AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/7/7-4)
Unit 7 · Topic 7.4
7.4 Reasoning Using Slope Fields
Once you have a slope field, you can sketch the solution curve through a given point, match a field to its differential equation, and use the equation itself to find tangent lines and concavity of solutions. Different starting points lead to different curves in the same family.
Key terms
- slope field
- solution curve
- initial condition
- family of solutions
- zero slope
Sketching a solution curve
Start at the given point (the initial condition). Draw a smooth curve that moves through the field, staying tangent to the segments it passes and following their direction both to the right and to the left. The curve should never cut sharply across segments.
For the well-behaved equations in this course, solution curves don't cross each other: at each point the slope is fixed, and through each point in the field there's exactly one solution curve. Through a different point you'd usually get a different curve. That's the family of solutions you saw in 7.2.
Matching a field to its equation
When asked which equation matches a slope field, test features one at a time and eliminate choices.
- Where are the segments horizontal? On a vertical line (like x = 0), a horizontal line (like y = 1) or a slanted line (like y = −x)?
- Are the segments in each column parallel (depends only on x) or in each row parallel (depends only on y)?
- In which regions are slopes positive or negative?
- Pick an easy point, like (1, 1), and compare the slope you see with each equation's value there.
Using the equation: tangent lines and concavity
The equation gives the slope of the solution at any point it passes through. With a point and that slope, you can write a tangent line and use it to estimate nearby values (local linearity, 4.6).
To find concavity, differentiate the equation implicitly to get d²y/dx². Wherever y appears, its derivative is dy/dx, which you then replace with the original expression. The sign of d²y/dx² at the point tells you whether the solution is concave up (tangent-line estimates are too low) or concave down (tangent-line estimates are too high).
Equilibrium and long-run behavior
If dy/dx = 0 along an entire horizontal line y = c, then the constant function y = c is a solution. Solution curves near that line may approach it or move away from it, which you can see from the field. For example, in dy/dx = 3 − y, solutions above y = 3 decrease toward it and solutions below increase toward it. So every solution approaches y = 3 as x → ∞.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Matching a slope field
A slope field has horizontal segments along both the x-axis and the y-axis. Its slopes are positive in Quadrants I and III and negative in Quadrants II and IV. Which equation matches: (A) dy/dx = x, (B) dy/dx = y, (C) dy/dx = x + y or (D) dy/dx = xy?
Show the solutionHide the solution
- Step 1: (A) is zero only when x = 0, so its segments aren't horizontal along the x-axis. Eliminate.
- Step 2: (B) is zero only when y = 0, so its segments aren't horizontal along the y-axis. Eliminate.
- Step 3: (C) is zero along the line y = −x, not along the axes. Eliminate.
- Step 4: (D): xy = 0 on both axes. In Quadrant I, x > 0 and y > 0, so xy > 0; in Quadrant III both are negative, so xy > 0; in Quadrants II and IV the signs differ, so xy < 0. Everything matches.
Answer: (D) dy/dx = xy
- Example 2
Tangent line and concavity from the equation
Let y = f(x) be the solution to dy/dx = 2x − y with f(1) = 1. (a) Write the tangent line at x = 1 and use it to estimate f(1.1). (b) Find d²y/dx² at (1, 1). Is the estimate too high or too low? (c) Show that y = 2x − 2 is also a solution of the differential equation.
Show the solutionHide the solution
- Step 1: (a) Slope at (1, 1): 2(1) − 1 = 1. Tangent line: y = 1 + 1(x − 1). At x = 1.1: f(1.1) ≈ 1.1.
- Step 2: (b) Differentiate dy/dx = 2x − y with respect to x: d²y/dx² = 2 − dy/dx = 2 − (2x − y).
- Step 3: At (1, 1): d²y/dx² = 2 − 1 = 1 > 0, so the solution is concave up near x = 1.
- Step 4: A concave-up curve lies above its tangent lines, so the estimate 1.1 is an underestimate.
- Step 5: (c) If y = 2x − 2, then dy/dx = 2, and 2x − y = 2x − (2x − 2) = 2. Both sides are 2, so it's a solution. In the slope field, it appears as a line that the segments lie along.
Answer: (a) y = x, so f(1.1) ≈ 1.1. (b) d²y/dx² = 1 at (1, 1); the estimate is an underestimate. (c) Both sides equal 2.
Common mistakes
- Drawing a solution curve only to the right of the starting point. Unless told otherwise, extend it both ways.
- Differentiating dy/dx = 2x − y and writing d²y/dx² = 2. The y term has derivative dy/dx, which you can't ignore.
- Leaving d²y/dx² in terms of dy/dx without substituting the expression for dy/dx.
- Matching a field by checking only one feature when two choices share it.
On the exam
- Free-response questions about differential equations often go: sketch a slope field, find a tangent line estimate, decide if it's an over- or underestimate using d²y/dx², then solve by separation of variables.
- When justifying an over- or underestimate, name the concavity and why: “d²y/dx² > 0 near the point, so the graph is concave up and lies above the tangent line.”
Connected topics
Videos
Check yourself
4 questions on 7.4 Reasoning Using Slope Fields. Pick an answer to see if you got it, and why.
A slope field for a differential equation has these features. Along any vertical line x = c, all the segments have the same slope. The segments are horizontal along the y-axis, have positive slope when x > 0 and negative slope when x < 0, and get steeper as |x| increases. Which of the following could be the differential equation?
Let y = f(x) be the solution to the differential equation dy/dx = (y − 1)(y − 4) with initial condition f(0) = 2. What is lim (x→∞) f(x)?
Consider the differential equation dy/dx = y² − 4. For which of the following initial conditions does the solution satisfy lim (x→∞) y = −2? I. y(0) = −3 II. y(0) = 1 III. y(0) = 3
Let y = f(x) be the solution to the differential equation dy/dx = x − y with f(1) = 3. Which of the following describes the graph of f at the point (1, 3)?
0 of 4 answered