How to Use This AB Free-Response Practice Set
The AP Calculus AB free-response section asks for complete, justified work, not just final answers. This page has three questions in that style: one calculator-active question and two no-calculator questions. Each has four parts, and each part has a solution that you can open after you have written your own answer. Allow about thirty minutes for the three questions if you want to simulate exam pacing, and write full sentences for every justification.
Question 1 allows a graphing calculator. Questions 2 and 3 should be done by hand, exactly as on the exam.
Question 1 (Calculator Active)
A tank initially holds \(50\) gallons of water. For \(0\le t\le10\) minutes, water flows into the tank at the rate \(R(t)=20e^{-0.1t}\) gallons per minute, and water is pumped out at the rate \(L(t)=3+0.6t\) gallons per minute.
(a) How many gallons of water flow into the tank during the first \(10\) minutes? Give your answer to three decimal places.
Show solution to (a)
The total inflow is the integral of the inflow rate: \(\displaystyle\int_0^{10}20e^{-0.1t}\,dt=200\left(1-e^{-1}\right)\approx126.424\) gallons.
Scoring: one point for the integral with correct limits, one point for the value.
(b) Is the amount of water in the tank increasing or decreasing at \(t=5\)? Give a reason.
Show solution to (b)
The rate of change of the amount \(A(t)\) is \(A'(t)=R(t)-L(t)\). At \(t=5\), \(R(5)=20e^{-0.5}\approx12.131\) and \(L(5)=3+3=6\), so \(A'(5)\approx6.131>0\). The amount is increasing at \(t=5\).
Scoring: one point for \(A'(5)\) with the correct difference, one point for the conclusion with the sign as the reason.
(c) How many gallons of water are in the tank at \(t=10\)?
Show solution to (c)
Use the initial amount plus the net change: \(A(10)=50+\int_0^{10}\left(R(t)-L(t)\right)dt\). The outflow integral is \(\int_0^{10}(3+0.6t)\,dt=3(10)+0.3(100)=60\). Therefore \(A(10)\approx50+126.424-60=116.424\) gallons.
Scoring: one point for using the initial amount with the net-change integral, one point for the answer.
(d) At what time \(t\), for \(0\le t\le10\), is the amount of water in the tank greatest? Justify your answer.
Show solution to (d)
The amount can change direction only where \(A'(t)=R(t)-L(t)=0\). The calculator gives \(20e^{-0.1t}=3+0.6t\) at \(t\approx8.811\). Since \(R\) is decreasing and \(L\) is increasing, \(R-L\) is positive for \(t<8.811\) and negative afterward, so \(A\) rises and then falls and the root is a maximum. Compare it with the endpoints: \(A(0)=50\), \(A(8.811)\approx117.412\), and \(A(10)\approx116.424\). The amount is greatest at \(t\approx8.811\) minutes.
Scoring: one point for setting \(A'=0\), one point for the critical time, one point for the justification using the sign change or the candidate comparison.
Question 2 (No Calculator)
Let \(f(x)=xe^{-x}\) for \(x\ge0\). Answer the following without a calculator.
(a) Find \(f'(x)\), and find the \(x\)-coordinate of the relative maximum of \(f\). Justify that it is a relative maximum.
Show solution to (a)
By the product rule, \(f'(x)=e^{-x}-xe^{-x}=(1-x)e^{-x}\). Since \(e^{-x}>0\), the derivative is zero only at \(x=1\). For \(0\le x<1\), \(f'(x)>0\), and for \(x>1\), \(f'(x)<0\). The sign change from positive to negative shows that \(f\) has a relative maximum at \(x=1\), where \(f(1)=\tfrac1e\).
(b) Find \(f''(x)\), and find the \(x\)-coordinate of the point of inflection of the graph of \(f\). Justify your answer.
Show solution to (b)
Differentiate \(f'(x)=(1-x)e^{-x}\): \(f''(x)=-e^{-x}-(1-x)e^{-x}=(x-2)e^{-x}\). It equals zero at \(x=2\). For \(x<2\), \(f''<0\) (concave down), and for \(x>2\), \(f''>0\) (concave up). Because the concavity changes, the graph has a point of inflection at \(x=2\).
(c) Write an equation for the line tangent to the graph of \(f\) at \(x=0\). Use it to approximate \(f(0.1)\), and state whether the approximation is an overestimate or an underestimate. Give a reason.
Show solution to (c)
Here \(f(0)=0\) and \(f'(0)=1\), so the tangent line is \(y=x\). Then \(f(0.1)\approx0.1\). Since \(f''(0)=-2<0\), the graph is concave down near \(x=0\), so the graph lies below its tangent line and the approximation is an overestimate.
(d) Find \(\displaystyle\int_0^1f(x)\,dx\).
Show solution to (d)
Integrate by parts with \(u=x\) and \(dv=e^{-x}dx\): \(\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx=-(x+1)e^{-x}\). Then \(\displaystyle\int_0^1xe^{-x}dx=\left[-(x+1)e^{-x}\right]_0^1=-\frac2e+1=1-\frac2e\).
Question 3 (No Calculator)
Consider the differential equation \(\dfrac{dy}{dx}=\dfrac{3x^2}{y}\), with the initial condition \(y(1)=2\).
(a) Write an equation for the line tangent to the solution curve at \((1,2)\), and use Euler's method with one step of size \(0.5\), starting at \((1,2)\), to approximate \(y(1.5)\).
Show solution to (a)
At \((1,2)\), \(\dfrac{dy}{dx}=\dfrac{3(1)^2}{2}=\dfrac32\), so the tangent line is \(y-2=\tfrac32(x-1)\). One Euler step gives \(y(1.5)\approx2+0.5\cdot\tfrac32=2.75\).
(b) Find \(\dfrac{d^2y}{dx^2}\) in terms of \(x\) and \(y\), and evaluate it at \((1,2)\). Is the Euler approximation in part (a) an overestimate or an underestimate? Give a reason.
Show solution to (b)
Differentiate \(\dfrac{dy}{dx}=3x^2y^{-1}\) with respect to \(x\): \(\dfrac{d^2y}{dx^2}=6xy^{-1}-3x^2y^{-2}\dfrac{dy}{dx}=\dfrac{6x}{y}-\dfrac{3x^2}{y^2}\cdot\dfrac{3x^2}{y}\). At \((1,2)\) this is \(\dfrac62-\dfrac{3}{4}\cdot\dfrac32=3-\dfrac98=\dfrac{15}8>0\). The solution is concave up there, so it lies above its tangent line and the Euler approximation \(2.75\) is an underestimate.
(c) Find the particular solution \(y=f(x)\) to the differential equation with \(y(1)=2\).
Show solution to (c)
Separate variables: \(y\,dy=3x^2\,dx\). Integrate: \(\dfrac{y^2}{2}=x^3+C\). Use \(y(1)=2\): \(2=1+C\), so \(C=1\). Then \(y^2=2x^3+2\), and because \(y(1)=2>0\) we take the positive root: \(y=\sqrt{2x^3+2}\).
(d) Use the particular solution to find the exact value of \(y(1.5)\), and compare it with the Euler approximation.
Show solution to (d)
\(y(1.5)=\sqrt{2(1.5)^3+2}=\sqrt{6.75+2}=\sqrt{8.75}=\dfrac{\sqrt{35}}{2}\approx2.958\). The Euler value \(2.75\) is less than \(2.958\), which agrees with the underestimate predicted in part (b).
Common Pitfalls on These Questions
Rate problems. In Question 1, the quantity in the tank changes at the rate inflow minus outflow, and the amount at a later time is the starting amount plus the accumulated net change. A frequent error is to integrate the net rate but forget the initial \(50\) gallons. Another is to find the time when the net rate is zero and stop there, without comparing the candidate with the endpoints of the interval.
Curve analysis. In Question 2, a relative extremum needs a sign change in \(f'\), and an inflection point needs a sign change in \(f''\). Finding a zero of the derivative is not enough. State the intervals or the test values you used, and say what the sign means for the graph.
Differential equations. In Question 3, separate the variables before integrating, add the constant immediately after the antiderivatives, and use the initial condition to solve for it. Points are not awarded for a solution that skips the constant, and the justification in part (b) must link the sign of the second derivative to the position of the curve relative to its tangent line.
Self-Scoring and Reflection
These questions also reward organization. Label each part, show the formula before substituting numbers, and keep exact values until the last line. Readers award points for visible reasoning, so a clean layout protects you when a late arithmetic slip occurs. For each part, check three things: whether your setup matches the one in the solution, whether your final value or conclusion agrees, and whether your justification names the specific fact that proves the claim. A solution that has the right answer but no reasoning would lose the justification point on the real exam. Count a part as fully correct only if all three are present.
After scoring, write one sentence on the most expensive mistake you made. Typical patterns are forgetting a constant of integration before applying an initial condition, using an open interval where a closed one is needed, or declaring a maximum without testing endpoints. Naming the pattern makes it far less likely to repeat on exam day. If a part was left blank, write down the first line you would have tried; on the real exam, a correct first step is often worth a point even when the rest is unfinished.
Check Your Understanding
- In which part did you most need to justify a conclusion with a sign argument, and what sign did you cite?
- Which setup could you have written down even if you could not finish the computation?
- How would your answer change if the problem asked for an exact value instead of a decimal approximation?
