Differential equations are BC-exclusive territory, and this unit rewards knowing a handful of standard models cold: Newton's Law of Cooling (a quantity approaching a limiting value at a rate proportional to how far away it still is), exponential decay (a quantity shrinking at a rate proportional to itself), and logistic growth (a population that grows fastest at half its carrying capacity and levels off as it approaches that capacity). Alongside these models, the exam tests Euler's method as a numerical way to approximate a solution when no clean closed form exists, and slope fields as a visual way to read a differential equation's behavior directly off a picture.
Euler's method works by repeatedly taking a small step in the direction the differential equation says the solution is currently heading: ynew = yold + (slope at the current point)×Δ x, then moving to the new point and repeating. It's an approximation, not the exact solution, so a natural follow-up question is always how far off that approximation actually is compared to solving the equation exactly. For logistic growth dP/dt = kP(1−P/L), the single fact worth memorizing is that the growth rate is a maximum exactly when P=L/2 — before that point growth is accelerating, after it growth is decelerating even though the population is still increasing. For a slope field, the key reading skill is noticing which variable the slope actually depends on: if the pattern of dashes looks identical across every column (doesn't change with x) but does change from row to row, the differential equation must be a function of y alone, and vice versa.
Below are 11 AP Calculus BC-style practice problems covering Euler's method and its error, Newton's Law of Cooling, exponential decay, logistic growth, reading a slope field, and motion under constant gravitational acceleration. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus BC QBank.
If you use Euler's method with Δx = 0.1 for the d.e. dy/dx = 2x, with initial value y(1)=4, then when x=1.2, y is approximately
Start at (1,4). Step 1: slope =2(1)=2. New y: 4+0.1(2)=4.2. New x: 1.1.
Step 2: at (1.1, 4.2), slope =2(1.1)=2.2. New y: 4.2+0.1(2.2)=4.2+0.22=4.42. New x: 1.2.
After two steps, y(1.2) ≈ 4.42. The correct answer is C.
The error in using Euler's method in Question 1 is
Since dy/dx=2x is separable and simple, solve it exactly: y = x2+C. Using y(1)=4: 4=1+C → C=3, so y=x2+3.
The exact value at x=1.2: y(1.2) = 1.44+3 = 4.44.
Compare to the Euler's method estimate from Question 1 (4.42): error = |4.44−4.42| = 0.02.
The correct answer is B.
The table shows selected values of the derivative for a differentiable function f.
| x | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| f′(x) | 3.0 | 1.5 | 0.5 | −1.0 | −2.0 | 1.0 |
Given that f(3)=80, use Euler's method with a step size of 2 to estimate f(7).
Start at (3, 80). Step 1 (step size 2, using the slope at x=3): f′(3)=1.5. New value: 80+2(1.5) = 83. New x: 5.
Step 2 (using the slope at x=5): f′(5)=−1.0. New value: 83+2(−1.0) = 81. New x: 7.
The estimate for f(7) is 81. The correct answer is C.
Keep the momentum going.
Explore 1,500+ Free AP Calculus BC ProblemsA cup of coffee at temperature 190°F is placed on a table in a room at 70°F. The d.e. for its temperature at time t (in minutes) is dy/dt = −0.15(y−70); y(0)=190. After 8 minutes, the temperature (in °F) of the coffee is approximately
This is Newton's Law of Cooling, with solution y = 70+Ce−0.15t. Using y(0)=190: 190=70+C → C=120, so y=70+120e−0.15t.
Evaluate at t=8: y(8) = 70+120e−1.2.
Compute e−1.2 ≈ 0.3012, so 120(0.3012) ≈ 36.14.
y(8) ≈ 70+36.14 ≈ 106. The correct answer is C.
A cup of coffee at temperature 190°F is placed on a table in a room at 70°F. The d.e. for its temperature at time t (in minutes) is dy/dt = −0.15(y−70); y(0)=190. Approximately how long does it take for the temperature of the coffee to drop to 80°F?
Using the solution from Question 4, y=70+120e−0.15t, set y=80 and solve for t: 80=70+120e−0.15t → 10=120e−0.15t.
e−0.15t = 1/12 → −0.15t = ln(1/12) = −ln 12.
t = ln(12)/0.15. Since ln 12 ≈ 2.485: t ≈ 2.485/0.15 ≈ 16.57.
Rounding, this is approximately 17 minutes. The correct answer is B.
The concentration of a medication injected into the bloodstream drops at a rate proportional to the existing concentration. If the factor of proportionality is 25% per hour, in approximately how many hours will the concentration be one-tenth of the initial concentration?
Since the rate of decrease is proportional to the current concentration, this is exponential decay: C(t) = C0e−0.25t.
Set C(t) = 0.1C0 and solve for t: 0.1 = e−0.25t → ln(0.1) = −0.25t → t = ln(10)/0.25.
Since ln 10 ≈ 2.3026: t ≈ 2.3026/0.25 ≈ 9.21.
This is approximately 9 1/5 hours. The correct answer is C.
6 down, 5 to go — over halfway there.
Explore 1,500+ Free AP Calculus BC ProblemsWhich of the following statements characterize(s) the logistic growth of a population whose limiting value is K, and whose initial value is less than K/2?
Logistic growth follows dP/dt = kP(1−P/K). Since the growth rate itself is a function of P, differentiate it with respect to P: d/dP[kP(1−P/K)] = k(1−2P/K), which is positive when P<K/2.
So while P<K/2, the growth rate is increasing — Statement I is true. This derivative equals zero exactly at P=K/2, which is where the growth rate is maximized — Statement II is true.
As P approaches K, the factor (1−P/K) approaches 0, so dP/dt → 0 — Statement III is true.
The correct answer is D.
Which of the following differential equations is not logistic?
A logistic equation always has the form kP(L−P) (or its expanded version) — a product of the variable and its distance from the limiting value, which is why it's quadratic in the dependent variable.
Choice A factors as 0.02P(100−P) (logistic). Choice B is already given in logistic form. Choice C factors as 0.003x(200−x) (logistic).
Choice D, dR/dt=0.2(250−R), is linear in R, not quadratic — this is Newton's Law of Cooling, not logistic growth (the rate is proportional only to the distance from the limit, not to the variable itself as well).
The correct answer is D.
Suppose P(t) denotes the size of an animal population at time t, and its growth is described by the d.e. dP/dt = 0.003P(800−P). If the initial population is 150, then the population is growing fastest
This is logistic growth with limiting value (carrying capacity) L=800. The growth rate is always maximized at exactly half the carrying capacity, regardless of the initial population.
L/2 = 800/2 = 400.
The correct answer is B. (Choice D describes when the growth rate is still increasing, not when it's at its peak — that happens throughout the entire range P<400, not at a single instant.)
9 down, 2 to go — almost there.
Explore 1,500+ Free AP Calculus BC ProblemsWhich differential equation has the slope field shown?
The dashes look identical across every column of the grid — the pattern doesn't change as x varies. This rules out any equation containing x, eliminating choices A and B.
Between the remaining choices, check the behavior at y=0: choice C gives 4(0)/(4+0)=0 (a flat, horizontal dash), while choice D gives 4/(4+0)=1 (a dash at a 45° angle). The picture shows the steepest dashes exactly along the x-axis (y=0), not flat ones.
Choice D also correctly predicts the dashes flattening out symmetrically as |y| grows: at y=±2, the slope is 4/(4+4)=0.5, and at y=±4, it's 4/(4+16)=0.2 — matching the picture's gradual flattening in both directions away from the x-axis.
The correct answer is D.
Almost done — 1 more.
Explore 1,500+ Free AP Calculus BC ProblemsA stone is thrown straight up from the ground. What should the initial velocity be if you want the stone to reach a height of 144 feet?
Under constant gravitational acceleration g=32 ft/sec2 (downward), velocity satisfies dv/dt=−32, so v(t)=v0−32t, and integrating again, s(t)=v0t−16t2 (starting from the ground, s(0)=0).
The stone reaches its maximum height when v=0: v0−32t=0 → t=v0/32. Substitute into s(t) to find the max height in terms of v0 alone: smax = v0(v0/32)−16(v0/32)2 = v02/32−v02/64 = v02/64.
Set this equal to 144 and solve for v0: v02/64=144 → v02=9216 → v0=√9216.
= 96. The correct answer is D.
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