2023 AP AP Calculus AB Practice Paper with Answers

Part (a): 3 points
- 1 point for interpretation with units (must reference total cubic meters/volume of water filtered and the time interval t = 2 to t = 10 hours)
- 1 point for the form of the left Riemann sum: (18)(3) + (26)(3) + (22)(2)
- 1 point for the numerical answer: 176

Part (b): 2 points
- 1 point for presenting R(10) - R(2) = 0 or (18 - 18)/8 = 0 (or stating R(2) = R(10))
- 1 point for answer with justification (must explicitly state that R is continuous because it is differentiable, and reference MVT or Rolle's Theorem)

Part (c): 2 points
- 1 point for the average value integral setup: (1 / 12) * \int_{0}^{12} W(t) dt
- 1 point for the correct answer: 23.107 (or 23.106)

Part (d): 2 points
- 1 point for the value of W'(7): -0.766 (or -0.767)
- 1 point for correct interpretation in context including units (cubic meters per hour per hour or m^3/hr^2) and time t = 7 hours.

### Part (a)
The vehicle changes direction when its velocity changes sign.
Set \(v(t) = 0\) for \(0 < t < 10\):
\[3.5e^{-0.08t}\sin\left(\frac{\pi}{4}t\right) = 0 \implies \sin\left(\frac{\pi}{4}t\right) = 0\]
\[\frac{\pi}{4}t = \pi \implies t = 4\]
\[\frac{\pi}{4}t = 2\pi \implies t = 8\]
Since \(v(t) > 0\) on \((0, 4)\), \(v(t) < 0\) on \((4, 8)\), and \(v(t) > 0\) on \((8, 10)\), the velocity changes sign at \(t = 4\) seconds and \(t = 8\) seconds.
Therefore, the vehicle changes direction at \(t = 4\) and \(t = 8\) seconds.

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### Part (b)
Acceleration is given by \(a(t) = v'(t)\).
Using a calculator:
\[a(3) = v'(3) \approx -1.685\text{ meters per second per second (or m/s}^2\text{)}.\]
To determine if the vehicle is speeding up or slowing down at \(t = 3\), check the signs of velocity and acceleration:
\[v(3) \approx 1.947 > 0\]
\[a(3) \approx -1.685 < 0\]
Because \(v(3)\) and \(a(3)\) have opposite signs, the vehicle is slowing down at \(t = 3\) seconds.

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### Part (c)
The position at \(t = 6\) is given by:
\[x(6) = x(0) + \int_0^6 v(t)\,dt\]
\[x(6) = 5 + \int_0^6 3.5e^{-0.08t}\sin\left(\frac{\pi}{4}t\right)\,dt\]
Using a calculator to evaluate the definite integral:
\[\int_0^6 v(t)\,dt \approx 4.689\]
\[x(6) = 5 + 4.689 = 9.689\text{ meters (or } 9.688\text{)}.\]

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### Part (d)
The total distance traveled over \(0 \le t \le 10\) is the integral of the vehicle's speed:
\[\text{Total distance} = \int_0^{10} |v(t)|\,dt\]
\[= \int_0^{10} \left| 3.5e^{-0.08t}\sin\left(\frac{\pi}{4}t\right) \right|\,dt\]
Evaluating using a calculator:
\[\text{Total distance} \approx 15.266\text{ meters}.\]

### Part (a) [2 points]
- 1 point for considering \(v(t) = 0\) or the sign of \(v(t)\).
- 1 point for correct answers (\(t = 4\) and \(t = 8\)) with reason.
- Scoring note: A response that states \(t = 4, 8\) without justification receives 0 out of 2 points. The reason must mention that \(v(t)\) changes sign at these times.

### Part (b) [3 points]
- 1 point for setup and numerical value of \(a(3) = v'(3) \approx -1.685\).
- 1 point for acceleration units (\(\text{m/s}^2\) or meters per second per second).
- 1 point for conclusion (slowing down) with valid reason referencing the opposite signs of \(v(3)\) and \(a(3)\).
- Scoring note: The connection \(a(t) = v'(t)\) or \(a(3) = v'(3)\) must be explicitly shown.

### Part (c) [2 points]
- 1 point for integral setup \(\int_0^6 v(t)\,dt\) or \(5 + \int_0^6 v(t)\,dt\).
- 1 point for correct final answer: \(9.689\) (or \(9.688\)).

### Part (d) [2 points]
- 1 point for integral setup \(\int_0^{10} |v(t)|\,dt\) or equivalent sum of integrals: \(\int_0^4 v(t)\,dt - \int_4^8 v(t)\,dt + \int_8^{10} v(t)\,dt\).
- 1 point for correct final answer: \(15.266\) (or \(15.265\)).

(a)
At \(t = 0\), \(W = 90\).
\[\left.\frac{dW}{dt}\right|_{t=0} = -\frac{1}{8}(90 - 10) = -\frac{1}{8}(80) = -10\]
An equation for the tangent line at \((0, 90)\) is:
\[y - 90 = -10(t - 0) \implies y = 90 - 10t\]
Using the tangent line to approximate \(W(2)\):
\[W(2) \approx 90 - 10(2) = 70^\circ\text{C}\]

(b)
Differentiating \(\frac{dW}{dt} = -\frac{1}{8}(W - 10)\) with respect to \(t\):
\[\frac{d^2W}{dt^2} = -\frac{1}{8}\left(\frac{dW}{dt}\right) = -\frac{1}{8}\left(-\frac{1}{8}(W - 10)\right) = \frac{1}{64}(W - 10)\]
Because \(W(t) > 10\) for all \(t \ge 0\), \(\frac{d^2W}{dt^2} = \frac{1}{64}(W - 10) > 0\).
Since the second derivative is positive, the graph of \(W\) is concave up on the interval \([0, 2]\). Therefore, the tangent line lies below the curve, meaning the tangent line approximation \(W(2) \approx 70^\circ\text{C}\) is an underestimate.

(c)
Separating variables:
\[\frac{1}{W - 10}\, dW = -\frac{1}{8}\, dt\]
Integrating both sides:
\[\int \frac{1}{W - 10}\, dW = \int -\frac{1}{8}\, dt\]
\[\ln|W - 10| = -\frac{1}{8}t + C\]
Since \(W(t) > 10\), \(|W - 10| = W - 10\):
\[\ln(W - 10) = -\frac{1}{8}t + C\]
Apply the initial condition \(W(0) = 90\):
\[\ln(90 - 10) = -\frac{1}{8}(0) + C \implies C = \ln(80)\]
Substitute \(C\) back into the equation:
\[\ln(W - 10) = -\frac{1}{8}t + \ln(80)\]
Exponentiating both sides:
\[W - 10 = e^{-\frac{1}{8}t + \ln(80)} = 80e^{-t/8}\]
\[W(t) = 10 + 80e^{-t/8}\]

(d)
\[\lim_{t \to \infty} W(t) = \lim_{t \to \infty} \left(10 + 80e^{-t/8}\right) = 10 + 0 = 10\]
As time increases without bound, the temperature of the metal block approaches \(10^\circ\text{C}\) (the ambient chamber temperature).

Part (a): 2 points
- 1 point for the tangent line equation or slope: \(\left.\frac{dW}{dt}\right|_{t=0} = -10\).
- 1 point for the approximation: \(W(2) \approx 70\).

Part (b): 2 points
- 1 point for \(\frac{d^2W}{dt^2} = \frac{1}{64}(W - 10)\) (or equivalent in terms of \(W\)).
- 1 point for concluding underestimate with justification mentioning \(\frac{d^2W}{dt^2} > 0\) (or graph of \(W\) is concave up) because \(W(t) > 10\).

Part (c): 4 points
- 1 point for separation of variables: \(\frac{1}{W - 10}\, dW = -\frac{1}{8}\, dt\).
- 1 point for correct antiderivatives: \(\ln|W - 10| = -\frac{1}{8}t\).
- 1 point for constant of integration and using initial condition \(W(0) = 90\).
- 1 point for solving for \(W(t)\): \(W(t) = 10 + 80e^{-t/8}\).

Note: A response with no separation of variables earns 0/4 in part (c).

Part (d): 1 point
- 1 point for finding \(\lim_{t \to \infty} W(t) = 10\) and interpreting that the temperature of the metal block approaches \(10^\circ\text{C}\) as \(t \to \infty\).

(a) At \(x = -1\), \(f'(x) > 0\) on \((-4, -1)\) and \(f'(x) < 0\) on \((-1, 3)\). Since \(f'(x)\) changes from positive to negative at \(x = -1\), \(f\) has a relative maximum at \(x = -1\).

(b) The graph of \(f\) is concave down where \(f'\) is decreasing.
From the graph of \(f'\):
- On \((-4, -1)\), \(f'\) decreases from \(3\) to \(0\).
- On \((-1, 1)\), the lower semicircle decreases from \(0\) to \(-2\).
- On \((1, 3)\), the lower semicircle increases from \(-2\) to \(0\).
- On \((3, 6)\), \(f'\) increases from \(0\) to \(3\).
Therefore, the graph of \(f\) is concave down on the open intervals \((-4, -1)\) and \((-1, 1)\) (or \((-4, 1)\)) because \(f'\) is decreasing on these intervals.

(c) Because \(f\) is differentiable on \([-4, 6]\), \(f\) is continuous at \(x = 3\), so \(\lim_{x \to 3} f(x) = f(3) = 4\).
Evaluating the limits of the numerator and denominator separately:
\[\lim_{x \to 3} (f(x) - x - 1) = 4 - 3 - 1 = 0\]
\[\lim_{x \to 3} (x^2 - 9) = 3^2 - 9 = 0\]
Because the limit produces the indeterminate form \(\frac{0}{0}\), L'Hôpital's Rule can be applied:
\[\lim_{x \to 3} \frac{f(x) - x - 1}{x^2 - 9} = \lim_{x \to 3} \frac{f'(x) - 1}{2x} = \frac{f'(3) - 1}{2(3)} = \frac{0 - 1}{6} = -\frac{1}{6}.\]

(d) The critical points of \(f\) on \([-4, 6]\) occur where \(f'(x) = 0\), which are \(x = -1\) and \(x = 3\).
The candidates for the absolute minimum are the endpoints \(x = -4, 6\) and the critical points \(x = -1, 3\).
Using the Fundamental Theorem of Calculus with the initial condition \(f(3) = 4\):
- \(f(3) = 4\)
- \(f(6) = f(3) + \int_3^6 f'(t)\,dt = 4 + \frac{1}{2}(3)(3) = 4 + 4.5 = 8.5\)
- \(f(-1) = f(3) - \int_{-1}^3 f'(t)\,dt = 4 - \left(-\frac{1}{2}\pi(2^2)\right) = 4 + 2\pi \approx 10.28\)
- \(f(-4) = f(-1) - \int_{-4}^{-1} f'(t)\,dt = (4 + 2\pi) - \frac{1}{2}(3)(3) = 2\pi - 0.5 \approx 5.78\)

Comparing the candidate values, the absolute minimum value of \(f\) on \([-4, 6]\) is \(f(3) = 4\).

Part (a): 1 point
- 1 point for the correct answer with justification based on the sign change of \(f'\).

Part (b): 2 points
- 1 point for identifying the intervals \((-4, -1)\) and \((-1, 1)\) (or \((-4, 1)\)).
- 1 point for the reason that \(f'\) is decreasing.

Part (c): 3 points
- 1 point for showing limits of numerator and denominator are both 0 separately.
- 1 point for correctly applying L'Hôpital's Rule.
- 1 point for the final answer of \(-\frac{1}{6}\).

Part (d): 3 points
- 1 point for considering \(f'(x) = 0\).
- 1 point for justification (evaluating candidates/endpoints or eliminating relative maximums).
- 1 point for identifying the absolute minimum value as \(4\).

(a) By the chain rule, $p'(x) = f'(g(x)) \cdot g'(x)$.
Using values from the table:
$$p'(5) = f'(g(5)) \cdot g'(5) = f'(1) \cdot 6 = (-2) \cdot 6 = -12$$

(b) To determine the concavity of $k$, find $k''(x)$ using the product rule and chain rule:
$$k''(x) = g'(x) \cdot (f(x))^3 + g(x) \cdot 3(f(x))^2 \cdot f'(x)$$
Evaluate at $x = 3$:
$$k''(3) = g'(3) \cdot (f(3))^3 + 3 g(3) (f(3))^2 f'(3)$$
$$k''(3) = (-1)(2)^3 + 3(8)(2)^2(5) = (-1)(8) + 24(4)(5) = -8 + 480 = 472$$
Because $k''(3) = 472 > 0$ (and $k''$ is continuous), the graph of $k$ is concave up at $x = 3$.

(c) By the Fundamental Theorem of Calculus, $\int_{1}^{3} f'(t)\,dt = f(3) - f(1)$.
$$M(3) = 4(3)^2 + \int_{1}^{3} f'(t)\,dt = 4(9) + (f(3) - f(1))$$
$$M(3) = 36 + (2 - 6) = 36 - 4 = 32$$

(d) Applying the Fundamental Theorem of Calculus to differentiate $M(x)$:
$$M'(x) = \frac{d}{dx}\left[4x^2 + \int_{1}^{x} f'(t)\,dt\right] = 8x + f'(x)$$
Evaluate at $x = 3$:
$$M'(3) = 8(3) + f'(3) = 24 + 5 = 29$$
Because $M'(3) = 29 > 0$, the function $M$ is increasing at $x = 3$.

Part (a): 2 points
- 1 point for applying the chain rule: $p'(x) = f'(g(x)) \cdot g'(x)$ or $p'(5) = f'(g(5)) \cdot g'(5)$
- 1 point for the correct answer: $-12$ with supporting work

Part (b): 3 points
- 1 point for correct derivative form using product and chain rules: $k''(x) = g'(x)(f(x))^3 + 3g(x)(f(x))^2 f'(x)$
- 1 point for computing $k''(3) = 472$ (or an unsimplified equivalent numeric expression)
- 1 point for concluding 'concave up' with a reason consistent with a positive value of $k''(3)$

Part (c): 2 points
- 1 point for applying FTC: $\int_{1}^{3} f'(t)\,dt = f(3) - f(1)$
- 1 point for the correct answer: $32$ (or $36 + (2 - 6)$)

Part (d): 2 points
- 1 point for finding $M'(x) = 8x + f'(x)$ and evaluating $M'(3) = 29$
- 1 point for answer 'increasing' with justification that $M'(3) > 0$

(a) Differentiating both sides of \( 2x^2 - 2xy + y^2 = 10 \) implicitly with respect to \( x \):
\[ \frac{d}{dx}(2x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(10) \]
\[ 4x - \left(2y + 2x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \]
\[ 4x - 2y + (2y - 2x)\frac{dy}{dx} = 0 \]
\[ (2y - 2x)\frac{dy}{dx} = 2y - 4x \]
\[ \frac{dy}{dx} = \frac{2(y - 2x)}{2(y - x)} = \frac{y - 2x}{y - x} \]

(b) A tangent line is horizontal when \( \frac{dy}{dx} = 0 \) and the denominator is nonzero. Thus, we set \( y - 2x = 0 \implies y = 2x \) (with \( y \neq x \)).
Substituting \( y = 2x \) into the original equation \( 2x^2 - 2xy + y^2 = 10 \):
\[ 2x^2 - 2x(2x) + (2x)^2 = 10 \]
\[ 2x^2 - 4x^2 + 4x^2 = 10 \implies 2x^2 = 10 \implies x^2 = 5 \implies x = \pm\sqrt{5} \]
When \( x = \sqrt{5} \), \( y = 2\sqrt{5} \).
When \( x = -\sqrt{5} \), \( y = -2\sqrt{5} \).
At both points, the denominator \( y - x \neq 0 \).
Therefore, the points with a horizontal tangent are \( (\sqrt{5}, 2\sqrt{5}) \) and \( (-\sqrt{5}, -2\sqrt{5}) \).

(c) A tangent line is vertical when the denominator of \( \frac{dy}{dx} \) is zero and the numerator is nonzero. Thus, \( y - x = 0 \implies y = x \) (with \( y \neq 2x \)).
Substituting \( y = x \) into the original equation \( 2x^2 - 2xy + y^2 = 10 \):
\[ 2x^2 - 2x(x) + x^2 = 10 \]
\[ 2x^2 - 2x^2 + x^2 = 10 \implies x^2 = 10 \implies x = \pm\sqrt{10} \]
When \( x = \sqrt{10} \), \( y = \sqrt{10} \).
When \( x = -\sqrt{10} \), \( y = -\sqrt{10} \).
At both points, the numerator \( y - 2x = -x = \mp\sqrt{10} \neq 0 \).
Therefore, the points with a vertical tangent are \( (\sqrt{10}, \sqrt{10}) \) and \( (-\sqrt{10}, -\sqrt{10}) \).

(d) Differentiating \( 2x^2 - 2xy + y^2 = 10 \) implicitly with respect to \( t \):
\[ 4x\frac{dx}{dt} - 2y\frac{dx}{dt} - 2x\frac{dy}{dt} + 2y\frac{dy}{dt} = 0 \]
\[ (4x - 2y)\frac{dx}{dt} + (2y - 2x)\frac{dy}{dt} = 0 \]
Substituting \( x = 1 \), \( y = 4 \), and \( \frac{dx}{dt} = 6 \):
\[ (4(1) - 2(4))(6) + (2(4) - 2(1))\frac{dy}{dt} = 0 \]
\[ (4 - 8)(6) + (8 - 2)\frac{dy}{dt} = 0 \]
\[ -24 + 6\frac{dy}{dt} = 0 \implies 6\frac{dy}{dt} = 24 \implies \frac{dy}{dt} = 4 \]
Alternatively, using the chain rule \( \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} \):
\[ \left.\frac{dy}{dx}\right|_{(1,4)} = \frac{4 - 2(1)}{4 - 1} = \frac{2}{3} \]
\[ \frac{dy}{dt} = \left(\frac{2}{3}\right)(6) = 4 \text{ units per second}. \]

Part (a): 2 points
• 1 pt: Correct implicit differentiation of \( 2x^2 - 2xy + y^2 = 10 \)
• 1 pt: Verification to isolate \( \frac{dy}{dx} = \frac{y-2x}{y-x} \)

Part (b): 2 points
• 1 pt: Sets \( y - 2x = 0 \) (or \( y = 2x \))
• 1 pt: Coordinates of both points \( (\sqrt{5}, 2\sqrt{5}) \) and \( (-\sqrt{5}, -2\sqrt{5}) \)

Part (c): 2 points
• 1 pt: Sets \( y - x = 0 \) (or \( y = x \))
• 1 pt: Coordinates of both points \( (\sqrt{10}, \sqrt{10}) \) and \( (-\sqrt{10}, -\sqrt{10}) \)

Part (d): 3 points
• 1 pt: Chain rule expression \( \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} \) OR differentiation with respect to \( t \)
• 1 pt: Evaluation of \( \frac{dy}{dx} = \frac{2}{3} \) at \( (1,4) \) OR substitution of known values into related rates equation
• 1 pt: Answer \( \frac{dy}{dt} = 4 \)