Unit 7 · Topic 7.4 · about 30 minutes · 7 parts · 13 scored checks, 4 of them multiple choice

AP Calculus AB 7.4: Reasoning Using Slope Fields

Q: Why does the second derivative have dy/dx in it?

Because y is a function of x. Differentiating a y term with respect to x uses the chain rule, which brings out dydx. You then replace dydx with the right side of the differential equation, so the answer is in terms of x and y.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 7: 5 to 10% of the multiple-choice section

What you'll be able to do

Estimate solutions to differential equations from a slope field and from the equation itself: trace the solution through a point, tell where it increases, decreases or stays constant, find its concavity, and approximate it with a tangent line.

Key vocabulary

Term	What it means	The exam trap
Solution of a differential equation	A function $y = f (x)$ whose derivative satisfies the equation at every $x$ in an interval.	Test a candidate by substituting it into both sides; a curve that looks right on a picture can still fail.
Particular solution	The one solution of a differential equation that also passes through a given point, such as $f (0) = 2$ .	An initial condition picks one member of a family; keep track of which point you were given.
Slope field	A picture of a differential equation: at each point of a grid, a short segment with the slope the equation gives there.	A solution curve runs along the segments it passes; it does not have to pass through the grid points.
Equilibrium solution	A constant function $y = c$ that solves a differential equation, so its slope field shows a full row of horizontal segments at height $c$ .	Write it as y = c, a function of x, and test it in the equation like any other candidate.
Tangent line approximation	An estimate of $f (a + h)$ from the line through $(a, f (a))$ with slope $f^{'} (a)$ : $f (a + h) \approx f (a) + f^{'} (a) h$ .	Use the slope at the given point, and keep the sign of the step h when you move to the left.

The slope field of dy/dx = x - y for x from -3 to 3 and y from -2 to 3, with three solution curves that follow the segments. A blue curve through (0, 2) falls to a lowest point near (1.1, 1.1) and then rises. A green straight line through (0, -1) is the solution y = x - 1. A red curve through (0, -2.5) rises and bends toward the green line from below.

Part 1 Think first Two cups of coffee, one room

Make a guess before the lesson starts. Committing to an answer, even a wrong one, helps the right one stick.

Predict first

Two cups of coffee sit in a room kept at 70 degrees Fahrenheit. One starts at 180 degrees and the other at 150 degrees. Each cup's temperature $T$ changes at the rate $\frac{d T}{d t} = - 0.1 (T - 70)$ degrees per minute, so the hotter cup is cooling at 11 degrees per minute at the start and the cooler one at 8. Which describes the two temperature graphs over the next hour?

In short. A solution of a differential equation is a function, and the equation usually has a whole family of them; on a slope field each one is a curve that runs along the segments, and one point picks out one curve.

The sign of $\frac{d y}{d x}$ at a point tells you whether the solution through it is increasing or decreasing, and a constant $y = c$ is a solution only when $\frac{d y}{d x}$ is 0 all along that row.
Differentiate the equation with respect to $x$ , treating $y$ as a function of $x$ , to get $\frac{d^{2} y}{d x^{2}}$ and the concavity of a solution.
The tangent line $y = b + m (x - a)$ at a known point estimates a particular solution nearby, and concavity tells you whether that estimate is too high or too low.

Part 2 Learn it Four short ideas, each checked as you go

4a. A solution is a function, and there is a family of them

A differential equation such as $\frac{d y}{d x} = x - y$ gives a slope at every point. A solution is a function $y = f (x)$ whose derivative obeys that rule at every $x$ . Test a candidate by substituting it.

Try $y = x - 1$ . Its derivative is 1, and $x - y = x - (x - 1) = 1$ . The two sides match for every $x$ , so $y = x - 1$ is a solution. It is the green line in the picture at the top.

It is not the only one. The picture is the slope field of this equation: a short segment at each grid point with slope $x - y$ . Start anywhere and run along the segments you pass, and you trace a solution. Start at $(0, 2)$ and you get the blue curve. Start at $(0, - 2.5)$ and you get the red one. Every point starts a curve, so the solutions form a family of functions. The one through a given point, such as $f (0) = 2$ , is a particular solution.

The members are not vertical shifts of one another. Moving $y = x - 1$ up 3 gives $y = x + 2$ , whose slope is 1 everywhere, but at $(0, 2)$ the equation calls for slope $0 - 2 = - 2$ . When the right side involves $y$ , the slope depends on the height, so a shifted curve gets the wrong slopes.

Quick check · Multiple choice

No calculator. Which of the following functions is a solution of the differential equation $\frac{d y}{d x} = x - y$ ?

4b. Rising, falling and constant solutions

You can describe a solution without any formula for it. Wherever $\frac{d y}{d x} > 0$ , the solution through that point is increasing, and wherever $\frac{d y}{d x} < 0$ , it is decreasing. So find the sign of the right side of the equation at the point.

Take $\frac{d y}{d x} = 6 - 2 y$ . Below the line $y = 3$ the right side is positive, so every solution there rises. Above it the right side is negative, so every solution there falls. On the row $y = 3$ itself, $6 - 2 (3) = 0$ for every $x$ .

That makes the constant function $y = 3$ a solution: its derivative is 0, and the equation gives 0 at every point of it. A constant solution is called an equilibrium solution. On a slope field it is a whole row of horizontal segments.

A few horizontal segments are not enough. For $\frac{d y}{d x} = x - y$ , every segment along the line $y = x$ is horizontal, yet $y = x$ is not a solution: its slope is 1, and the equation gives 0 there. The blue curve only crosses that line, at its lowest point. Test a constant $y = c$ by asking whether the right side is 0 at every point of that row, whatever $x$ is.

Quick check · Sort it

No calculator. For the differential equation $\frac{d y}{d x} = (y - 1) (y - 4)$ , tap each card, then tap what the solution through that point is doing there.

Increasing there

Decreasing there

Constant, an equilibrium solution

4c. Concavity straight from the equation

The second derivative of a solution comes from the differential equation itself. Differentiate both sides with respect to $x$ , remembering that $y$ is a function of $x$ , and then put the equation back in wherever $\frac{d y}{d x}$ appears.

For $\frac{d y}{d x} = x - y$ :

\frac{d^{2} y}{d x^{2}} = 1 - \frac{d y}{d x} = 1 - (x - y) = 1 - x + y .

At $(0, 2)$ this is $1 - 0 + 2 = 3 > 0$ , so the blue solution is concave up there, as the picture shows. Below the green line, where $y < x - 1$ , the expression $1 - x + y$ is negative, so solutions there are concave down. The red curve bends that way.

A term with $y$ in it needs the chain rule or the product rule:

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}, \frac{d}{d x} (x y) = y + x \frac{d y}{d x} .

Treating $y$ as a constant drops those $\frac{d y}{d x}$ factors, and the sign you get can be wrong.

The reading is the one from Unit 5: $\frac{d^{2} y}{d x^{2}} > 0$ means the solution is concave up there, and $\frac{d^{2} y}{d x^{2}} < 0$ means concave down.

Quick check · Calculate

No calculator. Let $y = f (x)$ be the solution of $\frac{d y}{d x} = x + y^{2}$ with $f (1) = - 1$ . What is the value of $\frac{d^{2} y}{d x^{2}}$ at the point $(1, - 1)$ ?

Your answer

4d. A tangent line estimates a particular solution

A particular solution $y = f (x)$ through the point $(a, b)$ has slope $m$ there, the value of $\frac{d y}{d x}$ at $(a, b)$ . Its tangent line is

y = b + m (x - a) .

Near $x = a$ the line stays close to the curve, so $f (a + h) \approx b + m h$ . The step $h$ can be negative.

For $\frac{d y}{d x} = x - y$ with $f (0) = 2$ , the slope is $m = 0 - 2 = - 2$ , the tangent line is $y = 2 - 2 x$ , and $f (0.1) \approx 2 - 2 (0.1) = 1.8$ .

Is 1.8 too high or too low? Concavity decides. A solution that is concave up near the point lies above its tangent line there, so the line gives an underestimate. Concave down gives an overestimate. Here $\frac{d^{2} y}{d x^{2}} = 1 - x + y$ , which is 3 at $(0, 2)$ and stays positive for $0 \leq x \leq 0.1$ , where $y$ is near 2. So 1.8 is an underestimate.

The sign of the slope does not decide this. A falling curve can sit above its tangent line or below it, and so can a rising one.

Quick check · Calculate

No calculator. Let $y = f (x)$ be the particular solution of $\frac{d y}{d x} = x^{2} - 2 y$ with $f (1) = 3$ . Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.2)$ .

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Moving a known solution up or down to get another one.	When the right side of the equation involves $y$ , the slope depends on the height, so a shifted curve has the wrong slopes at its new height.	Substitute the candidate into both sides of the equation, or trace the new solution along the slope field from its own point.
Calling any horizontal segment, or any line of them, an equilibrium solution.	A constant $y = c$ is a solution only if $\frac{d y}{d x} = 0$ at every point of the row $y = c$ . Level segments along $y = x$ , or down a column such as $x = 1$ , do not make a solution.	Set the right side equal to 0 and keep only the constant values of $y$ that make it 0 for every $x$ .
Treating $y$ as a constant when finding $\frac{d^{2} y}{d x^{2}}$ .	$y$ changes with $x$ , so every term with $y$ in it has its own derivative, carrying a factor of $\frac{d y}{d x}$ .	Use the chain rule or the product rule on each $y$ term, then replace $\frac{d y}{d x}$ with the right side of the equation.

Part 3 Watch it done Estimate a solution, then judge the estimate

Watch each step, then finish the last one yourself.

Let $y = f (x)$ be the particular solution of $\frac{d y}{d x} = 2 x - y$ with $f (0) = 3$ . Use the line tangent to the graph of $f$ at $x = 0$ to approximate $f (0.2)$ , and decide whether the approximation is an overestimate or an underestimate.

Predict first

Before the steps open: will the tangent line come out too high or too low?

Step 1 · Find the slope at the point
At $(0, 3)$ , $\frac{d y}{d x} = 2 (0) - 3 = - 3$ .
Step 2 · Write the tangent line and use it
The tangent line is $y = 3 - 3 (x - 0) = 3 - 3 x$ , so $f (0.2) \approx 3 - 3 (0.2) = 2.4$ .
Step 3 · Find the second derivative from the equation
$\frac{d^{2} y}{d x^{2}} = 2 - \frac{d y}{d x} = 2 - (2 x - y) = 2 - 2 x + y$ . At $(0, 3)$ it is $2 - 0 + 3 = 5$ . For $0 \leq x \leq 0.2$ the solution stays near 3, so $2 - 2 x + y$ stays positive.
Why this step?
Why look at the second derivative instead of the slope, which is negative?
The slope says which way the solution moves, not how it bends. Whether a curve sits above or below its tangent line depends on its concavity, and the second derivative is what measures that.
Your step · Decide and justify
No calculator. Finish it. For the particular solution of $\frac{d y}{d x} = 2 x - y$ with $f (0) = 3$ , the tangent line at $x = 0$ gives $f (0.2) \approx 2.4$ , and $\frac{d^{2} y}{d x^{2}} = 2 - 2 x + y$ . Is 2.4 an overestimate or an underestimate of $f (0.2)$ ?
Which reason supports your answer?

Part 4 Try it Sketch a solution from the equation alone

Try it · Sketch the graph

No calculator. Let $y = f (x)$ be the solution of $\frac{d y}{d x} = - \frac{y}{2}$ with $f (0) = 8$ . Sketch a possible graph of $f$ for $x$ from 0 to 2. The first point is fixed at 8.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

Part 5 Explain it Write it as a free-response answer

Free response on the exam is written by hand, so work this on paper, including the sketch in part (a). Nothing you type here leaves the page. When you are done, open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. Consider the differential equation $\frac{d y}{d x} = x (y - 1)$ .

(a) On paper, draw axes and sketch the slope field for the given differential equation at the nine points with $x = - 1, 0, 1$ and $y = 0, 1, 2$ . Then sketch the solution curve that passes through the point $(0, 2)$ .

(b) Let $y = f (x)$ be the particular solution to the differential equation with the initial condition $f (1) = 3$ . Write an equation for the line tangent to the graph of $f$ at $x = 1$ , and use it to approximate $f (1.2)$ .

(c) Find $\frac{d^{2} y}{d x^{2}}$ in terms of $x$ and $y$ . Is the approximation in part (b) an overestimate or an underestimate of $f (1.2)$ ? Justify your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) The slopes are $x (y - 1)$ : 0 at all three points with $x = 0$ and all three with $y = 1$ ; 1 at $(1, 2)$ and $(- 1, 0)$ ; and $- 1$ at $(- 1, 2)$ and $(1, 0)$ . The solution through $(0, 2)$ decreases to a minimum at $(0, 2)$ , where it is level, then increases, staying above the row $y = 1$ , as in the figure.

(b) At $(1, 3)$ , $\frac{d y}{d x} = 1 (3 - 1) = 2$ . The tangent line is $y = 3 + 2 (x - 1)$ , so $f (1.2) \approx 3 + 2 (0.2) = 3.4$ .

(c) $\frac{d^{2} y}{d x^{2}} = (y - 1) + x \frac{d y}{d x} = (y - 1) + x^{2} (y - 1) = (y - 1) (1 + x^{2})$ . At $(1, 3)$ it is $(2) (2) = 4 > 0$ . For $1 \leq x \leq 1.2$ , $\frac{d y}{d x} = x (y - 1) > 0$ , so $f$ increases from 3 and $y > 1$ , which keeps $\frac{d^{2} y}{d x^{2}} > 0$ . The graph of $f$ is concave up there and lies above its tangent line, so 3.4 is an underestimate.

Slope segments at the nine points with x = -1, 0, 1 and y = 0, 1, 2 for dy/dx = x(y - 1). The segments at y = 1 and at x = 0 are horizontal, those at (1, 2) and (-1, 0) have slope 1, and those at (-1, 2) and (1, 0) have slope -1. A solution curve comes down from about 2.65 at x = -1, decreases to a minimum at (0, 2), where it is level, then increases to about 2.65 at x = 1, staying above y = 1. — Part (a): the nine segments and the solution through (0, 2).

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

In (a), draws zero slopes at the points with $x = 0$ and with $y = 1$ , positive slopes at $(1, 2)$ and $(- 1, 0)$ , and negative slopes at $(- 1, 2)$ and $(1, 0)$ .
In (a), draws a solution curve through $(0, 2)$ that follows the segments, decreases to a minimum at $(0, 2)$ , where it is level, then increases, staying above $y = 1$ .
In (b), finds the slope 2 at $(1, 3)$ , writes the tangent line $y = 3 + 2 (x - 1)$ , and gets $f (1.2) \approx 3.4$ .
In (c), uses the product rule with $y$ as a function of $x$ and gets $\frac{d^{2} y}{d x^{2}} = (y - 1) (1 + x^{2})$ , in terms of $x$ and $y$ only.
In (c), answers underestimate, because $\frac{d^{2} y}{d x^{2}} > 0$ on the interval, so $f$ is concave up and lies above its tangent line.

Part 6 Check yourself Reason from a slope field

The first three questions use this slope field and its equation. Read each slope from the equation; the picture shows the pattern.

The figure shows the slope field for the differential equation $\frac{d y}{d x} = (x + 1) (y - 2)$ .

The slope field of dy/dx = (x + 1)(y - 2) for x from -3 to 2 and y from -1 to 4. Every segment on the row y = 2 and on the column x = -1 is horizontal. Above y = 2 the segments slope down to the left of x = -1 and up to its right; below y = 2 they slope up to the left of x = -1 and down to its right, getting steeper farther from those lines. — The slope field of dy/dx = (x + 1)(y - 2).

Case question 1 · Multiple choice

No calculator. Which statement best describes the solution curve through the point $(- 1, 3)$ at $x = - 1$ ?

Case question 2 · Calculate

No calculator. Let $y = g (x)$ be the particular solution of $\frac{d y}{d x} = (x + 1) (y - 2)$ with $g (0) = 4$ . Use the line tangent to the graph of $g$ at $x = 0$ to approximate $g (- 0.5)$ .

Your answer

Case question 3 · Decide and justify

No calculator. For the same particular solution $g$ with $g (0) = 4$ , $\frac{d^{2} y}{d x^{2}} = (y - 2) + (x + 1) \frac{d y}{d x}$ , which is $(y - 2) (1 + (x + 1)^{2})$ . A classmate says the approximation $g (- 0.5) \approx 3$ from the tangent line is an overestimate. Is the classmate right?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the concavity of the solution of $\frac{d y}{d x} = y - x^{2}$ through the point $(1, 4)$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. Tap a differential equation, then its constant solutions.

Part 7 Keep it Bring back earlier ideas

Reasoning with a slope field leans on drawing one and on reading concavity from a second derivative. Pulling both back up is what keeps them.

Bring it back · Multiple choice

No calculator. In the slope field for $\frac{d y}{d x} = y - 2 x$ , along which line are all of the segments horizontal?

Bring it back · Multiple choice

No calculator. A function $g$ is twice differentiable for all $x$ , and $g^{''} (x) = (x - 1) (x + 3)^{2}$ . At which values of $x$ does the graph of $g$ have a point of inflection?

In three lines

A solution of a differential equation is a function, and a slope field shows the whole family; one point picks one curve.
The sign of dy/dx says rising or falling, a level row is an equilibrium solution, and the second derivative found from the equation gives concavity.
The tangent line at a known point estimates a particular solution, an underestimate where the solution is concave up and an overestimate where it is concave down.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a slope field question, draw each segment with the right sign and roughly the right steepness, and make zero slopes clearly flat. When you sketch a solution through a point, let it follow the segments and never cross an equilibrium row. For an overestimate or underestimate question, write $\frac{d^{2} y}{d x^{2}}$ from the equation, give its sign near the point, and say what that sign means for the curve and its tangent line.

Questions students ask

Can two solution curves of the same differential equation cross?

For the equations in this course, no. Where two solutions met they would share a point, and the equation gives one slope there and one path onward, so they would be the same curve. That is also why a solution starting above an equilibrium row stays above it.

Does a solution curve have to pass through the points where segments are drawn?

No. The segments are samples. A solution curve runs between them and follows the pattern the nearby segments show, the way a river follows the current.

How do I find the equilibrium solutions?

Set the right side of the equation equal to 0 and look for constant values of $y$ that make it 0 for every $x$ . For $\frac{d y}{d x} = (y - 1) (y - 4)$ those are $y = 1$ and $y = 4$ . A zero that depends on $x$ alone, such as $x = 1$ , is not one.

Why does the second derivative have dy/dx in it?

Because $y$ is a function of $x$ . Differentiating a $y$ term with respect to $x$ uses the chain rule, which brings out $\frac{d y}{d x}$ . You then replace $\frac{d y}{d x}$ with the right side of the differential equation, so the answer is in terms of $x$ and $y$ .

Is it enough to check the second derivative only at the starting point?

It tells you the concavity at that point. A full justification says the second derivative keeps its sign over the interval you step across, which usually follows from what the solution does there, as in the model answer above.

Course alignment, for teachers

AP Calculus AB topic 7.4, Unit 7: Differential Equations.

Skill 4.D: Use appropriate graphing techniques.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first Two cups of coffee, one room

Part 2 Learn it Four short ideas, each checked as you go

4a. A solution is a function, and there is a family of them

4b. Rising, falling and constant solutions

4c. Concavity straight from the equation

4d. A tangent line estimates a particular solution

Part 3 Watch it done Estimate a solution, then judge the estimate

Part 4 Try it Sketch a solution from the equation alone

Part 5 Explain it Write it as a free-response answer

Part 6 Check yourself Reason from a slope field

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic