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# Slope Fields for AP Calculus

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Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

A slope field (or a direction field ) for first-order differential equations is a graphic representation of the slopes of a family of curves. It consists of a set of short line segments drawn on a pair of axes. These line segments are the tangents to a family of solution curves for the differential equation at various points. The tangents show the direction in which the solution curves will follow. Slope fields are useful in sketching solution curves without having to solve a differential equation algebraically.

### Example 1

If =0.5x, draw a slope field for the given differential equation.

Step 1: Set up a table of values forfor selected values of x.

Note that since=0.5x, the numerical value of is independent of the value of y. For example, at the points (1, –1), (1, 0), (1, 1), (1, 2), (1, 3), and at all the points whose x -coordinates are 1, the numerical value ofis 0.5 regardless of their y-coordinates. Similarly, for all the points, whose x-coordinates are 2 (e.g., (2,–1), (2, 0), (2, 3), etc. ), =1. Also, remember that represents the slopes of the tangent lines to the curve at various points. You are now ready to draw these tangents.

Step 2: Draw short line segments with the given slopes at the various points. The slope field for the differential equation =0.5x is shown in Figure 13.5-1.

### Example 2

Figure 13.5-2 shows a slope field for one of the differential equations given below. Identify the equation.

1. =2x
2. − 2x
3. = y
4. = − y
5. =x + y

### Solution:

If you look across horizontally at any row of tangents, you'll notice that the tangents have the same slope. (Points on the same row have the same y-coordinate but different x -coordinates.)

Therefore, the numerical value of (which represents the slope of the tangent) depends solely on they -coordinate of a point and it is independent of the x -coordinate. Thus, only choice (c) and choice (d) satisfy this condition. Also notice that the tangents have a negative slope when y > 0 and have a positive slope when y < 0.

Therefore, the correct choice is (c) = − y.

### Example 3

A slope field for a differential equation is shown in Figure 13.5-3. Draw a possible graph for the particular solution y = f (x) to the differential equation function, if (a) the initial condition is f (0)= − 2 and (b) the initial condition is f (0)=0.

### Solution:

Begin by locating the point (0, –2) as given in the initial condition. Follow the flow of the field and sketch the graph of the function. Repeat the same procedure with the point (0, 0). See the curves as shown in Figure 13.5-4.

### Example 4

Given the differential equation = − x y.

1. Draw a slope field for the differential equation at the 15 points indicated on the provided set of axes in Figure 13.5-5.
2. Sketch a possible graph for the particular solution y = f (x) to the differential equation with the initial condition f (0)=3.
3. Find, algebraically, the particular solution y = f (x)to the differential equation with the initial condition f (0)=3.

### Solution:

1. Set up a table of values for at the 15 given points.
2. Then sketch the tangents at the various points as shown in Figure 13.5-6.

3. Locate the point (0, 3) as indicated in the initial condition. Follow the flow of the field and sketch the curve as shown Figure 13.5-7.
4. Step 1: Rewrite = –xy as = − xdx.
5. Step 2: Integrate both sides xdx and obtain ln
6. Step 3: Apply the exponential function to both sides and obtain
7. Step 4: Simplify the equation and get
Let k =ec and you have
8. Step 5: Substitute initial condition (0, 3) and obtain k =3. Thus you have

Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

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