# Slope Fields

Differential Equation 2 – Slope Fields

Of course, we always want to see the graph of an equation we are studying. The graph of a differential equation is a slope field.  A first derivative expressed as a function of x and y gives the slope of the tangent line to the solution curve that goes through any point in the plane.

Slope fields make use of this by imposing a grid of points evenly spaced across the Cartesian plane. At each point the value of the derivative is calculated and a short segment with that slope is drawn. These segments graphed together form the slope field.

Here is the slope field for the differential equation $\displaystyle \frac{dy}{dx}=-\frac{x}{y}$ A good way to introduce slope fields to your class is to put or project a coordinate system on the board. Give each student one or two points (1, –3), (1, –2), (1, –1), (1, 1), (1, 2), etc. Have them calculate the derivative at their point(s) and then come to the board and draw a short segment through their point(s) with the slope they calculated. The result will be a slope field.  (This is, in fact, a common free-response question on the AP exams. Students are given a graph with 9 – 12 points plotted and they are asked to use them to draw a slope field for a given differential equation.)

The big idea with slope fields is to use them to get an idea of what the solutions look like, especially if the differential equation cannot be solved.  The solutions are lurking in the slope field. What do they look like in the figure? Circles of course. We solved this differential equation in the last post. The general solution is ${{x}^{2}}+{{y}^{2}}={{C}^{2}}$. Of course they are not all that simple.

Since the solution graphs are lurking in the slope field, the next thing to do is to use the slope field to sketch a particular solution.  After plotting the initial condition (a point) students should draw a curve through the point that follows the slope field from edge to edge in both directions.

In the previous post (example 2) we found the particular solution of $\displaystyle \frac{dy}{dx}=-\frac{x}{y}$ with the initial condition point (4, –3) to be $y=-\sqrt{25-{{x}^{2}}}$. This is shown drawn on the slope field in the next graph. The black dot is the point (4, –3). Notice how the solution graph follows the slope field, but does not necessarily hit any of the segments. The solution will touch a segment only if the midpoint of the segment happens to be on the solution – this is not usually the case. Slope fields are tedious and time-consuming to draw by hand. It’s a job for computers. There are various graphing calculator programs available on the internet. Calculator screen are not the best for seeing slope fields; they are too small and you should be sure to always be in a square window (or the slopes will not look right).

There are many websites that will draw slope fields and solution curves for you.  You can try this one. The figures in this post were done with Winplot (of course, my favorite). The good ones let you draw and animate solution curves over the slope field. (I have not found a good slope field generator app for iPads; if anyone makes apps, consider this a hint.)

Here is a brief example that shows how powerful an animated graph can be. In Winplot, follow the path Window > 2-dim > Equa > differential > dy/dx.  Enter the differential equation in the box and adjust the other settings as necessary.  Be sure you are in a square window (CTRL+Q).

The example below is from the 2002 BC exam question 5: $dy/dx=2y-4x$. Notice that this equation is not separable; students were not expected to solve it. They were asked to draw the solution curves through the two points (0, 1) and (0, –1) shown here in blue. These points are marked on the graph (Equa > point > (x,y)). The general solution, found by CAS, is $y=C{{e}^{2x}}+2x+1$. Enter this (Equa > 1.Explicit) and open the C slider (Anim > individual > C).

In the video below the C values go from –5 to 5. They stop momentarily at the two initial condition points (C = –2 for (0,–1) and C = 0 for (0, 1)). These are what the students were expected to sketch. The point is to see how the different values of C effect the equation, each giving its own particular solution, and to see the different solution hiding in the slope field.

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