AP-CALCBC-7-FRQ

U7 FRQ Practice

Master AP Calculus BC Unit 7 free-response questions: slope fields, Euler's method, separation of variables, and logistic models with a step-by-step attack plan.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U7 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

What this lesson covers

Unit 7 free-response questions reward students who can move fluidly between representations of a differential equation — a slope field, a numerical estimate, and an exact solution. The College Board loves multi-part FRQs that begin with a tangent line or slope field sketch, march through an Euler's method estimate, and finish with a full separation-of-variables solution complete with an initial condition. This guide does not re-teach each technique; instead it shows you how to string them together under time pressure, where points are won and lost, and how to phrase answers so a reader awards every point available.

How Unit 7 FRQs Are Structured

A typical Unit 7 free-response prompt gives you a differential equation, usually of the form dydx=f(x,y)\frac{dy}{dx} = f(x,y), and an initial condition such as y(0)=2y(0) = 2. The parts then escalate in difficulty and each one tests a different Unit 7 skill.
PartSkill testedWhat graders want
(a)Slope field / tangent lineCorrect slopes plotted or a tangent-line approximation
(b)Second derivative / concavityd2ydx2\frac{d^2y}{dx^2} found by implicit differentiation
(c)Euler's methodA numbered table of steps with correct step size
(d)Separation of variablesFull solution with +C+C solved from the initial condition
Because the parts are graded independently, a mistake in part (c) will not cost you points in part (d) as long as you restart cleanly. Never skip a part just because an earlier one stumped you. Read the whole question first so you know which technique each part demands, then budget roughly 3–4 minutes per part on the calculator-inactive sections.

Winning the Slope Field and Tangent Line Parts

Many Unit 7 FRQs open by asking you to sketch short segments at given lattice points or to write the equation of the tangent line at the initial point. For the tangent line, plug the initial point into dydx\frac{dy}{dx} to get the slope, then use point-slope form: yy0=f(x0,y0)(xx0)y - y_0 = f(x_0,y_0)(x - x_0).

A frequent follow-up asks you to use that tangent line to approximate a nearby value, then to decide whether the estimate is an overestimate or underestimate. This is where concavity matters. Compute d2ydx2\frac{d^2y}{dx^2} by differentiating dydx\frac{dy}{dx} implicitly. If the second derivative is positive near the point, the curve is concave up and the tangent line lies below it, so the linear approximation is an underestimate.

Common misconception: students confuse the sign of the slope with concavity. The tangent line's accuracy depends on concavity, not on whether the function is increasing. Always justify overestimate/underestimate claims by referencing the sign of the second derivative — a bare answer with no justification typically earns zero on that reasoning point.

Euler's Method and Separation of Variables Under Pressure

When a part says "use Euler's method with two steps," show a table. Each row applies ynew=yold+dydxΔxy_{new} = y_{old} + \frac{dy}{dx}\cdot \Delta x, where dydx\frac{dy}{dx} is evaluated at the current point. Graders look for the step size, the slope at each step, and the running yy-value. Writing the arithmetic explicitly protects you: even one correct step often earns a point.

For separation of variables, the workflow is fixed: get all yy-terms with dydy on one side and all xx-terms with dxdx on the other, integrate both sides, add a single constant CC, then immediately substitute the initial condition to solve for CC before simplifying. A classic error is solving for CC after exponentiating incorrectly. Substitute early.
StepAction
1Separate: dyg(y)=f(x)dx\frac{dy}{g(y)} = f(x)\,dx
2Integrate both sides, add CC
3Apply initial condition to find CC
4Solve explicitly for yy, keep the correct sign/branch
Don't forget to state the domain restriction if the problem asks for it, and always end with y=y = an explicit function of xx when a solution is requested.

Logistic and Exponential Model Parts

BC-specific parts often layer a logistic model onto the FRQ. If a population satisfies dPdt=kP(1PL)\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right), you should instantly recognize LL as the carrying capacity and know limtP(t)=L\lim_{t\to\infty} P(t) = L. The population grows fastest when P=L2P = \frac{L}{2}, a fact graders frequently test with a short-answer part.

For exponential models governed by dydt=ky\frac{dy}{dt} = ky, the solution is y=y0ekty = y_0 e^{kt}; you may be asked to solve for kk from a data point. A common trap: the question asks for the fastest rate of change, which for logistic growth is not at the start but at the inflection point where P=L2P = \frac{L}{2}.

You generally do NOT need to solve the full logistic differential equation on an FRQ — that closed-form solution is not required knowledge. Instead, use qualitative facts: carrying capacity, the half-capacity maximum growth, and limiting behavior. Reading these features off the equation quickly banks easy points before you tackle any heavier computation.

Time Management and Justification Language

The scoring guidelines are keyword-driven. When a prompt says "explain" or "justify," you must connect a numeric fact to a conclusion in a sentence. For instance: "Because d2ydx2>0\frac{d^2y}{dx^2} > 0 on this interval, the graph is concave up, so the tangent-line approximation is an underestimate."

Budget your time so no single part consumes more than its share. If separation of variables gets algebraically messy, leave your antiderivative and constant clearly displayed — partial credit lives there. Circle or box final answers, label units when a context demands them, and never erase work that might contain an earned point. A blank space earns nothing; a partial attempt often earns something.

Finally, keep an eye on whether the section is calculator-active. On calculator-inactive parts, present exact values like ln3\ln 3 rather than decimals, and only convert to a decimal when the problem explicitly asks.

Key terms

Slope field.
A grid of short line segments whose slopes equal dydx\frac{dy}{dx} at each point, giving a visual picture of solution curves without solving the equation.
Euler's method.
A step-by-step numerical approximation using ynew=yold+dydxΔxy_{new} = y_{old} + \frac{dy}{dx}\cdot \Delta x to estimate solution values from an initial condition.
Separation of variables.
A solution technique that rearranges a differential equation so each variable appears with its own differential, then integrates both sides.
Carrying capacity.
The limiting value LL in a logistic model dPdt=kP(1PL)\frac{dP}{dt}=kP(1-\frac{P}{L}) that the population approaches as tt\to\infty.
Concavity test for approximation.
Using the sign of d2ydx2\frac{d^2y}{dx^2} to decide whether a tangent-line estimate is an overestimate (concave down) or underestimate (concave up).
Initial condition.
A given point y(x0)=y0y(x_0)=y_0 used to solve for the constant of integration and pin down a particular solution.
Logistic differential equation.
A model of constrained growth, dPdt=kP(1PL)\frac{dP}{dt}=kP(1-\frac{P}{L}), whose growth is fastest at P=L2P=\frac{L}{2}.

Worked example

Consider the differential equation dydx=x(y1)\frac{dy}{dx} = x(y-1) with initial condition y(0)=3y(0)=3. (a) Write the equation of the tangent line at x=0x=0 and use it to approximate y(0.2)y(0.2). (b) Determine whether the approximation is an over- or underestimate. (c) Find the particular solution y=f(x)y=f(x).
Part (a): At (0,3)(0,3), dydx=0(31)=0\frac{dy}{dx} = 0\cdot(3-1)=0. The tangent line is y=3+0(x0)=3y = 3 + 0(x-0) = 3. So the approximation gives y(0.2)3y(0.2)\approx 3.

Part (b): Differentiate dydx=x(y1)\frac{dy}{dx}=x(y-1) implicitly. Using the product rule, d2ydx2=(y1)+xdydx=(y1)+xx(y1)\frac{d^2y}{dx^2} = (y-1) + x\frac{dy}{dx} = (y-1) + x\cdot x(y-1). At (0,3)(0,3) this is (31)+0=2>0(3-1)+0 = 2 > 0, so the graph is concave up near x=0x=0. A concave-up curve lies above its tangent line, so the estimate is an underestimate.

Part (c): Separate variables: dyy1=xdx\frac{dy}{y-1} = x\,dx. Integrate: lny1=x22+C\ln|y-1| = \frac{x^2}{2} + C. Apply the initial condition y(0)=3y(0)=3: ln31=0+C\ln|3-1| = 0 + C, so C=ln2C = \ln 2. Exponentiate: y1=ex2/2eln2=2ex2/2|y-1| = e^{x^2/2}\cdot e^{\ln 2} = 2e^{x^2/2}. Since y=3>1y=3>1 at the start, y1>0y-1>0, so y=1+2ex2/2y = 1 + 2e^{x^2/2}.

Practice questions

A population PP satisfies dPdt=0.05P(1P800)\frac{dP}{dt}=0.05P\left(1-\frac{P}{800}\right). At what population is the population growing fastest?
  1. P=0P = 0
  2. P=400P = 400
  3. P=800P = 800
  4. P=40P = 40

Answer: P=400P = 400

For a logistic model dPdt=kP(1PL)\frac{dP}{dt}=kP(1-\frac{P}{L}), the growth rate is maximized at the inflection point where P=L2P=\frac{L}{2}. Here L=800L=800, so the fastest growth occurs at P=400P=400. Note P=800P=800 is the carrying capacity where growth slows to zero, a common distractor.
Given dydx=2xy\frac{dy}{dx}=2xy with y(0)=1y(0)=1, use Euler's method with two steps of size Δx=0.5\Delta x = 0.5 to approximate y(1)y(1).

Answer: y(1)1.5y(1)\approx 1.5

Start at (0,1)(0,1). Step 1: slope =2(0)(1)=0=2(0)(1)=0, so y(0.5)1+0(0.5)=1y(0.5)\approx 1 + 0(0.5) = 1. Step 2: at (0.5,1)(0.5,1) slope =2(0.5)(1)=1=2(0.5)(1)=1, so y(1)1+1(0.5)=1.5y(1)\approx 1 + 1(0.5) = 1.5. Showing each slope evaluation and the update ynew=yold+slopeΔxy_{new}=y_{old}+\text{slope}\cdot\Delta x earns the computational points even if arithmetic slips occur.
Solve the differential equation dydx=xy\frac{dy}{dx}=\frac{x}{y} with initial condition y(0)=2y(0)=2, and give the explicit solution.

Answer: y=x2+4y=\sqrt{x^2+4}

Separate: ydy=xdxy\,dy = x\,dx. Integrate: y22=x22+C\frac{y^2}{2}=\frac{x^2}{2}+C. Apply y(0)=2y(0)=2: 42=0+C\frac{4}{2}=0+C, so C=2C=2. Then y2=x2+4y^2=x^2+4. Since y(0)=2>0y(0)=2>0, take the positive root: y=x2+4y=\sqrt{x^2+4}. Choosing the correct branch (positive root) is essential — dropping it loses a point.

FAQ

Do I need to memorize the closed-form solution of the logistic equation for BC FRQs?
No. The AP exam does not require the full logistic solution formula. You should know the qualitative facts: the carrying capacity LL, that limtP=L\lim_{t\to\infty}P=L, and that growth is fastest when P=L2P=\frac{L}{2}. These points are what FRQs actually test.
How much work do I need to show for Euler's method?
Show a step-by-step table or written computation with the step size, the slope evaluated at each current point, and the updated yy-value. Graders award points for the correct process even if a final arithmetic answer is slightly off, so never just write a single number.
How do I decide if a tangent-line approximation is an overestimate or underestimate?
Compute the second derivative near the point of tangency. If d2ydx2>0\frac{d^2y}{dx^2}>0 the curve is concave up and lies above its tangent, making the estimate an underestimate; if d2ydx2<0\frac{d^2y}{dx^2}<0 it is an overestimate. Always justify with the sign of the second derivative.
What is the most common mistake on separation-of-variables FRQ parts?
Forgetting the constant CC or solving for it too late. Add +C+C right after integrating, then substitute the initial condition before exponentiating or simplifying. Also remember to choose the correct sign branch when you take a square root or remove absolute value bars.

Learn this with a teacher, not a page

The Crimsora tutor teaches U7 FRQ Practice live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.