points (-1, -1) and (1, 1). y = ln(c) * 1/x vertically stretches the graph of y = 1/x by factor ln(c); the resulting graph passes through (-1, -ln(c)) and (1, ln(c)). For c = 1, 2, 3, 4 we have ln(c) = 0, .70, 1.10, 1.39 respectively. For c = 2, 3, 4 the corresponding level curves are as described above, passing through points (-1, -.70) and (1, .70); (-2, -1.10) and (2, 1.10); (-3, -1.39) and (3, 1.39). For c = 0 we have ln(c) = 0 so the curve is y = 0 * 1/x = 0; this curve coincides with the x axis. For c = 1/2, 1/3 and 1/4 the values of ln(c) are -.70, -1.10 and -1.39. [ note that ln(1/c) = ln(1) - ln(c) = 0 - ln(c) = -ln(c) ]. So the level curves 'flip' about the x axis, with the curves respectively passing through points (-1, .70) and (1, -.70); (-2, 1.10) and (2, -1.10); (-3, 1.39) and (3, -1.39). The family of curves in the first quadrant descends from the y axis, each curve gradually changing direction until it becomes asymptotic to the x axis. Each curve 'dips down' toward the origin; for c = 1, 2, 3, 4 the curves 'dip down' less and less gradually, each succeeding curve lying a bit higher (in the y direction) than the one preceding it and representing an altitude above the x-y plane which is 1 unit higher than the one before. The c = 1 curve actually coincides with the positive x and y axes, making a right angle at the origin. In the third quadrant the c = 1, 2, 3, 4 curves are symmetric with respect to the origin with those in the first quadrant, and as we move away from the origin the altitudes above the xy plane again increase by 1 with every new curve. The c = 1/2, 1/3, 1/4, ... curves in the second and fourth quadrants are reflections through the x axis of the curves from the first and third quadrants. They represent altitudes of 1/2, 1/3, 1/4, ... above the xy plane.
course Mth 272 7/31 2:30 pm Query 27*********************************************
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Given Solution: `a The level curves are the curves for which e^(xy) = c. Solving e^(xy) = c for y in terms of x: first take the natural log of both sides of the equation to get xy = ln(c). Then solve for y, obtaining y = ln(c) / x. Think of this as y = ln(c) * 1/x. The graph of 1/x has vertical asymptotes at the y axis, approaching +infinity as we approach x = 0 from the right and -infinity as we approach x = 0 from the left, approaching the x axis as x -> infinity or x -> -infinity and passing through the points (-1, -1) and (1, 1). y = ln(c) * 1/x vertically stretches the graph of y = 1/x by factor ln(c); the resulting graph passes through (-1, -ln(c)) and (1, ln(c)). For c = 1, 2, 3, 4 we have ln(c) = 0, .70, 1.10, 1.39 respectively. For c = 2, 3, 4 the corresponding level curves are as described above, passing through points (-1, -.70) and (1, .70); (-2, -1.10) and (2, 1.10); (-3, -1.39) and (3, 1.39). For c = 0 we have ln(c) = 0 so the curve is y = 0 * 1/x = 0; this curve coincides with the x axis. For c = 1/2, 1/3 and 1/4 the values of ln(c) are -.70, -1.10 and -1.39. [ note that ln(1/c) = ln(1) - ln(c) = 0 - ln(c) = -ln(c) ]. So the level curves 'flip' about the x axis, with the curves respectively passing through points (-1, .70) and (1, -.70); (-2, 1.10) and (2, -1.10); (-3, 1.39) and (3, -1.39). The family of curves in the first quadrant descends from the y axis, each curve gradually changing direction until it becomes asymptotic to the x axis. Each curve 'dips down' toward the origin; for c = 1, 2, 3, 4 the curves 'dip down' less and less gradually, each succeeding curve lying a bit higher (in the y direction) than the one preceding it and representing an altitude above the x-y plane which is 1 unit higher than the one before. The c = 1 curve actually coincides with the positive x and y axes, making a right angle at the origin. In the third quadrant the c = 1, 2, 3, 4 curves are symmetric with respect to the origin with those in the first quadrant, and as we move away from the origin the altitudes above the xy plane again increase by 1 with every new curve. The c = 1/2, 1/3, 1/4, ... curves in the second and fourth quadrants are reflections through the x axis of the curves from the first and third quadrants. They represent altitudes of 1/2, 1/3, 1/4, ... above the xy plane.