2025 AP AP Calculus AB Practice Paper with Answers

Part A
The average mass of the compound over the interval \(0 \le t \le 6\) is given by the average value formula:
$$\text{Average mass} = \frac{1}{6 - 0} \int_0^6 M(t)\, dt = \frac{1}{6} \int_0^6 9.6 \arctan(0.25t)\, dt$$
Evaluating the definite integral using a graphing calculator:
$$\frac{1}{6}(33.97874) = 5.663123$$
From time \(t = 0\) to \(t = 6\) hours, the average mass of the compound in the bioreactor was \(5.663\) grams.

Part B
The average rate of change of \(M\) over the interval \([0, 6]\) is:
$$\frac{M(6) - M(0)}{6 - 0} = \frac{9.6 \arctan(1.5) - 0}{6} = 1.572470$$
To find when the instantaneous rate of change equals the average rate of change, set \(M'(t)\) equal to this value:
$$M'(t) = \frac{38.4}{16 + t^2} = 1.572470$$
Using a calculator to solve for \(t\) on \(0 \le t \le 6\):
$$16 + t^2 = 24.420177 \implies t^2 = 8.420177 \implies t = 2.901754$$
The instantaneous rate of change of \(M\) equals the average rate of change at time \(t = 2.902\) hours.

Part C
The end behavior of the rate of change of the mass is described by the limit as \(t \to \infty\) of \(M'(t)\):
$$\lim_{t \to \infty} M'(t) = \lim_{t \to \infty} \frac{38.4}{16 + t^2} = 0$$

Part D
By the Fundamental Theorem of Calculus:
$$B'(t) = M'(t) - 0.15 \ln(t^2 + 1) = \frac{38.4}{16 + t^2} - 0.15 \ln(t^2 + 1)$$
Setting \(B'(t) = 0\) on the interval \(6 \le t \le 24\):
$$\frac{38.4}{16 + t^2} = 0.15 \ln(t^2 + 1) \implies t = 7.023547$$
Testing critical points and endpoints on \(6 \le t \le 24\):
- \(B(6) = M(6) = 9.434820\)
- \(B(7.023547) = M(7.023547) - \int_6^{7.023547} 0.15 \ln(x^2 + 1)\, dx \approx 10.117178 - 0.603099 = 9.514079\)
- \(B(24) = M(24) - \int_6^{24} 0.15 \ln(x^2 + 1)\, dx \approx 13.504104 - 15.688267 = -2.184163\)

Alternatively, \(B'(t) > 0\) for \(6 \le t < 7.024\) and \(B'(t) < 0\) for \(7.024 < t \le 24\). Therefore, \(B\) attains its absolute maximum at time \(t = 7.024\) (or \(7.023\)) hours.

Part A (2 points):
- Point 1 (P1): Average value formula (correct definite integral with evidence of division by 6).
- Point 2 (P2): Answer (5.663 or 5.664).

Part B (2 points):
- Point 3 (P3): Average rate of change expression or value (\(\frac{M(6)-M(0)}{6}\) or 1.572).
- Point 4 (P4): Answer with supporting equation (\(t = 2.902\)).

Part C (2 points):
- Point 5 (P5): Limit expression (\(\lim_{t \to \infty} M'(t)\) or \(\lim_{t \to \infty} \frac{38.4}{16 + t^2}\)).
- Point 6 (P6): Value (0).

Part D (3 points):
- Point 7 (P7): Considers \(B'(t) = 0\) (or \(M'(t) - 0.15 \ln(t^2 + 1) = 0\)).
- Point 8 (P8): Justification (Candidates test evaluating \(B(t)\) at \(t = 6, 7.024, 24\) OR global sign analysis of \(B'(t)\)).
- Point 9 (P9): Answer with supporting work (\(t = 7.024\) or \(7.023\)).

A. The intersection occurs where \(p(x) = q(x)\), which gives \(3 + 2\sin x = e^{0.4x}\). Using a graphing calculator, the positive intersection point is \(k = 2.988533\approx 2.989\).

The area of \(M\) is given by:
\[\text{Area} = \int_0^k (p(x) - q(x))\,dx = \int_0^{2.988533} \left(3 + 2\sin x - e^{0.4x}\right)dx \approx 7.180 \text{ (or } 7.179\text{)}.\]

B. The area of an isosceles right triangle with leg length \(b(x) = p(x) - q(x)\) is \(A(x) = \frac{1}{2}(b(x))^2 = \frac{1}{2}(p(x) - q(x))^2\).

The volume of the solid is:
\[\text{Volume} = \frac{1}{2}\int_0^k (p(x) - q(x))^2\,dx = \frac{1}{2}\int_0^{2.988533} \left(3 + 2\sin x - e^{0.4x}\right)^2dx \approx 7.126 \text{ (or } 7.125\text{)}.\]

C. The outer radius is \(R(x) = 5 - q(x)\) and the inner radius is \(r(x) = 5 - p(x)\).

The volume of the solid of revolution is:
\[\text{Volume} = \pi \int_0^k \left((5 - q(x))^2 - (5 - p(x))^2\right)dx\]
(or \(\pi \int_0^{2.989} \left((5 - e^{0.4x})^2 - (2 - 2\sin x)^2\right)dx\)).

D. The vertical distance between the two curves is given by \(D(x) = p(x) - q(x)\) for \(0 \le x \le k\).

\(D'(x) = p'(x) - q'(x) = 2\cos x - 0.4e^{0.4x}\).

Setting \(D'(x) = 0\) gives \(2\cos x - 0.4e^{0.4x} = 0\). Solving on the interval \(0 < x < k\) yields \(x = 1.237372 \approx 1.237\) (or \(1.238\)).

Justification:
- At \(x = 0\), \(D(0) = p(0) - q(0) = 3 - 1 = 2\).
- At \(x = 1.237372\), \(D(1.237372) \approx 4.890 - 1.640 = 3.250\).
- At \(x = k\), \(D(k) = 0\).

Alternatively, \(D'(x) > 0\) for \(0 < x < 1.237\) and \(D'(x) < 0\) for \(1.237 < x < k\). Because \(D'(x)\) changes from positive to negative at \(x = 1.237\) and this is the only critical point on the interval, the vertical distance attains its absolute maximum at \(x = 1.237\) (or \(1.238\)).

Part A (2 points):
- Point 1 (P1): Correct integrand and limits in a definite integral setup \(\int_0^k (p(x) - q(x))\,dx\).
- Point 2 (P2): Correct numerical answer of \(7.180\) (or \(7.179\)) with \(k = 2.989\) (or \(2.988\)).

Part B (2 points):
- Point 3 (P3): Integrand of the form \(\frac{1}{2}(p(x) - q(x))^2\) in a definite integral.
- Point 4 (P4): Correct numerical answer of \(7.126\) (or \(7.125\)).

Part C (3 points):
- Point 5 (P5): Form of integrand showing difference of squared radii \(R^2 - r^2\).
- Point 6 (P6): Correct integrand \((5 - q(x))^2 - (5 - p(x))^2\).
- Point 7 (P7): Limits \(0\) to \(k\), constant \(\pi\), and differential \(dx\).

Part D (2 points):
- Point 8 (P8): Sets \(D'(x) = 0\) (or \(p'(x) = q'(x)\)).
- Point 9 (P9): Correct answer \(x = 1.237\) (or \(1.238\)) with valid justification.

Part A:
Using the average rate of change of \( R \) on \([3, 7]\):
\[ R'(5) \approx \frac{R(7) - R(3)}{7 - 3} = \frac{30 - 18}{4} = \frac{12}{4} = 3 \text{ gallons per hour per hour (or } \text{gal/hr}^2\text{)} \]

Part B:
Because \( R \) is differentiable, \( R \) is continuous on the interval \([0, 12]\).
Since \( R(0) = 12 \) and \( R(12) = 38 \), and \( 12 < 25 < 38 \) (or since \( R(3) = 18 < 25 < 30 = R(7) \)), by the Intermediate Value Theorem (IVT), there must exist at least one value \( c \) in \( (0, 12) \) such that \( R(c) = 25 \).

Part C:
Using a trapezoidal sum with the subintervals \([0, 3]\), \([3, 7]\), and \([7, 12]\):
\[ \int_0^{12} R(t)\,dt \approx \frac{R(0) + R(3)}{2}(3 - 0) + \frac{R(3) + R(7)}{2}(7 - 3) + \frac{R(7) + R(12)}{2}(12 - 7) \]
\[ = \frac{12 + 18}{2}(3) + \frac{18 + 30}{2}(4) + \frac{30 + 38}{2}(5) \]
\[ = 15(3) + 24(4) + 34(5) = 45 + 96 + 170 = 311 \]

Part D:
The total amount of water drained from time \( t = 0 \) to \( t = 6 \) is given by:
\[ \int_0^6 D(t)\,dt = \int_0^6 \left( \frac{1}{4}t^2 - 2t + 20 \right) dt \]
\[ = \left[ \frac{1}{12}t^3 - t^2 + 20t \right]_0^6 \]
\[ = \left( \frac{1}{12}(6)^3 - (6)^2 + 20(6) \right) - \left( 0 - 0 + 0 \right) \]
\[ = \left( \frac{216}{12} - 36 + 120 \right) = 18 - 36 + 120 = 102 \text{ gallons} \]

Part A (2 points):
- Point 1 (P1): Earned for the correct difference quotient setup with table values, e.g., \( \frac{30 - 18}{7 - 3} \), leading to \( 3 \).
- Point 2 (P2): Earned for the correct units: gallons per hour per hour, \(\text{gallons}/\text{hour}^2\), or \(\text{gal}/\text{hr}^2\).

Part B (2 points):
- Point 3 (P3): Earned for explicitly stating that \( R \) is continuous because \( R \) is differentiable.
- Point 4 (P4): Earned for justification applying the Intermediate Value Theorem with inequality showing \( 25 \) is between two function values (e.g., \( R(0) < 25 < R(12) \) or \( R(3) < 25 < R(7) \)) and concluding "yes".

Part C (2 points):
- Point 5 (P5): Earned for the correct form of the trapezoidal sum with three subintervals.
- Point 6 (P6): Earned for the correct answer of \( 311 \) supported by work.

Part D (3 points):
- Point 7 (P7): Earned for setting up the integral \( \int_0^6 D(t)\,dt \) or \( \int_0^6 \left(\frac{1}{4}t^2 - 2t + 20\right) dt \).
- Point 8 (P8): Earned for the correct antiderivative \( \frac{1}{12}t^3 - t^2 + 20t \).
- Point 9 (P9): Earned for the correct numerical answer of \( 102 \) (simplification not required if left in equivalent arithmetic form).

A. By the Fundamental Theorem of Calculus, \( h'(x) = f(x) \) and \( h''(x) = f'(x) \).
For \( 3 \le x \le 6 \), the graph of \( f \) is a line segment with slope \( m = \frac{-3 - 3}{6 - 3} = -2 \).
The equation of the line segment is \( f(x) = 3 - 2(x - 3) = 9 - 2x \).
Therefore, \( h'(4) = f(4) = 9 - 2(4) = 1 \).
Since the slope of the line segment is \(-2\), \( h''(4) = f'(4) = -2 \).

B. The graph of \( h \) has a point of inflection where \( h''(x) = f'(x) \) changes sign, which corresponds to where \( f \) changes from increasing to decreasing or from decreasing to increasing.
- On \((-4, -2)\), \( f \) is decreasing, and on \((-2, 3)\), \( f \) is increasing, so \( f \) changes from decreasing to increasing at \( x = -2 \).
- On \((-2, 3)\), \( f \) is increasing, and on \((3, 6)\), \( f \) is decreasing, so \( f \) changes from increasing to decreasing at \( x = 3 \).
- On \((3, 6)\), \( f \) is decreasing, and on \((6, 8)\), \( f \) is increasing, so \( f \) changes from decreasing to increasing at \( x = 6 \).
Thus, the points of inflection of the graph of \( h \) occur at \( x = -2 \), \( x = 3 \), and \( x = 6 \).

C. \( h(-4) = \int_{0}^{-4} f(t)\,dt = -\int_{-4}^{0} f(t)\,dt \).
The region between the graph of \( f \) and the \( x \)-axis from \( x = -4 \) to \( x = 0 \) is a semicircle of radius \( 2 \) located below the \( x \)-axis. Its signed area is \(-\frac{1}{2}\pi(2^2) = -2\pi\).
Thus, \( h(-4) = -(-2\pi) = 2\pi \).

\( h(6) = \int_{0}^{6} f(t)\,dt = \int_{0}^{3} f(t)\,dt + \int_{3}^{4.5} f(t)\,dt + \int_{4.5}^{6} f(t)\,dt \).
- From \( t = 0 \) to \( t = 3 \), the region is a triangle with base \( 3 \) and height \( 3 \), so the integral is \( \frac{1}{2}(3)(3) = \frac{9}{2} \).
- From \( t = 3 \) to \( t = 4.5 \), the region is a triangle above the axis with base \( 1.5 \) and height \( 3 \), area \( \frac{1}{2}(1.5)(3) = \frac{9}{4} \).
- From \( t = 4.5 \) to \( t = 6 \), the region is a triangle below the axis with base \( 1.5 \) and height \( 3 \), signed area \( -\frac{9}{4} \).
Thus, \( h(6) = \frac{9}{2} + \frac{9}{4} - \frac{9}{4} = \frac{9}{2} \).

D. The function \( h \) is continuous on \([-4, 8]\), so its absolute minimum occurs at an endpoint or a critical point where \( h'(x) = f(x) = 0 \).
In the interval \([-4, 8]\), \( f(x) = 0 \) at \( x = -4, 0, 4.5, 7.5 \).
Evaluating \( h(x) \) at candidates:
- \( h(-4) = 2\pi \approx 6.283 \)
- \( h(0) = 0 \)
- \( h(4.5) = \frac{9}{2} + \frac{9}{4} = \frac{27}{4} = 6.75 \)
- \( h(7.5) = h(6) - \frac{1}{2}(1.5)(3) = \frac{9}{2} - \frac{9}{4} = \frac{9}{4} = 2.25 \)
- \( h(8) = h(7.5) + \frac{1}{2}(0.5)(1) = \frac{9}{4} + \frac{1}{4} = \frac{5}{2} = 2.5 \)

Comparing values, the absolute minimum value of \( h \) on \([-4, 8]\) is \( 0 \), which occurs at \( x = 0 \).

Part A (2 points):
- Point 1 (P1): Considers \( h'(x) = f(x) \) and finds \( h'(4) = 1 \).
- Point 2 (P2): Finds \( h''(4) = f'(4) = -2 \).

Part B (2 points):
- Point 3 (P3): Identifies all three values \( x = -2 \), \( x = 3 \), and \( x = 6 \) (and no other values in the interval).
- Point 4 (P4): Provides correct reasoning tied to \( f \) changing between increasing and decreasing (or slope of \( f \) changing sign) at each of the three points.

Part C (2 points):
- Point 5 (P5): Finds \( h(-4) = 2\pi \).
- Point 6 (P6): Finds \( h(6) = \frac{9}{2} \) (or \( 4.5 \)).

Part D (3 points):
- Point 7 (P7): Considers \( h'(x) = f(x) = 0 \) (identifies critical points \( x = 0, 4.5, 7.5 \)).
- Point 8 (P8): Justification via a global argument (e.g., evaluating candidates \( x = -4, 0, 4.5, 7.5, 8 \) or sign analysis of \( h' \)).
- Point 9 (P9): Identifies \( x = 0 \) as the location of the absolute minimum.

Part A:
The velocity of particle \(A\) is the derivative of its position function:
\[v_A(t) = x_A'(t) = \frac{d}{dt}\left[(3 - t)e^{2t}\right] = (-1)e^{2t} + (3 - t)(2e^{2t}) = (5 - 2t)e^{2t}\]
At \(t = 2\):
\[v_A(2) = (5 - 2(2))e^{2(2)} = (1)e^4 = e^4\]

Part B:
To determine the direction of motion, find when each velocity is positive or negative.
For particle \(A\):
\[v_A(t) = (5 - 2t)e^{2t} = 0 \implies t = \frac{5}{2}\]
Since \(e^{2t} > 0\) for all \(t\):
- \(v_A(t) > 0\) (moving right) for \(0 < t < \frac{5}{2}\)
- \(v_A(t) < 0\) (moving left) for \(\frac{5}{2} < t < 4\)

For particle \(B\):
\[v_B(t) = 6t^2(t^3 - 8) = 0 \implies t = 0 \text{ or } t = 2\]
On the interval \(0 < t < 4\):
- \(v_B(t) < 0\) (moving left) for \(0 < t < 2\)
- \(v_B(t) > 0\) (moving right) for \(2 < t < 4\)

Comparing the signs:
- For \(0 < t < 2\): \(v_A(t) > 0\) and \(v_B(t) < 0\) (opposite directions)
- For \(2 < t < \frac{5}{2}\): \(v_A(t) > 0\) and \(v_B(t) > 0\) (same direction)
- For \(\frac{5}{2} < t < 4\): \(v_A(t) < 0\) and \(v_B(t) > 0\) (opposite directions)

Therefore, particles \(A\) and \(B\) are moving in opposite directions on the open intervals \(0 < t < 2\) and \(\frac{5}{2} < t < 4\).

Part C:
The acceleration of particle \(B\) is given by:
\[a_B(t) = v_B'(t) = \frac{d}{dt}\left[6t^5 - 48t^2\right] = 30t^4 - 96t\]
At \(t = 1\):
\[a_B(1) = 30(1)^4 - 96(1) = -66\]
\[v_B(1) = 6(1)^2(1^3 - 8) = 6(-7) = -42\]
Because \(v_B(1) < 0\) and \(a_B(1) < 0\), the velocity and acceleration have the same sign at \(t = 1\). Therefore, the speed of particle \(B\) is increasing at \(t = 1\).

Part D:
The position of particle \(B\) at \(t = 2\) is:
\[x_B(2) = x_B(0) + \int_0^2 v_B(t)\,dt = -5 + \int_0^2 6t^2(t^3 - 8)\,dt\]
Using the antiderivative:
\[\int_0^2 (6t^5 - 48t^2)\,dt = \left[ t^6 - 16t^3 \right]_0^2 = (2^6 - 16(2^3)) - 0 = 64 - 128 = -64\]
\[x_B(2) = -5 + (-64) = -69\]

Part A (2 points):
- Point 1 (P1): Considers \(x_A'(t)\) using product rule.
- Point 2 (P2): Answer \(e^4\) (or \((5 - 4)e^4\)).

Part B (3 points):
- Point 3 (P3): Considers zeros/sign of \(v_A(t)\) or \(v_B(t)\).
- Point 4 (P4): Correct sign analysis for both particles on \(0 < t < 4\).
- Point 5 (P5): Correct intervals \((0, 2)\) and \(\left(\frac{5}{2}, 4\right)\) with justification.

Part C (2 points):
- Point 6 (P6): Finds \(a_B(1) = -66\).
- Point 7 (P7): Conclusion "speed is increasing" with reason comparing signs of \(v_B(1)\) and \(a_B(1)\).

Part D (2 points):
- Point 8 (P8): Integrand and correct antiderivative \(t^6 - 16t^3\) (or \((t^3 - 8)^2\) with substitution).
- Point 9 (P9): Correct position \(-69\) utilizing initial condition \(x_B(0) = -5\).

### Part A
Differentiate both sides of the equation \( 2y^3 - 3y^2 - 12y + x^2 = 3 \) implicitly with respect to \( x \):
\[ \frac{d}{dx}\left(2y^3 - 3y^2 - 12y + x^2\right) = \frac{d}{dx}(3) \]
\[ 6y^2\frac{dy}{dx} - 6y\frac{dy}{dx} - 12\frac{dy}{dx} + 2x = 0 \]
\[ (6y^2 - 6y - 12)\frac{dy}{dx} = -2x \]
\[ \frac{dy}{dx} = \frac{-2x}{6y^2 - 6y - 12} = \frac{-x}{3(y^2 - y - 2)} \]

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### Part B
Evaluate the derivative at the point \( (4, 1) \):
\[ \left.\frac{dy}{dx}\right|_{(4,1)} = \frac{-4}{3(1^2 - 1 - 2)} = \frac{-4}{3(-2)} = \frac{2}{3} \]

The equation of the tangent line at \( (4, 1) \) is:
\[ y - 1 = \frac{2}{3}(x - 4) \implies y = 1 + \frac{2}{3}(x - 4) \]

Approximating the \( y \)-coordinate when \( x = 4.3 \):
\[ y \approx 1 + \frac{2}{3}(4.3 - 4) = 1 + \frac{2}{3}(0.3) = 1 + 0.2 = 1.2 \]

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### Part C
For \( x > 0 \), the tangent line to curve \( C \) is vertical when the denominator of \( \frac{dy}{dx} \) is equal to \( 0 \) and the numerator is non-zero:
\[ 3(y^2 - y - 2) = 0 \]
\[ 3(y - 2)(y + 1) = 0 \implies y = 2 \text{ or } y = -1 \]

Since \( y > 0 \), it follows that \( y = 2 \).

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### Part D
Differentiate both sides of \( x^2 y + 2\ln y = 16 \) implicitly with respect to \( t \):
\[ \frac{d}{dt}\left(x^2 y + 2\ln y\right) = \frac{d}{dt}(16) \]
\[ 2x\frac{dx}{dt}y + x^2\frac{dy}{dt} + \frac{2}{y}\frac{dy}{dt} = 0 \]

Substitute \( x = 4 \), \( y = 1 \), and \( \frac{dx}{dt} = -3 \):
\[ 2(4)(-3)(1) + (4)^2\frac{dy}{dt} + \frac{2}{1}\frac{dy}{dt} = 0 \]
\[ -24 + 16\frac{dy}{dt} + 2\frac{dy}{dt} = 0 \]
\[ 18\frac{dy}{dt} = 24 \implies \frac{dy}{dt} = \frac{24}{18} = \frac{4}{3} \]

### Part A (2 points)
- Point 1 (P1): Earned for correct implicit differentiation: \( 6y^2\frac{dy}{dx} - 6y\frac{dy}{dx} - 12\frac{dy}{dx} + 2x = 0 \) (or equivalent).
- Point 2 (P2): Earned for verification/solving for \( \frac{dy}{dx} \) to obtain the given expression.

### Part B (2 points)
- Point 3 (P3): Earned for evaluating \( \left.\frac{dy}{dx}\right|_{(4, 1)} = \frac{2}{3} \).
- Point 4 (P4): Earned for tangent line approximation value: \( 1 + \frac{2}{3}(4.3 - 4) = 1.2 \) (or unsimplified equivalent).

### Part C (2 points)
- Point 5 (P5): Earned for setting the denominator equal to zero: \( 3(y^2 - y - 2) = 0 \) or \( y^2 - y - 2 = 0 \).
- Point 6 (P6): Earned for concluding \( y = 2 \) and discarding \( y = -1 \) because \( y > 0 \).

### Part D (3 points)
- Point 7 (P7): Earned for an eligible attempt at implicit differentiation with respect to \( t \) (at most one error).
- Point 8 (P8): Earned for a completely correct differentiated equation: \( 2x\frac{dx}{dt}y + x^2\frac{dy}{dt} + \frac{2}{y}\frac{dy}{dt} = 0 \).
- Point 9 (P9): Earned for substituting values and finding \( \frac{dy}{dt} = \frac{4}{3} \) (or unsimplified equivalent such as \( \frac{24}{18} \)).