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Differential Equations & Slope Fields AP Calculus AB FRQ: Explained with Practice & Scoring Tips

You will typically be asked to sketch a slope field, write an equation for a tangent line to approximate a value, and use separation of variables to find a specific solution y = f(x) given an initial condition.

Slope Field (Differential Equation) $\frac{dy}{dx} = f(x, y)$
When to useAt each given point (x, y), calculate the numerical value of the derivative and draw a short line segment with that slope.
Separation of Variables $\int \frac{1}{g(y)} dy = \int f(x) dx$
When to useUsed to solve $\frac{dy}{dx} = f(x)g(y)$ . You MUST move all y terms to the dy side and all x terms to the dx side before integrating.
Tangent Line Approximation $y - y_1 = m(x - x_1)$
When to useUse the given point $(x_1, y_1)$ and calculate $m = \frac{dy}{dx}|_{(x_1, y_1)}$ to find the linear approximation $L(x)$ .
Particular Solution $y = f(x)$
When to useAfter integrating, use the initial condition $(x_0, y_0)$ to solve for the constant of integration $C$ .

Step-by-Step SOP

1
1. Sketching the Slope Field
Plug the coordinates of each indicated dot into the differential equation. Draw segments. Remember: slope 0 is horizontal; undefined slope (like 1/0) means no segment or a vertical dash (though usually left blank in AP).
2
2. Linear Approximation
Identify the point of tangency. Compute the slope at that point. Write the tangent line equation and plug in the 'nearby' x-value to estimate y.
3
3. Separate, Integrate, and Solve (The Big Step)
Move the y-terms to the left (by division/multiplication) and dx to the right. Integrate both sides. **Immediately** add '+ C' to the side with x.
4
4. Solve for C and Find the Domain
Plug in the initial condition $(x_0, y_0)$ to find the value of C. Then, isolate y to get the final function. Check if the solution is valid for the interval containing $x_0$ .

Practice Questions

1 2

Question 1

2024 AP Calculus AB Exam — FRQ 3

Medium

The depth of seawater at a location can be modeled by the function

H

that satisfies the differential equation

\frac{dH}{dt} = \frac{1}{2}(H-1)\cos(\frac{t}{2})

, where

H(t)

is measured in feet and

t

is measured in hours after noon (

t=0

). It is known that

H(0) = 4

Slope field for dH/dt = 1/2(H-1)cos(t/2) — Slope field for the differential equation modeling seawater depth H(t).

APart A

Easy

A portion of the slope field for the differential equation is provided. Sketch the solution curve,

y = H(t)

, through the point (0, 4).

Need a Hint?

Start at the point (0, 4) and follow the flow of the slope segments. Ensure the curve is smooth and extends across the requested interval.

Show Solution

Plot the Initial Condition

Identify and mark the point

(0, 4)

on the y-axis.

Sketch the Curve

Draw a curve passing through

(0, 4)

that remains tangent to the nearby slope lines. The curve should increase, reach a peak near

t = \pi

, and then decrease.

Graph of solution of differential equation

AP Scoring Note — 1 Point

Earned for a solution curve that passes through

(0, 4)

, extends to at least

t = 4.5

, and follows the general direction of the slope segments without obvious conflicts.

BPart B

Medium

For

0 < t < 5

, it can be shown that

H(t) > 1

. Find the value of

t

, for

0 < t < 5

, at which

H

has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither. Justify your answer.

Need a Hint?

Critical points occur where

\frac{dH}{dt} = 0

. Since

H(t) > 1

, focus on where

\cos(\frac{t}{2}) = 0

. Use the first derivative test to justify the relative extremum.

Show Solution

Find the Critical Point

Set

\frac{dH}{dt} = 0

. Since

H-1 \neq 0

(given

H>1

), we solve

\cos(\frac{t}{2}) = 0

. In the interval

0 < t < 5

, this occurs when

\frac{t}{2} = \frac{\pi}{2}

, so

t = \pi

Perform the First Derivative Test

For

0 < t < \pi

\cos(\frac{t}{2}) > 0

, so

\frac{dH}{dt} > 0

(H is increasing). For

\pi < t < 5

\cos(\frac{t}{2}) < 0

, so

\frac{dH}{dt} < 0

(H is decreasing).

Conclusion and Justification

Since

\frac{dH}{dt}

changes from positive to negative at

t = \pi

H

has a relative maximum at

t = \pi

AP Scoring Note — 3 Points

P1 (Sign Analysis): Considering the sign of

dH/dt

\cos(t/2)

.

P2 (Identify t): Correctly finding

t = \pi

.

P3 (Justification): Providing a correct justification and conclusion of 'relative maximum'.

CPart C

Hard

Use separation of variables to find

y = H(t)

, the particular solution to the differential equation

\frac{dH}{dt} = \frac{1}{2}(H-1)\cos(\frac{t}{2})

with initial condition

H(0) = 4

Need a Hint?

Separate

H

terms to the left and

t

terms to the right. Integrate both sides, add

+C

, and solve for

C

using

(0, 4)

before isolating

H

Show Solution

Separate Variables

\frac{1}{H-1} \, dH = \frac{1}{2}\cos(\frac{t}{2}) \, dt

Integrate Both Sides

\int \frac{1}{H-1} \, dH = \int \frac{1}{2}\cos(\frac{t}{2}) \, dt \implies \ln|H-1| = \sin(\frac{t}{2}) + C

Solve for Constant C

Using

H(0) = 4

\ln|4-1| = \sin(0) + C \implies \ln(3) = 0 + C \implies C = \ln(3)

Solve for H(t)

\ln(H-1) = \sin(\frac{t}{2}) + \ln(3) \implies H-1 = e^{\sin(t/2) + \ln(3)} \implies H-1 = 3e^{\sin(t/2)}

Final Expression

H(t) = 1 + 3e^{\sin(t/2)}

AP Scoring Note — 5 Points

P1 (Separation): 1 pt for correct separation.

P2 & P3 (Antiderivatives): 1 pt for each correct side.

P4 (C and Initial Condition): 1 pt for

+C

and substituting

(0, 4)

.

P5 (Answer): 1 pt for the final function. **Note: No separation = 0/5.**

Question 2

2023 AP Calculus AB Exam — FRQ 3

Medium

A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function

M

models the temperature of the milk at time

t

, where

M(t)

is measured in degrees Celsius (

^{\circ}C

) and

t

is the number of minutes since the bottle was placed in the pan.

M

satisfies the differential equation

\frac{dM}{dt} = \frac{1}{4}(40-M)

. At time

t = 0

, the temperature of the milk is

5^{\circ}C

. It can be shown that

M(t) < 40

for all values of

t

Slope field for dM/dt = 1/4(40-M) — Slope field for the differential equation modeling milk temperature M(t).

APart A

Easy

A slope field for the differential equation is shown. Sketch the solution curve through the point (0, 5).

Need a Hint?

Start at (0, 5). Since dM/dt is positive but decreasing as M approaches 40, the curve should be concave down and level off at the horizontal asymptote M = 40.

Show Solution

Initial Point

Identify the point

(0, 5)

on the vertical axis.

Sketching the Curve

Draw a smooth curve that follows the slope segments, increasing from

(0, 5)

and asymptotically approaching the dashed line at

M = 40

Graph of solution of dM/dt = 1/4(40-M) with initial condition M(0)=5

AP Scoring Note — 1 Point

The curve must pass through

(0, 5)

, follow the slope lines, and stay entirely below the

M = 40

line.

BPart B

Easy

Use the line tangent to the graph of

M

t = 0

to approximate

M(2)

, the temperature of the milk at time

t = 2

minutes.

Need a Hint?

Find the slope at t=0 by plugging M=5 into the differential equation. Then use the point-slope form.

Show Solution

Calculate the Slope

t=0

M=5

\frac{dM}{dt}|_{t=0} = \frac{1}{4}(40-5) = \frac{35}{4}

Tangent Line Equation

y - 5 = \frac{35}{4}(t - 0) \implies y = 5 + \frac{35}{4}t

Approximate M(2)

M(2) \approx 5 + \frac{35}{4}(2) = 5 + \frac{35}{2} = 22.5^{\circ}C

AP Scoring Note — 2 Points

1 pt for the correct slope (

35/4

) and 1 pt for the final approximation (

22.5

CPart C

Medium

Write an expression for

\frac{d^2M}{dt^2}

in terms of

M

. Use it to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of

M(2)

. Give a reason for your answer.

Need a Hint?

Differentiate the first derivative with respect to t. Remember that dM/dt itself is 1/4(40-M).

Show Solution

Find the Second Derivative

\frac{d^2M}{dt^2} = \frac{1}{4}(-\frac{dM}{dt}) = -\frac{1}{4}(\frac{1}{4}(40-M)) = -\frac{1}{16}(40-M)

Determine Concavity

Since

M(t) < 40

(40-M)

is always positive. Therefore,

\frac{d^2M}{dt^2} = -\frac{1}{16}(\text{positive}) < 0

Conclusion

Because

\frac{d^2M}{dt^2} < 0

, the graph of

M

is concave down. A tangent line approximation for a concave down graph is an **overestimate**.

AP Scoring Note — 2 Points

1 pt for the correct expression of

\frac{d^2M}{dt^2}

and 1 pt for the correct conclusion with reason (concavity).

DPart D

Hard

Use separation of variables to find an expression for

M(t)

, the particular solution to the differential equation with initial condition

M(0) = 5

Need a Hint?

This is the 'Big Step'. Move (40-M) to the dM side and dt to the other. Watch the negative sign during integration!

Show Solution

Separate Variables

\frac{1}{40-M} \, dM = \frac{1}{4} \, dt

Integrate

-\ln|40-M| = \frac{1}{4}t + C

Use Initial Condition (0, 5)

-\ln|40-5| = \frac{1}{4}(0) + C \implies C = -\ln(35)

Solve for M

\ln(40-M) = -\frac{1}{4}t + \ln(35) \implies 40-M = e^{-t/4 + \ln(35)} \implies 40-M = 35e^{-t/4}

Final Particular Solution

M(t) = 40 - 35e^{-t/4}

AP Scoring Note — 4 Points

1 pt: Separation; 1 pt: Antiderivatives; 1 pt: Constant

C

and initial condition; 1 pt: Solving for

M

. No separation = 0/4.

Timed Practice

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Take a 20-minute timed mock exam to test your understanding.

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Common Pitfalls

⚠
The '+ C' CatastropheIf you forget '+ C' during integration, you lose almost all points for that part (often 0/5 or 0/6). You must add it the moment you integrate.
⚠
Failure to Separate VariablesIf you don't separate variables (e.g., trying to integrate $\frac{dy}{dx} = xy$ as $y = \int xy dx$ ), you get 0 points for the entire section. Separation is the first priority.
⚠
The Absolute Value in ln|y|When $\int \frac{1}{y} dy = \ln|y|$ , don't forget the absolute value. When solving for y, use the initial condition to determine if the expression inside is positive or negative (i.e., whether $y = e^{...}$ or $y = -e^{...}$ ).
⚠
Algebraic Errors in Isolating yBe careful with exponentiation. $e^{f(x) + C}$ becomes $C \cdot e^{f(x)}$ (where the new C is $e^{\text{old C}}$ ), not $e^{f(x)} + e^C$ .
⚠
Slope Field PrecisionEnsure that slopes of 0 are perfectly horizontal and that positive/negative slopes are clearly distinguished. Comparison matters: a slope of 2 should look steeper than a slope of 1.

On this page

1. Sketching the Slope Field

2. Linear Approximation

3. Separate, Integrate, and Solve (The Big Step)

4. Solve for C and Find the Domain

Practice Questions

Plot the Initial Condition

Sketch the Curve

AP Scoring Note — 1 Point

Find the Critical Point

Perform the First Derivative Test

Conclusion and Justification

AP Scoring Note — 3 Points

Separate Variables

Integrate Both Sides

Solve for Constant C

Solve for H(t)

Final Expression

AP Scoring Note — 5 Points

Initial Point

Sketching the Curve

AP Scoring Note — 1 Point

Calculate the Slope

Tangent Line Equation

Approximate M(2)

AP Scoring Note — 2 Points

Find the Second Derivative

Determine Concavity

Conclusion

AP Scoring Note — 2 Points

Separate Variables

Integrate

Use Initial Condition (0, 5)

Solve for M

Final Particular Solution

AP Scoring Note — 4 Points

Ready to challenge yourself?