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CLAREMONT MCKENNA COLLEGE
Solved Exercises on Slope Fields
PART 1: TRADITIONAL METHOD
1) Sketch the slope field over the region [-5,5] x [-5,5]. Include the isocline in your
sketch.
y
′
= x
2
Next, trace the solution that satisfies 𝑦( 1 ) = 0
Solution:
a) Understanding y′ (the Derivative):
• y′ represents the slope of the tangent line to the solution curve at any given point
(x,y).
• For each point (x,y) in the plane, we compute y′=f(x,y).
b) Determine Slopes for Various Points:
- At each point (x,y), draw a small line segment with the slope given by y′.
- If y′=0 at a point, the line segment is horizontal.
- As y′ increases, the line segment becomes steeper.
c) Plotting Slopes:
- Choose a grid of points in the (x,y) plane.
- Calculate the slope y′ at each grid point using the differential equation.
- For example, using a table of values for y’, we substitute an arbitrary value for x
(both negative and positive value) to determine the direction and behavior of the
slope field.
x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5
y 0 0 0 0 0 0 0 0 0 0 0
y’ 27 18 11 6 3 2 3 6 11 18 27
d) Plot Slope Segments:
- At each grid point, draw a short line segment with the slope calculated.
- If y′=0y′=0, draw a horizontal segment.
- If y′>0y′>0, draw a segment with a positive slope. The greater y′, the steeper the segment.
- If y′<0y′<0, draw a segment with a negative slope. The more negative y′, the steeper the
segment downward.
To visualize what the graph looks like virtually, a trace is plotted using the website
GeoGebra using y’ over the region [-5,5] x [-5,5]. The point is (1,0) based on the given in
c) Plot Slope Segments:
o Recall how we draw line segments for slope fields using the earlier example and
try doing the slope field on your own.
o To check, we can use GeoGebra again for the correct plot. Trace the point (1,0)
based on the given in #2.
PART 2: USING TRACING SOFTWARE (MAXIMA)
a) Install and Open Maxima:
- Make sure you have Maxima installed on your computer. Open the Maxima
console.
- Here’s how to install it on windows: https://maxima.sourceforge.io/windows-
install.html
b) Define the Differential Equation:
• Let us use some given function y’ where,
′
= (y + 2 )
2
(y − 4 )
• Plot this in Maxima using the command plotdf as shown below. Disregard the
command for trajectory for a while.
c) Create a Grid of Points and Compute Slopes:
- Define a range for x and y. For example:
o x_min: - 5;
o x_max: 5;
o y_min: - 5;
o y_max: 5;
o step: 0.5;
d) Find Equilibrium Solutions:
• Equilibrium solutions occur where y' = 0. In this case,
′
= (y + 2 )
2
(y − 4 )
y + 2
2
y − 4
• The trace of y=-2 and y=4 is presented below using the command
[trajectory_at, x, y]. In this case, x=0 while y will be the values which we will
plug in.
- 𝑥 = 0 , 𝑦 = −
- 𝑥 = 0 , 𝑦 =
e) Determine Stability:
• Next, we evaluate the sign of y' near the equilibrium points to determine
stability. Try clicking near some points on the graph for each y value. Note
that you may have to type the command plotdf for different trajectories each
time.
• Observations:
o For 𝑦 < − 2 , ( y + 2 )
2
(y − 4 ) > 0 , which means it is a positive slope
and the solution moves away.
o For 𝑦 > − 2 , ( y + 2 )
2
(y − 4 ) < 0 , which means it is a negative slope
and the solution moves toward.
o Thus, y = - 2 is SEMI-STABLE, as only one side of the solution is
asymptotic (For a review on the stability of the trace points, check
out this link).
• Based from these behaviors of slope fields, do you think two distinct
solutions to the equation intersect?
o It’s possible. The derivative F(x,y) is well-defined and has a unique
value at any given point in the region under consideration. If we
assume the derivative is well-behaved everywhere and the curves
cannot cross at nonzero angles, then the solution curve passing
through any point in 2D space must be tangent to a single, distinct
slope.
o Therefore, while it is possible for not well-behaved solution curves to
intersect at a point, such intersections are rare.
