7.4 - Reasoning Using Slope Fields

Sketching Particular Solutions from Slope Fields

A particular solution is a specific curve that passes through a given point on a slope field. Unlike the general solution, which represents all the possible curves, a particular solution corresponds to one specific value of

C

Either way, a solution to a differential equation is a function. The general solution is a family of functions (one curve for each value of

C

), and on a slope field it shows up as a whole set of curves, each one following the hash marks. A particular solution is one function from that family.

Some solutions are just horizontal lines. If

\frac{dy}{dx} = 0

at every point on a line

y = c

, then

y = c

is itself a solution, called an equilibrium solution. For example,

\frac{dy}{dx} = -0.5y

0

all along

y = 0

, so

y = 0

is a solution.

Drawing Solution Curves

To sketch a particular solution through a point, follow the “current” created by the slope field hash marks. Think of dropping a leaf in a river. The hash marks show which way the water flows, and they guide the leaf along a path.

Example: Consider the slope field for

\frac{dy}{dx} = 2 - xy

Point A: Sketch the particular solution through

(0, 1)

Starting at

(0, 1)

and following the hash marks: - Moving left, in the negative

x

direction, the curve drops quickly (the slopes there are positive, so going left means going down). It bottoms out a little below

y = -1

, between

x = -2

and

x = -1.5

, where the hash marks level off, then rises slowly toward the

x

-axis, which looks like a horizontal asymptote - Moving right, in the positive

x

direction, the curve rises at first, reaches a maximum where the hash marks go horizontal, then curves downward

Point B: Sketch the particular solution through

(2, -1)

Starting at

(2, -1)

, the hash marks show: - Moving left, the hash marks get very steep, so the curve drops fast and leaves the bottom of the window around

x = 1.5

. It can look like a vertical asymptote, but it isn’t one: the curve just keeps dropping below the window before it turns around - Moving right, the curve rises quickly at first, but the slopes get smaller as it climbs. It crosses the

x

-axis between

x = 2

and

x = 2.5

, levels off at a maximum just above the

x

-axis (around

x = 3.2

), then drifts slowly back down toward it - Near

B

, the curve comes up steeply from the bottom edge, passes through

B

, peaks just above the

x

-axis, and then flattens out

Matching Slope Fields to Solution Equations

Given a slope field and a specific point, you might have to identify which equation is the particular solution passing through that point.

Problem: The slope field below represents a differential equation. Which function is the particular solution passing through

(0, 0)

(A)

y = \frac{x}{x^2 - 9}

(B)

y = \frac{x^2}{x + 3}

(C)

y = \tan(x) - 3

(D)

y = e^{x+3} - 1

Strategy: Use Multiple Tests

Test 1: Does it pass through the given point?

Substitute

(0, 0)

(A): $\frac{0}{0^2 - 9} = \frac{0}{-9} = 0$ ✓

(B): $\frac{0^2}{0 + 3} = \frac{0}{3} = 0$ ✓

(C): $\tan(0) - 3 = 0 - 3 = -3$ ✗

(D): $e^{0+3} - 1 = e^3 - 1 \approx 19.09$ ✗

Eliminate (C) and (D).

Test 2: Identify the vertical asymptotes

The slope field shows a vertical asymptote at

x = -3

, and only there.

(A): the denominator $x^2 - 9 = (x-3)(x+3)$ equals zero when $x = 3$ or $x = -3$ , so this has vertical asymptotes at both $x = 3$ and $x = -3$ ✗

(B): the denominator $x + 3$ equals zero only when $x = -3$ , so this has a vertical asymptote at $x = -3$ only ✓

Test 3: Check the behavior near the asymptote

For (B),

y = \frac{x^2}{x+3}

. The numerator

x^2

is always positive, so the sign depends entirely on

(x + 3)

x \to -3^+

, the denominator approaches

0^+

, so

y \to +\infty

x \to -3^-

, the denominator approaches

0^-

, so

y \to -\infty

That matches the steep hash marks on both sides of

x = -3

in the slope field.

Answer: (B)

y = \frac{x^2}{x + 3}

General Approach

Verify the point. Substitute the coordinates into each option

Check the asymptotes. Vertical asymptotes come from zero denominators, and horizontal asymptotes come from end behavior

Examine special behaviors. Periodic functions like $\tan(x)$ have repeating vertical asymptotes, and exponential functions have no vertical asymptotes

Use graphing technology if you have it, to confirm

Finding Points with Specific Slope Values

A common question asks: “for which points does

\frac{dy}{dx}

equal a specific value?”

Problem: Given the slope field for

\frac{dy}{dx} = \frac{x - 2}{y}

, describe all the points where

\frac{dy}{dx} = 1

Algebraic Solution

Set the differential equation equal to the slope you want:

\frac{x - 2}{y} = 1

Solve for the relationship between

x

and

y

x - 2 = y \quad \Rightarrow \quad y = x - 2

Answer: every point on the line

y = x - 2

has a slope of

1

, except

(2, 0)

. There

y = 0

, so

\frac{x - 2}{y}

is undefined.

Visual Confirmation

On the slope field, find all the hash marks with slope

1

, meaning diagonal lines going up and to the right at 45°. They form a diagonal line pattern along

y = x - 2

Another Example

Problem: For the differential equation

\frac{dy}{dx} = x + y

, find all the points where

\frac{dy}{dx} = 0

x + y = 0 \quad \Rightarrow \quad y = -x

Answer: every point on the line

y = -x

has a derivative of zero, meaning horizontal tangent lines.

Determining if a Slope Field Matches a Differential Equation

Sometimes you have to explain why a given slope field does NOT match a differential equation.

Problem: Explain why the slope field below does NOT represent

\frac{dy}{dx} = -0.5y

Analysis Method

For

\frac{dy}{dx} = -0.5y

, notice that the derivative depends only on $y$ , not on

x

Key principle: when

\frac{dy}{dx}

depends only on

y

, every point with the same

y

-coordinate has to have an identical slope.

Test this property. Along the horizontal line

y = 2

: - The derivative should be

-0.5(2) = -1

everywhere - So every hash mark at

y = 2

should have the same slope of

-1

Looking at the slope field: the hash marks at

y = 2

have different slopes at different

x

-values, which violates the requirement.

Alternative Explanation

Test a specific point. When

y = 0

, the differential equation gives:

\frac{dy}{dx} = -0.5(0) = 0

So every point along the

x

-axis, where

y = 0

, has to have slope

0

, meaning horizontal hash marks.

Looking at the slope field: the hash marks along the

x

-axis are not horizontal, since they have nonzero slopes.

Conclusion: the slope field doesn’t match

\frac{dy}{dx} = -0.5y

, because the slopes vary along horizontal lines, and specifically, the slopes at

y = 0

are nonzero when they should be zero.

Summary of Reasoning Techniques

For Sketching Curves

Start at the given point

Follow the flow of the hash marks like a river current

Look for patterns. Where do the curves flatten (horizontal marks)? Where are the asymptotes (steep or missing marks)?

For Matching Equations

Eliminate options that don’t pass through the given point

Identify and count the vertical asymptotes

Check for periodic vs. non-periodic behavior

Consider end behavior and horizontal asymptotes

For Describing Points with Specific Slopes

Set $\frac{dy}{dx}$ equal to the value you want

Solve algebraically for the relationship between $x$ and $y$

The result is usually a curve or a line

For Verifying Matches

If $\frac{dy}{dx}$ depends only on $x$ , the slopes are constant along vertical lines

If $\frac{dy}{dx}$ depends only on $y$ , the slopes are constant along horizontal lines

Check the special cases, like where the derivative should be zero

Key Takeaways

Particular solution: a specific curve through a given point, which you sketch by following the slope field hash marks

Families of functions: the general solution is a family of functions (one for each $C$ ). A horizontal line $y = c$ where $\frac{dy}{dx} = 0$ all along it is an equilibrium solution

Matching solutions to slope fields: use point verification, asymptote identification, and behavior analysis

Finding points with specific slopes: set $\frac{dy}{dx}$ equal to the value and solve algebraically

Constant slopes along lines: if $\frac{dy}{dx} = f(y)$ only, the slopes are identical along horizontal lines, and if $\frac{dy}{dx} = f(x)$ only, they’re identical along vertical lines

Asymptote detection: vertical asymptotes appear where the hash marks are extremely steep or missing, so check the denominators of the candidate functions

Special value verification: test where $\frac{dy}{dx} = 0$ (horizontal marks) or is undefined (vertical marks or no marks)

Elimination strategy: use multiple tests to systematically knock out the wrong answer choices

Visual plus algebraic: combine slope field observation with equation manipulation for a complete understanding

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