You will not be asked to graph isoclines like this on lab exams, but if you'd like to see the Mathematica notebook that was used to make the above plot, click on the icon to the left. It's still a little tough to see clearly, but try to verify from this image that as you trace along any particular one of the red hyperbolic isoclines, the slopes of the vectors which intersect it are all the same. Slope Field for dy/dx = x y with superimposed isoclines x y = C It is actually possible to achieve this effect with Mathematica, and the results end up looking something like this: Should find that all of the field marks intersecting the isocline have exactly the same slopes. Tracing our way along any of these superimposed curves we Superimpose some sample curves from the family x y = C on Would need to redo the problem with a much denser slope field, and then To be able to pick out the isoclines clearly we Now that you know that the isoclines are hyperbolas, can you see them? It's a hyperbola rotated 45 degrees from it's standard position.) Do you recognize it? It may take a littleĭigging in the deep recesses of your memory to come up with the fact that thisĬurve is a hyperbola for any non-zero value of C. So the equation x y = C describes the shape of the So let's rewrite the equation again as: constant = x y. The differential equation tells us: dy/dx = x y,īut we said a little earlier that isoclines are where slope isĬonstant. It easily just by looking at the graph as we've done in the past), look for to find an isocline, (especially if you can't find Word isocline-a curve in the slope field along which all of the field marks So what are they? Well, let's remind ourselves of the definition of the Isoclines of this field are not even straight lines! Slope field whose isoclines are neither vertical nor horizontal. For the first time in this laboratory we are witnessing a The slopes of the field marks get steeper the further away you go from This impression is simplyĪ result of the graph's y-axis symmetry. Slopes along vertical lines, but look more closely. Produced a picture that looked somewhat like the following:Īt first glance it may seem that the field marks once again have matching Give a geometric method of finding them, i.e.Slope Fields with Mathematica Making Slope Fields by Yourself A Post Exercise Discussion of dy/dx = x y, on the region -3 ≤ x ≤ 3,ĭon't cheat! If you didn't do the exercise in Mathematica before youĬame here to see the discussion, go back and do itĪssuming that you did the exercise correctly, you should have Use your description to explain why every nonzero complex number has exactly two square roots.Give a word description of the square function.Now increase p from Pi to 2Pi and study the resulting plots. Then plot the segment of the line connecting the points We will have to use other devices.į(z) = z 2 as a map of one copy of the complex plane into another.Įnter the following command to define a vertical line in the complex plane, Since both the domain and the range of such a function have dimension two - as a real space - the graph will be an object in real 4-space. In this part we examine various graphical tools for representing complex functions of a complex variable. Mathematica Tutor Part 14: Graphical representations of complex functions
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