How to Use This BC Free-Response Practice Set
The AP Calculus BC 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 particle moves in the \(xy\)-plane for \(0\le t\le2\). Its velocity components are \(\dfrac{dx}{dt}=2t\) and \(\dfrac{dy}{dt}=3-t^2\), and at time \(t=0\) the particle is at the point \((1,0)\).
(a) Find the speed and the acceleration vector of the particle at \(t=1\).
Show solution to (a)
At \(t=1\), the velocity is \(\langle2,2\rangle\), so the speed is \(\sqrt{4+4}=2\sqrt2\approx2.828\). The acceleration vector is the derivative of velocity, \(\langle2,-2t\rangle\), which at \(t=1\) equals \(\langle2,-2\rangle\).
(b) Find the position of the particle at \(t=2\).
Show solution to (b)
Integrate each component and use the initial position: \(x(2)=1+\int_0^22t\,dt=1+4=5\) and \(y(2)=0+\int_0^2(3-t^2)\,dt=6-\tfrac83=\tfrac{10}3\). The position is \(\left(5,\tfrac{10}{3}\right)\approx(5,3.333)\).
(c) At what time \(t\), for \(0 A horizontal tangent requires \(\dfrac{dy}{dt}=0\) with \(\dfrac{dx}{dt}\ne0\). Setting \(3-t^2=0\) gives \(t=\sqrt3\approx1.732\), where \(\dfrac{dx}{dt}=2\sqrt3\ne0\). The height is \(y(\sqrt3)=3\sqrt3-\dfrac{(\sqrt3)^3}{3}=3\sqrt3-\sqrt3=2\sqrt3\approx3.464\). This is the maximum height of the particle, since \(dy/dt\) changes from positive to negative there. (d) Find the total distance the particle travels from \(t=0\) to \(t=2\). Total distance is the integral of speed: \(\displaystyle\int_0^2\sqrt{(2t)^2+(3-t^2)^2}\,dt=\int_0^2\sqrt{t^4-2t^2+9}\,dt\approx6.144\). The straight-line distance from \((1,0)\) to \((5,\tfrac{10}3)\) is about \(5.2\), and the path cannot be shorter, so \(6.144\) passes the check. Let \(g(x)=x^2e^{-x^2}\). Answer the following without a calculator. (a) Write the first four nonzero terms and the general term of the Maclaurin series for \(g\). Start from \(e^u=1+u+\dfrac{u^2}{2!}+\dfrac{u^3}{3!}+\cdots\) with \(u=-x^2\): \(e^{-x^2}=1-x^2+\dfrac{x^4}{2}-\dfrac{x^6}{6}+\cdots\). Multiply by \(x^2\): \(g(x)=x^2-x^4+\dfrac{x^6}{2}-\dfrac{x^8}{6}+\cdots\). The general term is \(\dfrac{(-1)^n x^{2n+2}}{n!}\) for \(n=0,1,2,\dots\). (b) Find the radius of convergence of the series in part (a). By the ratio test, \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{x^{2n+4}/(n+1)!}{x^{2n+2}/n!}=\dfrac{x^2}{n+1}\to0\) for every \(x\). The limit is less than \(1\) for all \(x\), so the series converges everywhere and the radius of convergence is infinite. (c) Find \(g^{(6)}(0)\), the sixth derivative of \(g\) at \(x=0\). The coefficient of \(x^6\) in the series is \(\dfrac12\), and that coefficient equals \(\dfrac{g^{(6)}(0)}{6!}\). Therefore \(g^{(6)}(0)=720\cdot\dfrac12=360\). (d) Use the first two nonzero terms of the series for \(\int_0^x g(t)\,dt\) to approximate \(\displaystyle\int_0^{1/2}g(x)\,dx\), and use the Alternating Series Error Bound to bound the error. Integrating term by term, \(\displaystyle\int_0^x g(t)\,dt=\frac{x^3}{3}-\frac{x^5}{5}+\frac{x^7}{14}-\cdots\). At \(x=\tfrac12\), the terms are \(\tfrac1{24}\), \(\tfrac1{160}\), \(\tfrac1{1792}\), and so on, alternating in sign and decreasing in magnitude to \(0\). The approximation with two terms is \(\dfrac1{24}-\dfrac1{160}=\dfrac{17}{480}\approx0.0354\). By the Alternating Series Error Bound, the error is at most the first omitted term, \(\dfrac1{1792}<0.0006\). The graph of the polar curve \(r=1+\cos\theta\) for \(0\le\theta\le2\pi\) is a cardioid. Let \(R\) be the region inside the cardioid and outside the circle \(r=1\). (a) Find the values of \(\theta\) at which the cardioid and the circle intersect. Set \(1+\cos\theta=1\). Then \(\cos\theta=0\), so \(\theta=\dfrac\pi2\) and \(\theta=\dfrac{3\pi}2\) on \([0,2\pi]\), equivalently \(\theta=\pm\dfrac\pi2\). Between \(-\tfrac\pi2\) and \(\tfrac\pi2\), \(\cos\theta>0\), so the cardioid is outside the circle there. (b) Write and evaluate an expression for the area of \(R\). Use the difference of the squared radii: \(\text{Area}=\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}\left[(1+\cos\theta)^2-1^2\right]d\theta=\dfrac12\int_{-\pi/2}^{\pi/2}\left(2\cos\theta+\cos^2\theta\right)d\theta\). By symmetry this is \(\displaystyle\int_0^{\pi/2}\left(2\cos\theta+\cos^2\theta\right)d\theta=2+\dfrac\pi4\), because \(\int_0^{\pi/2}2\cos\theta\,d\theta=2\) and \(\int_0^{\pi/2}\cos^2\theta\,d\theta=\dfrac\pi4\). (c) Find \(\dfrac{dy}{dx}\) at \(\theta=\dfrac\pi2\). Write \(x=(1+\cos\theta)\cos\theta\) and \(y=(1+\cos\theta)\sin\theta\). Then \(\dfrac{dx}{d\theta}=-\sin\theta-2\sin\theta\cos\theta=-\sin\theta(1+2\cos\theta)\) and \(\dfrac{dy}{d\theta}=-\sin^2\theta+\cos\theta+\cos^2\theta\). At \(\theta=\tfrac\pi2\): \(\dfrac{dx}{d\theta}=-1\) and \(\dfrac{dy}{d\theta}=-1\), so \(\dfrac{dy}{dx}=1\). (d) Find the greatest \(x\)-coordinate of any point on the cardioid. Justify your answer. From part (c), \(\dfrac{dx}{d\theta}=-\sin\theta(1+2\cos\theta)\), which is zero at \(\theta=0,\ \dfrac{2\pi}3,\ \pi,\ \dfrac{4\pi}3,\ 2\pi\). The corresponding values of \(x=(1+\cos\theta)\cos\theta\) are \(2\), \(-\dfrac14\), \(0\), \(-\dfrac14\), and \(2\). The greatest value is \(2\), at \(\theta=0\) (and \(\theta=2\pi\)). Parametric motion. In Question 1, speed is the magnitude of the velocity vector, while acceleration is a vector. Distance traveled requires the integral of speed, not the straight-line distance between the endpoints and not the difference of the position coordinates. A horizontal tangent needs the vertical rate to be zero while the horizontal rate is not. Series. In Question 2, build new series from known ones instead of taking derivatives repeatedly. Substitute \(-x^2\) into the exponential series, multiply by \(x^2\), and integrate term by term. When you use the Alternating Series Error Bound, check three things in writing: the terms alternate, their magnitudes decrease, and they tend to zero. Polar curves. In Question 3, area uses \(\tfrac12 r^2\), and an area between two curves uses the difference of the squared radii, not the square of the difference. Find the limits of integration from the intersection points, and use symmetry only when you can state why it applies. For slopes, differentiate \(x\) and \(y\) with respect to \(\theta\) first. 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.Show solution to (c)
Show solution to (d)
Question 2 (No Calculator)
Show solution to (a)
Show solution to (b)
Show solution to (c)
Show solution to (d)
Question 3 (No Calculator)
Show solution to (a)
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Show solution to (c)
Show solution to (d)
Common Pitfalls on These Questions
Self-Scoring and Reflection
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