27 open query

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course Mth 272

11/17 6:45 pm EST

027.

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Question: `qQuery problem 7.3.38 level curves of e^(xy) for c = 1, 2, 3, 4, 1/2, 1/3, 1 / 4

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Your solution:

e^xy = c

ln e^xy = ln c

xy = ln c

y = ln c/x

level curves:

c = 1, y = 0/x = 0

c = 2, y = ln 2/x, y = 0.70/x

c = 3, y =ln 3/x, y = 1.10/x

c = 4, y = ln 4/x, y = 1.40/x

y = 0.70/x brings the curve closer than ln x to the origin of (0,0); as 1.10/x and 1.40/x move further away in both directions; quadrants 1 and 3.

level curves:

c = ½, y = ln 0.5/x, y = -0.70/x

c = 1/3, y = ln 0.33/x, y = -1.10/x

c = ¼, y = ln 0.25/x, y = -1.40/x

y = -0.70/x, -1.10/x, and -1.40/x exemplify the same results as above in comparison of location to graphed ln x; however, it is inverse in quadrants area, and the location is 2 and 4 quadrants.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

#### Ok, the information really helps here. I had a little bit of trouble applying my graphing here given dimension, but clear pertaining to concept, making a bit more progress.

You should be very familiar with the graph of y = 1 / x. Focus on the point (1, 1) and the fact that the graph in the right half-plane has asymptotes with the positive y and positive x axes.

If you multiply this function by ln(c) the point becomes (1, ln(c)). For c = 1, 2, 3, 4 the points rise higher and higher, but with less space between successive points. The level curves pass through these points, still with asymptotes at the x and y axes. Each curve is a bit 'higher' than the preceding.

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self-critique rating #$&*: 3

&#Good responses. See my notes and let me know if you have questions. &#