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Section 2.2

Section 2.2. THE GEOMETRY OF SYSTEMS. Some old geometry. We learned to represent a DE with a slope field, which is a type of vector field . Solutions to the DE run parallel to the vectors in the slope field. Can we do something similar with systems of DEs?. Some new geometry.

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Section 2.2

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  1. Section 2.2 THE GEOMETRY OF SYSTEMS

  2. Some old geometry We learned to represent a DE with a slope field, which is a type of vector field. Solutions to the DE run parallel to the vectors in the slope field. Can we do something similar with systems of DEs?

  3. Some new geometry Last time, we talked about how solutions to a system can be represented by parametric curves on the phase plane. How does one draw these pictures? This is the graph of the parametric equation (x,y) = (R(t), F(t)) for the IVP. R(0) = 4 F(0) = 1

  4. How can we find the slopes of solution curves in the xy-plane?

  5. The slope of a parametric curve If you have any continuous curve in the xy-plane, then its slope* is the limit as x goes to 0 of y/x. In other words, it’s dy/dx. Suppose our curve is the graph of a parametric equation (x(t), y(t)). We can find (dx/dt, dy/dt) easily, but what is the slope in terms of x and y? Use the Chain Rule: So we want to draw the vector (dx/dt, dy/dt), which has slope dy/dx. *I’m assuming that the slope exists (the curve doesn’t have a corner).

  6. Drawing the vector field for a DE Solution curves are tangent to the vectors. Let’s draw the vector field for Pick the point (0.5, 1). The vector at this point is (dx/dt, dy/dt) = (1, -0.5). Its slope is dy/dx = -1/2. Now pick more points…

  7. Direction field You could also sketch a direction field, which just has little lines indicating slopes instead of vectors. Sketch solution curves along these vectors.

  8. Note on numerical methods Section 2.4 covers Euler’s Method for systems of DEs. The method works the same way as Euler’s Method for single DEs. The idea: • You start at the point (x0, y0) = (x(t0), y(t0)). • Use the system of DEs to find dx/dt and dy/dt. • Pick a value of t. • Then use the fact that (x/t, y/t)  (dx/dt, dy/dt) to find x and y. (so x  (t · dx/dt), etc.) • Therefore (x1, y1) = (x0 + x, y0 + y) and t1 = t0 + t. • Repeat this process to find more points.

  9. Notation There is a lot of new notation in this section. Look at the bottom of p. 173. I’ll summarize it on the board.

  10. Putting it all together Now that you’re experts on graphing systems, we’re going to work through the system on p. 179 in all its gory detail!

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