2023 AP AP Calculus BC 模擬試題連答案詳解

(a)
\( \int_{20}^{100} R(t)\,dt \) represents the total amount (or volume) of filtered water, in cubic meters (\(\text{m}^3\)), that enters the clean-water holding basin from time \( t = 20 \) minutes to time \( t = 100 \) minutes.

Using a trapezoidal sum with subintervals \([20, 50]\), \([50, 80]\), and \([80, 100]\):
$$\int_{20}^{100} R(t)\,dt \approx \frac{R(20) + R(50)}{2}(50 - 20) + \frac{R(50) + R(80)}{2}(80 - 50) + \frac{R(80) + R(100)}{2}(100 - 80)$$
$$= \frac{5.8 + 7.4}{2}(30) + \frac{7.4 + 7.4}{2}(30) + \frac{7.4 + 6.1}{2}(20)$$
$$= (6.6)(30) + (7.4)(30) + (6.75)(20) = 198 + 222 + 135 = 555$$

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(b)
Because \( R \) is differentiable on \([0, 120]\), \( R \) is continuous on the closed interval \([50, 80]\) and differentiable on the open interval \((50, 80)\).

The average rate of change of \( R \) over \([50, 80]\) is:
$$\frac{R(80) - R(50)}{80 - 50} = \frac{7.4 - 7.4}{30} = \frac{0}{30} = 0$$

By the Mean Value Theorem (or Rolle's Theorem), there must exist at least one value \( c \), with \( 50 < c < 80 \), such that \( R'(c) = 0 \).

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(c)
The average rate of inflow of filtered water over the interval \( 0 \le t \le 120 \) is given by the average value formula:
$$\text{Average value} = \frac{1}{120 - 0} \int_{0}^{120} W(t)\,dt = \frac{1}{120} \int_{0}^{120} \left(5 + \frac{t}{60} + 3\sin\left(\frac{t}{20}\right)\right) dt$$

Evaluating the definite integral with a calculator (or analytically):
$$\int_{0}^{120} \left(5 + \frac{t}{60} + 3\sin\left(\frac{t}{20}\right)\right) dt = \left[ 5t + \frac{t^2}{120} - 60\cos\left(\frac{t}{20}\right) \right]_{0}^{120} = 780 - 60\cos(6) \approx 722.38978$$

$$\frac{1}{120} (722.38978) \approx 6.019915 \approx 6.020 \text{ (or } 6.019\text{)}$$

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(d)
Using a calculator (or differentiating directly, \( W'(t) = \frac{1}{60} + \frac{3}{20}\cos\left(\frac{t}{20}\right) \)):
$$W'(90) = \frac{1}{60} + 0.15\cos(4.5) \approx -0.014953 \approx -0.015$$

Interpretation: At time \( t = 90 \) minutes, the rate at which filtered water is flowing into the clean-water holding basin is decreasing at a rate of \( 0.015 \text{ m}^3/\text{min}^2 \) (or changing at a rate of \(-0.015 \text{ m}^3/\text{min}^2\)).

Part (a): 3 points
- 1 point for correct interpretation with units (must reference cubic meters of water added/pumped and the interval \( t = 20 \) to \( t = 100 \)).
- 1 point for the form of the trapezoidal sum.
- 1 point for the final numerical answer (\( 555 \)).

Scoring notes:
- An unsimplified correct sum such as \( \frac{5.8+7.4}{2}(30) + 7.4(30) + \frac{7.4+6.1}{2}(20) \) earns both the second and third points.

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(b): 2 points
- 1 point for presenting \( \frac{R(80)-R(50)}{80-50} = 0 \) or \( R(80) - R(50) = 0 \) or \( R(80) = R(50) \).
- 1 point for answer ("Yes") with justification including the statement that \( R \) is continuous because \( R \) is differentiable, and referencing the Mean Value Theorem or Rolle's Theorem.

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(c): 2 points
- 1 point for the average value integral setup \( \frac{1}{120 - 0}\int_{0}^{120} W(t)\,dt \).
- 1 point for the correct numerical answer (\( 6.020 \) or \( 6.019 \)).

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(d): 2 points
- 1 point for finding \( W'(90) \approx -0.015 \) (or \(-0.014\)).
- 1 point for interpretation with units in context (must state rate of inflow is decreasing at \( 0.015 \text{ m}^3/\text{min}^2 \) or changing at \(-0.015 \text{ m}^3/\text{min}^2\) at time \( t = 90 \) minutes).

(a)
\( x'(t) = \frac{d}{dt}\left[\ln(t^2 + 2)\right] = \frac{2t}{t^2 + 2} \)
\( x''(2) = \left. \frac{d}{dt}\left( \frac{2t}{t^2 + 2} \right) \right|_{t=2} = \left. \frac{2(t^2+2) - 2t(2t)}{(t^2+2)^2} \right|_{t=2} = \frac{4 - 2(2)^2}{(2^2+2)^2} = -\frac{4}{36} = -\frac{1}{9} \approx -0.111111 \)

\( y'(t) = \sin(e^{0.5 t}) \)
\( y''(2) = \left. \frac{d}{dt}\left( \sin(e^{0.5 t}) \right) \right|_{t=2} = \left. 0.5 e^{0.5 t} \cos(e^{0.5 t}) \right|_{t=2} = 0.5 e \cos(e) \approx -1.239181 \)

The acceleration vector at time \( t = 2 \) is:
\( \mathbf{a}(2) = \langle x''(2), y''(2) \rangle = \langle -0.111, -1.239 \rangle \) (or \((-0.111, -1.239)\)).

(b)
\( \text{Speed} = \sqrt{(x'(t))^2 + (y'(t))^2} = \sqrt{\left(\frac{2t}{t^2+2}\right)^2 + \left(\sin(e^{0.5 t})\right)^2} \)

Setting \( \sqrt{\left(\frac{2t}{t^2+2}\right)^2 + \left(\sin(e^{0.5 t})\right)^2} = 1.0 \) for \( 0 \le t \le 4 \):
Using a graphing calculator to solve for \( t \) on \( [0, 4] \), the first solution is:
\( t \approx 0.379198 \implies t = 0.379 \).

(c)
At \( t = 2 \):
\( \frac{dy}{dx} = \frac{y'(2)}{x'(2)} = \frac{\sin(e^{0.5 \cdot 2})}{\frac{2(2)}{2^2+2}} = \frac{\sin(e)}{\frac{4}{6}} = \frac{3\sin(e)}{2} \approx 0.616172 \)
The slope of the tangent line at \( t = 2 \) is \( 0.616 \).

The \(y\)-coordinate of the drone's position at \( t = 2 \) is:
\( y(2) = y(0) + \int_0^2 y'(t)\,dt = 3 + \int_0^2 \sin(e^{0.5 t})\,dt \approx 3 + 1.718566 = 4.718566 \approx 4.719 \) (or \( 4.718 \)).

(d)
\( \text{Total distance} = \int_0^4 \sqrt{(x'(t))^2 + (y'(t))^2}\,dt = \int_0^4 \sqrt{\left(\frac{2t}{t^2+2}\right)^2 + \left(\sin(e^{0.5 t})\right)^2}\,dt \approx 4.224 \).

Part (a): 2 points
- 1 point for \( x''(2) \) with setup
- 1 point for \( y''(2) \) with setup
Note: The acceleration vector may be written in component form \(\langle x''(2), y''(2) \rangle\) or as an ordered pair \((x''(2), y''(2))\).

Part (b): 2 points
- 1 point for setting \( \sqrt{(x'(t))^2 + (y'(t))^2} = 1.0 \) (speed equation)
- 1 point for answer \( t = 0.379 \)
Note: A response that gives only the final answer \( t = 0.379 \) without showing the setup earns 0 out of 2 points.

Part (c): 3 points
- 1 point for slope \( \frac{dy}{dx}\big|_{t=2} = \frac{y'(2)}{x'(2)} \approx 0.616 \) with supporting work
- 1 point for definite integral \( \int_0^2 \sin(e^{0.5 t})\,dt \)
- 1 point for \( y(2) \approx 4.719 \) (or \( 4.718 \))

Part (d): 2 points
- 1 point for correct integrand and limits in \( \int_0^4 \sqrt{(x'(t))^2 + (y'(t))^2}\,dt \)
- 1 point for answer \( 4.224 \)

(a) At \( t = 0 \), \( H(0) = 85 \).
\[ \left.\frac{dH}{dt}\right|_{t=0} = -\frac{1}{3}(85 - 10) = -\frac{75}{3} = -25 \]
The equation of the tangent line at \( t = 0 \) is:
\[ y = 85 - 25(t - 0) = 85 - 25t \]
Using the tangent line to approximate \( H(1) \):
\[ H(1) \approx 85 - 25(1) = 60^\circ\text{C} \]

(b) Differentiating both sides of the differential equation with respect to \( t \):
\[ \frac{d^2H}{dt^2} = -\frac{1}{3}\left(\frac{dH}{dt}\right) = -\frac{1}{3}\left(-\frac{1}{3}(H - 10)\right) = \frac{1}{9}(H - 10) \]
Since \( H(t) > 10 \) for all \( t \ge 0 \), \( \frac{d^2H}{dt^2} = \frac{1}{9}(H - 10) > 0 \).
Because the second derivative is positive, the graph of \( H \) is concave up for \( t \ge 0 \). Therefore, the tangent line lies below the curve, and the approximation \( H(1) \approx 60 \) is an underestimate.

(c) The step size is \( \Delta t = \frac{2 - 0}{2} = 1 \).
- Step 1 (from \( t = 0 \) to \( t = 1 \)):
\( (t_0, H_0) = (0, 85) \)
\( \left.\frac{dH}{dt}\right|_{(0,85)} = -\frac{1}{3}(85 - 10) = -25 \)
\( H_1 = 85 + (-25)(1) = 60 \)

- Step 2 (from \( t = 1 \) to \( t = 2 \)):
\( (t_1, H_1) = (1, 60) \)
\( \left.\frac{dH}{dt}\right|_{(1,60)} = -\frac{1}{3}(60 - 10) = -\frac{50}{3} \)
\( H_2 = 60 + \left(-\frac{50}{3}\right)(1) = \frac{180 - 50}{3} = \frac{130}{3} \)

Thus, \( H(2) \approx \frac{130}{3} \).

(d) Separating variables:
\[ \frac{1}{H - 10}\, dH = -\frac{1}{3}\, dt \]
Integrating both sides:
\[ \int \frac{1}{H - 10}\, dH = \int -\frac{1}{3}\, dt \]
\[ \ln|H - 10| = -\frac{1}{3}t + C \]
Using the initial condition \( H(0) = 85 \):
\[ \ln|85 - 10| = \ln(75) = -\frac{1}{3}(0) + C \implies C = \ln 75 \]
Since \( H(t) > 10 \), \( |H - 10| = H - 10 \):
\[ \ln(H - 10) = -\frac{1}{3}t + \ln 75 \]
Exponentiating both sides:
\[ H - 10 = e^{-\frac{1}{3}t + \ln 75} = 75e^{-t/3} \]
\[ H(t) = 10 + 75e^{-t/3} \]

Part (a): 2 points
- 1 point for finding \( \left.\frac{dH}{dt}\right|_{t=0} = -25 \) or tangent line equation
- 1 point for the approximation \( H(1) \approx 60 \)

Part (b): 2 points
- 1 point for \( \frac{d^2H}{dt^2} = \frac{1}{9}(H - 10) \)
- 1 point for 'underestimate' with valid justification referencing \( \frac{d^2H}{dt^2} > 0 \) or concavity

Part (c): 2 points
- 1 point for Euler's method setup with \( \Delta t = 1 \)
- 1 point for the final approximation \( \frac{130}{3} \) (or equivalent)

Part (d): 3 points
- 1 point for separation of variables and correct antiderivatives \( \ln|H - 10| = -\frac{1}{3}t \)
- 1 point for constant of integration and using initial condition \( H(0) = 85 \)
- 1 point for solving explicitly for \( H(t) = 10 + 75e^{-t/3} \)

(a) A horizontal tangent line occurs where \(g'(x) = 0\). On the interval \(-4 < x < 6\), \(g'(x) = 0\) at \(x = -2\) and \(x = 2\).
- At \(x = -2\), \(g'(x)\) changes sign from positive to negative, so \(g\) has a relative maximum at \(x = -2\).
- At \(x = 2\), \(g'(x)\) changes sign from negative to positive, so \(g\) has a relative minimum at \(x = 2\).

(b) The graph of \(g\) is concave down when \(g'\) is decreasing, which occurs on \((-4, 0)\) and \((4, 6)\).
The graph of \(g\) is decreasing when \(g'(x) < 0\), which occurs on \((-2, 2)\).
Therefore, the graph of \(g\) is both concave down and decreasing on the interval \((-2, 0)\) because \(g'(x) < 0\) and \(g'\) is decreasing on this interval.

(c) Since \(g\) is differentiable, \(g\) is continuous at \(x = 0\), so \(\lim_{x \to 0} g(x) = g(0) = 3\).
Evaluating the limits of the numerator and denominator separately:
\[\lim_{x \to 0} (g(x) - 3\cos x) = g(0) - 3\cos 0 = 3 - 3 = 0\]
\[\lim_{x \to 0} (x^2 + 2x) = 0\]
Since the limit produces the indeterminate form \(\frac{0}{0}\), L'Hospital's Rule applies:
\[\lim_{x \to 0} \frac{g(x) - 3\cos x}{x^2 + 2x} = \lim_{x \to 0} \frac{g'(x) + 3\sin x}{2x + 2} = \frac{g'(0) + 3\sin 0}{2(0) + 2} = \frac{-2 + 0}{2} = -1.\]

(d) The function \(g\) is continuous on \([-4, 6]\), so the candidates for the absolute extrema are the endpoints \(x = -4, 6\) and the critical points \(x = -2, 2\).
Using \(g(x) = g(0) + \int_0^x g'(t)\,dt = 3 + \int_0^x g'(t)\,dt\):
- \(g(-4) = 3 + \int_0^{-4} g'(t)\,dt = 3 - \int_{-4}^0 g'(t)\,dt = 3 - \left[\frac{1}{2}(2)(2) - \frac{1}{2}(2)(2)\right] = 3 - 0 = 3\)
- \(g(-2) = 3 + \int_0^{-2} g'(t)\,dt = 3 - \int_{-2}^0 g'(t)\,dt = 3 - \left[-\frac{1}{2}(2)(2)\right] = 3 - (-2) = 5\)
- \(g(2) = 3 + \int_0^2 g'(t)\,dt = 3 + \left[-\frac{1}{2}(2)(2)\right] = 3 - 2 = 1\)
- \(g(6) = g(2) + \int_2^6 g'(t)\,dt = 1 + \frac{1}{2}\pi(2)^2 = 1 + 2\pi\)

Comparing values:
- Absolute minimum value is \(1\) (at \(x = 2\)).
- Absolute maximum value is \(1 + 2\pi\) (at \(x = 6\)).

Part (a): 2 points
- 1 point for identifying \(x = -2\) and \(x = 2\) as locations of horizontal tangents.
- 1 point for correct classifications of relative maximum at \(x = -2\) and relative minimum at \(x = 2\) with justifications based on the sign change of \(g'\).

Part (b): 2 points
- 1 point for identifying the interval \((-2, 0)\).
- 1 point for valid reasoning that \(g'(x) < 0\) (or \(g\) is decreasing) and \(g'\) is decreasing (or \(g'' < 0\)).

Part (c): 2 points
- 1 point for verifying conditions for and applying L'Hospital's Rule (evaluating separate limits of numerator and denominator to show \(0/0\)).
- 1 point for the correct limit answer of \(-1\).

Part (d): 3 points
- 1 point for considering critical points (\(x = -2, 2\)) and endpoints (\(x = -4, 6\)).
- 1 point for correctly evaluating \(g(x)\) at the candidates using geometric areas.
- 1 point for identifying the correct absolute minimum value of \(1\) and absolute maximum value of \(1 + 2\pi\).

(a) The area of the region is given by the integral of the upper function minus the lower function:
\[ \text{Area} = \int_0^4 (f(x) - g(x))\,dx = \int_0^4 f(x)\,dx - \int_0^4 \frac{10}{2x+1}\,dx \]
Evaluating the integral of \( g(x) \):
\[ \int_0^4 \frac{10}{2x+1}\,dx = \left[ 5\ln|2x+1| \right]_0^4 = 5\ln(9) - 5\ln(1) = 5\ln 9 = 10\ln 3 \]
Therefore, the area is:
\[ \text{Area} = 24 - [5\ln 9 - 0] = 24 - 5\ln 9 = 24 - 10\ln 3 \]

(b) The improper integral is evaluated using a limit:
\[ \int_0^\infty (g(x))^2\,dx = \lim_{b\to\infty} \int_0^b \left( \frac{10}{2x+1} \right)^2\,dx = \lim_{b\to\infty} \int_0^b 100(2x+1)^{-2}\,dx \]
Finding the antiderivative:
\[ \int 100(2x+1)^{-2}\,dx = 100 \cdot \frac{(2x+1)^{-1}}{-1 \cdot 2} = -\frac{50}{2x+1} \]
Evaluating across limits:
\[ \lim_{b\to\infty} \left[ -\frac{50}{2x+1} \right]_0^b = \lim_{b\to\infty} \left( -\frac{50}{2b+1} - \left(-\frac{50}{1}\right) \right) = 0 + 50 = 50 \]
Thus, the integral converges to \( 50 \).

(c) To find \( \int_0^4 h(x)\,dx = \int_0^4 x f'(x)\,dx \), use integration by parts:
Let \( u = x \) and \( dv = f'(x)\,dx \).
Then \( du = dx \) and \( v = f(x) \).
\[ \int_0^4 x f'(x)\,dx = \left[ x f(x) \right]_0^4 - \int_0^4 f(x)\,dx \]
\[ = \left( 4 f(4) - 0 \cdot f(0) \right) - \int_0^4 f(x)\,dx \]
\[ = 4(5) - 0 - 24 = 20 - 24 = -4 \]

Part (a): 3 points
• 1 point for setting up the integrand \( f(x) - g(x) \) in a definite integral from 0 to 4.
• 1 point for finding the correct antiderivative of \( g(x) \), namely \( 5\ln|2x+1| \) or equivalent.
• 1 point for the correct numerical answer (e.g., \( 24 - 10\ln 3 \) or \( 24 - 5\ln 9 \)).

Part (b): 3 points
• 1 point for properly expressing the improper integral using limit notation: \( \lim_{b\to\infty} \int_0^b (g(x))^2\,dx \).
• 1 point for finding the correct antiderivative of \( (g(x))^2 \), namely \( -\frac{50}{2x+1} \).
• 1 point for the final answer of \( 50 \).

Part (c): 3 points
• 1 point for identifying parts \( u = x \) and \( dv = f'(x)\,dx \).
• 1 point for the correct integration by parts formulation: \( x f(x) - \int f(x)\,dx \).
• 1 point for the correct final value of \( -4 \).

(a) Differentiating \(f''(x) = 2x + \cos(3x) \cdot f'(x)\) with respect to \(x\) using the product rule and chain rule:
\[f'''(x) = \frac{d}{dx}\left[2x + \cos(3x) f'(x)\right] = 2 - 3\sin(3x)f'(x) + \cos(3x)f''(x)\]
Evaluating the derivatives at \(x = 0\):
\[f(0) = 4\]
\[f'(0) = -1\]
\[f''(0) = 2(0) + \cos(0) \cdot f'(0) = 0 + 1(-1) = -1\]
\[f'''(0) = 2 - 3\sin(0)f'(0) + \cos(0)f''(0) = 2 - 0 + 1(-1) = 1\]
The third-degree Taylor polynomial for \(f\) about \(x = 0\) is:
\[T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3\]
\[T_3(x) = 4 - x - \frac{1}{2}x^2 + \frac{1}{6}x^3\]

(b) By the Lagrange error bound:
\[|T_3(0.2) - f(0.2)| \le \frac{\max_{0 \le c \le 0.2} |f^{(4)}(c)|}{4!} |0.2 - 0|^4\]
Given \(|f^{(4)}(x)| \le 18\) for \(0 \le x \le 0.5\):
\[|T_3(0.2) - f(0.2)| \le \frac{18}{24} (0.2)^4 = \frac{3}{4} (0.0016) = 0.0012\]
Thus, the approximation is within \(0.0012\) of the exact value of \(f(0.2)\).

(c) Method 1 (Using Derivatives):
\[g(x) = f(x)\sin(2x) \implies g(0) = f(0)\sin(0) = 0\]
\[g'(x) = f'(x)\sin(2x) + 2f(x)\cos(2x)\]
\[g'(0) = f'(0)\sin(0) + 2f(0)\cos(0) = (-1)(0) + 2(4)(1) = 8\]
\[g''(x) = f''(x)\sin(2x) + 2f'(x)\cos(2x) + 2f'(x)\cos(2x) - 4f(x)\sin(2x) = f''(x)\sin(2x) + 4f'(x)\cos(2x) - 4f(x)\sin(2x)\]
\[g''(0) = f''(0)\sin(0) + 4f'(0)\cos(0) - 4f(0)\sin(0) = 0 + 4(-1)(1) - 0 = -4\]
The second-degree Taylor polynomial for \(g\) about \(x = 0\) is:
\[T_2(x) = g(0) + g'(0)x + \frac{g''(0)}{2!}x^2 = 8x + \frac{-4}{2}x^2 = 8x - 2x^2\]

Method 2 (Using Series Multiplication):
\[f(x) = 4 - x + \dots\]
\[\sin(2x) = 2x - \frac{(2x)^3}{3!} + \dots = 2x - \frac{4}{3}x^3 + \dots\]
\[g(x) = (4 - x + \dots)(2x + \dots) = 8x - 2x^2 + \dots\]
Thus, \(T_2(x) = 8x - 2x^2\).

Part (a): 4 points
- 1 point for applying the product rule and chain rule to find \(f'''(x)\).
- 1 point for the correct expression for \(f'''(x)\).
- 1 point for evaluating \(f''(0) = -1\) and \(f'''(0) = 1\).
- 1 point for the third-degree Taylor polynomial \(T_3(x) = 4 - x - \frac{1}{2}x^2 + \frac{1}{6}x^3\).

Part (b): 2 points
- 1 point for the form of the Lagrange error bound: \(\frac{\max |f^{(4)}(x)|}{4!} (0.2)^4\) or \(\frac{18}{4!} (0.2)^4\).
- 1 point for demonstrating that \(\text{Error} \le \frac{18}{24}(0.2)^4 = 0.0012\) (or \(\frac{3}{2500}\)).

Part (c): 3 points
- 1 point for finding \(g'(0) = 8\) or showing the first-degree term \(8x\).
- 1 point for finding \(g''(0) = -4\) or showing the second-degree coefficient via series multiplication.
- 1 point for the Taylor polynomial \(T_2(x) = 8x - 2x^2\).