course Mth 174
Can you help me understand how to draw the followingSketch the direction field for the differential equation dy / dx = .49 e^x / y. Sketch some solution curves corresponding to this field.
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I understand that it has something to do with the slope at a point. Do we need to plug in different numbers fo x and y and see what the slope is at them points.
First, as y -> 0 the value of dy/dx approaches +infinity; the slopes approach vertical. So a set of short vertical line segments through the x axis would be a good start.
For all x, e^x > 0 so whenever y is positive the slopes will be positive, and whenever y is negative the slopes will be negative.
As you move to the right, e^x becomes very large very fast, and e^x / y will also be very large (though division by y decreases the magnitude of the result, it's easy to make e^x much larger than y by just moving a little ways to the right).
So to the right, the slopes will again be nearly vertical.
As we move to the left (for negative x), e^x becomes very small very quickly, so except very near the x axis the slopes will approach zero. A set of horizontal lines toward the left edge of the graph would depict this behavior.
To see what happens in between these extremes, consider first the y axis, where x = 0. The slopes at the point (0, y) is .49 / y. So when y = .49, the slope is 1; when y = .98 the slope is 1/2; when y = 1.47 the slope is 1/3 and the slopes continue to decrease as you move up the y axis. Something similar happens along the negative y axis, but the slopes will be negative.
Along the vertical line x = 1 the slopes will be greater; when y = .49 the slope will be e = 2.718, approx., when y = 2 the slope will be 1.36 approx., and as y increases the slopes gradually approach 0.
If you similarly consider the vertical lines where x = -1, x = 2 and x = 3 you will get a pretty good idea how to map this slope field.
Another strategy might be to consider the set of points where .49 e^x / y is equal to a fixed value c:
If .49 e^x y = c, then y = c / (.49 e^x) = 2.04 c * e^-x.
For c = 1 this is the curve y = 2.04 e^-x. This curve can be easily sketched, and at points on this curve the slope is c = 1. A series of short segments with slope 1 would depict the direction field along this curve.
The c = 2 curve is y = 4.08 e^-x; this curve is also easy to sketch and a series of short segments with slope 2 can be placed at intervals along this curve.
It would be worthwhile to continue for the c = -1 and c = -2 curves, then consider the curves c = 3 and c = 4, and ask yourself what would happen for additional values of c. The direction field so constructed would be consistent with that previously constructed.
To answer a question of this nature you could either sketch the direction field or give a written description of the process.
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