# 7.4 Slope Fieldsap Calculus

## Calculus AB and Calculus BC

Calculus Examples. Popular Problems. The slope of the line is the value of, and the y-intercept is the value of. ## CHAPTER 9 Differential Equations

### B. SLOPE FIELDS

In this section we solve differential equations by obtaining a slope field or calculator picture that approximates the general solution. We call the graph of a solution of a d.e. a solution curve.

The slope field of a d.e. is based on the fact that the d.e. can be interpreted as a statement about the slopes of its solution curves.

EXAMPLE 1

The d.e. tells us that at any point (x, y) on a solution curve the slope of the curve is equal to its y-coordinate. Since the d.e. says that y is a function whose derivative is also y, we know that

Slope fields show up on both the AP Calculus AB and BC tests. While at first this topic might seem daunting, the questions on the test are actually quite straightforward. Just keep one thing in mind: Go with the flow! A slope field shows the direction of flow for solutions to a differential equation. 7B Slope of Curve 4 Definition: The slope of a function, f, at a point x = (x, f(x)) is given by m = f '(x) = f '(x) is called the derivative of f with respect to x. Other names for f '(x): slope instantaneous rate of change speed velocity EX 2 Find the derivative of f(x) = 4x - 1. Therefore by drawing a curve through consecutive slope lines, you can find a solution to the differential equation. Take the example of #dy/dx# at #(3, 4)#. Here we see that. #dy/dx = 3 - 4 =-1# So you would draw a line of slope #-1# at #(3,4)#. Repeat this for maybe 4 by 4 points to get the following slope field. Hopefully this helps! Asked. 07/26/19 Write the equation of the line that passes through 7, -4 and -1, 2 in slope intercept form.

y = ex

is a solution. In fact, y = Cex is a solution of the d.e. for every constant C, since y = Cex = y.

The d.e. y = y says that, at any point where y = 1, say (0, 1) or (1, 1) or (5, 1), the slope of the solution curve is 1; at any point where y = 3, say (0, 3), (ln 3,3), or (π, 3), the slope equals 3; and so on.

In Figure N9–1a we see some small line segments of slope 1 at several points where y = 1, and some segments of slope 3 at several points where y = 3. In Figure N9–1b we see the curve of y = ex with slope segments drawn in as follows:

FIGURE N9–1a

FIGURE N9–1b

Figure N9–1c is the slope field for the d.e. Slopes at many points are represented by small segments of the tangents at those points. The small segments approximate the solution curves. If we start at any point in the slope field and move so that the slope segments are always tangent to our motion, we will trace a solution curve. The slope field, as mentioned above, closely approximates the family of solutions.

FIGURE N9–1c

EXAMPLE 2

The slope field for the d.e. is shown in Figure N9–2.

(a) Carefully draw the solution curve that passes through the point (1, 0.5).

(b) Find the general solution for the equation.

FIGURE N9–2

SOLUTIONS:

(a) In Figure N9–2a we started at the point (1, 0.5), then moved from segment to segment drawing the curve to which these segments were tangent. The particular solution curve shown is the member of the family of solution curves

y = ln x + C

that goes through the point (1, 0.5).

FIGURE N9–2a

FIGURE N9–2b

(b) Since we already know that, if then we are assured of having found the correct general solution in (a).

In Figure N9–2b we have drawn several particular solution curves of the given d.e. Note that the vertical distance between any pair of curves is constant.

EXAMPLE 3

Match each slope field in Figure N9–3 with the proper d.e. from the following set. Find the general solution for each d.e. The particular solution that goes through (0,0) has been sketched in.

(A) y = cos x

(B)

(C)

(D)

FIGURE N9–3a

FIGURE N9–3b

FIGURE N9–3c

FIGURE N9–3d

SOLUTIONS:

(A) goes with Figure N9–3c. The solution curves in the family y = sin x + C are quite obvious.

(B) goes with Figure N9–3a. The general solution is the family of parabolas y = x2 + C.

For (C) the slope field is shown in Figure N9–3b. The general solution is the family of cubics y = x3 − 3x + C.

(D) goes with Figure N9–3d; the general solution is the family of lines y =

EXAMPLE 4

(a) Verify that relations of the form x2 + y2 = r2 are solutions of the d.e.

(b) Using the slope field in Figure N9–4 and your answer to (a), find the particular solution to the d.e. given in (a) that contains point (4, −3).

FIGURE N9–4

SOLUTIONS:

(a) By differentiating equation x2 + y2 = r2 implicitly, we get 2x + 2y from which which is the given d.e.

(b) x2 + y2 = r2 describes circles centered at the origin. For initial point (4,−3), (4)2 + (−3)2 = 25. So x2 + y2 = 25. However, this is not the particular solution.

A particular solution must be differentiable on an interval containing the initial point. This circle is not differentiable at (−5,0) and (5,0). (The d.e. shows undefined when y = 0, and the slope field shows vertical tangents along the x-axis.) Hence, the particular solution includes only the semicircle in quadrants III and IV.

Solving x2 + y2 = 25 for y yields The particular solution through point (4,−3) is with domain −5 < x< 5.

Derivatives of Implicitly Defined Functions

In Examples 2 and 3 above, each d.e. was of the form = f (x) or y = f (x). We were able to find the general solution in each case very easily by finding the antiderivative We now consider d.e.’s of the form where f (x,y) is an expression in x and y; that is, is an implicitly defined function. Example 4 illustrates such a differential equation. Here is another example.

EXAMPLE 5

Figure N9–5 shows the slope field for

At each point (x,y) the slope is the sum of its coordinates. Three particular solutions have been added, through the points

FIGURE N9–5

Slope fields show up on both the AP Calculus AB and BC tests. While at first this topic might seem daunting, the questions on the test are actually quite straightforward. Just keep one thing in mind: Go with the flow!

A slope field shows the direction of flow for solutions to a differential equation.

## What is a Slope Field?

A slope field is a visual representation of a differential equation of the form dy/dx = f(x, y). At each sample point (x, y), there is a small line segment whose slope equals the value of f(x, y).

That is, each segment on the graph is a representation of the value of dy/dx. (Check out AP Calculus Review: Differential Equations for more about differential equations on the AP Calculus exams.)

Because each segment has slope equal to the derivative value, you can think of the segments as small pieces of tangent lines. Any curve that follows the flow suggested by the directions of the segments is a solution to the differential equation.

Each curve represents a particular solution to a differential equation.

### Example — Building a Slope Field

Consider the differential equation dy/dx = xy. Let’s sketch a slope field for this equation. Although it takes some time to do it, the best way to understand what a slope field does is to construct one from scratch.

First of all, we need to decide on our sample points. For our purposes, I’m going to choose points within the window [-2, 2] × [-2, 2], and we’ll sample points in increments of 1. Just keep in mind that the window could be anything, and increments are generally smaller than 1 in practice.

Now plug in each sample point (x, y) into the (multivariable) function xy. We will keep track of the work in a table.

x - yx = -2x = -1x = 0x = 1x = 2
y = 2-2 - 2 = -4-1 - 2 = -30 - 2 = -21 - 2 = -12 - 2 = 0
y = 1-2 - 1 = -3-1 - 1 = -20 - 1 = -11 - 1 = 02 - 1 = 1
y = 0-2 - 0 = -2-1 - 0 = -10 - 0 = 01 - 0 = 12 - 0 = 2
y = -1-2 - (-1) = -1-1 - (-1) = 00 - (-1) = 11 - (-1) = 22 - (-1) = 3
y = -2-2 - (-2) = 0-1 - (-2) = 10 - (-2) = 21 - (-2) = 32 - (-2) = 4

Ok, now let’s draw the slope field. Remember, the values in the table above represent slopes — positive slopes mean go up; negative ones mean go down; and zero slopes are horizontal.

Spend some time matching each slope value from the table with its respective segment on the graph.

It’s important to realize that this is just a sketch. A more accurate picture would result from sampling many more points. For example, here is a slope field for dy/dx = xy generated by a computer algebra system. The viewing window is the same, but now there are 400 sample points (rather than the paltry 25 samples in the first graph).

Slope field for dy/dx = xy

## Analyzing Slope Fields

Now let’s get down to the heart of the matter. What do I need to know about slope fields on the AP Calculus AB or BC exams?

You’ll need to master these basic skills:

• Given a slope field, select the differential equation that best matches it.
• Given a slope field, estimate values of a solution with given initial condition.
• Sketch a slope field on indicated sample points, from a given differential equation.

We’ve already seen above how to sketch a slope field, so let’s get some practice with the first two skills in the list instead.

### Sample Problem 1

The slope field shown above corresponds to which of the following differential equations.

A.dy/dx = y2

B.dy/dx = sin y C.dy/dx = -sin y D.dy/dx = sin x

#### Solution

Look for the clues. The segments have the same slopes in any given row (left to right across the graph). Therefore, since the slopes do not change with respect to x, we can assume that dy/dx is a function of y alone. That eliminates choice D.

The horizontal segments occur when y = 0, π, and -π. However, the only point at which y2 equals 0 is y = 0 (not π or -π). That narrows it down to a choice between B and C.

Finally, notice that slopes are positive when 0 < y < π and negative when -π < y < 0. This pattern corresponds to the values of sin y. (The signs are opposite for -sin y, ruling out choice C).

The correct choice is B.

### Sample Problem 2

Suppose y = f(x) is a particular solution to the differential equation dy/dx = xy such that f(0) = 0. Use the slope field shown earlier to estimate the value of f(2).

A. -2.5

B. -1.3

C. 0.2

D. 1.1

#### Solution

Because f(0) = 0, the solution curve must begin at (0, 0). Then sketch the curve carefully following the directions of the segments. It helps to imagine that the segments are showing the currents in a river. Your solution should be like a raft carried along by the currents.

Then find the approximate value of f(2) on your solution curve.

The best choice from among the answers is: D. 1.1.

### Slope Fields and Euler’s Method

Often if you see a slope field problem in the Free Response section of the exam, one part of the problem might be to use Euler’s Method to estimate a value of a solution curve.

While the slope field itself can be used to estimate solutions, Euler’s Method is much more precise and does not rely on the visual representation. Check out this review article for practice using the method: AP Calculus BC Review: Euler’s Method.

## Summary

• Slope fields are visual representations of differential equations of the form dy/dx = f(x, y).
• At each sample point of a slope field, there is a segment having slope equal to the value of dy/dx.
• Any curve that follows the flow suggested by the directions of the segments is a solution to the differential equation.