2024 AP AP Calculus BC 模拟试题及答案详解

(a) The total amount of water entering the cistern for \(0 \le t \le 6\) is given by:
\[ \int_0^6 R(t)\,dt = \int_0^6 \left(10\sin\left(\frac{t^2}{12}\right) + 15\right)dt \approx 126.657 \text{ gallons} \]

(b) The rate of change of the amount of water in the cistern is \(A'(t) = R(t) - W(t)\).
At \(t = 2\):
\[ R(2) = 10\sin\left(\frac{4}{12}\right) + 15 \approx 18.272 \]
\[ W(2) = 3\sqrt{2^3 + 4} = 3\sqrt{12} \approx 10.392 \]
Since \(A'(2) = R(2) - W(2) = 18.272 - 10.392 = 7.880 > 0\), the amount of water in the cistern is increasing at \(t = 2\).

(c) The amount of water in the cistern at time \(t = 8\) is:
\[ A(8) = 80 + \int_0^8 \left(R(t) - W(t)\right)dt \]
\[ A(8) = 80 + \int_0^8 \left(10\sin\left(\frac{t^2}{12}\right) + 15 - 3\sqrt{t^3 + 4}\right)dt \approx 80 + 39.166 = 119.166 \text{ gallons} \]

(d) The absolute maximum of \(A(t)\) on \([0, 8]\) can occur at endpoints or critical points where \(A'(t) = R(t) - W(t) = 0\).
Setting \(R(t) - W(t) = 0\) on \(0 \le t \le 8\):
\[ 10\sin\left(\frac{t^2}{12}\right) + 15 - 3\sqrt{t^3 + 4} = 0 \implies t \approx 3.998 \]
Since \(A'(t) = R(t) - W(t) > 0\) for \(0 \le t < 3.998\) and \(A'(t) = R(t) - W(t) < 0\) for \(3.998 < t \le 8\), the amount of water in the cistern is increasing on \([0, 3.998]\) and decreasing on \([3.998, 8]\).
Therefore, the amount of water in the cistern reaches its absolute maximum at time \(t = 3.998\) hours (or \(t = 3.997\)).

(a) 2 points:
- 1 point for definite integral setup \(\int_0^6 R(t)\,dt\)
- 1 point for answer \(126.657\) (or \(126.658\))

(b) 1 point:
- 1 point for conclusion with reason (evaluates \(R(2) - W(2) > 0\) or compares \(R(2) > W(2)\))

(c) 3 points:
- 1 point for integrand \(R(t) - W(t)\) in a definite integral
- 1 point for uses initial condition (adds \(80\))
- 1 point for final answer \(119.166\)

(d) 3 points:
- 1 point for setting \(R(t) - W(t) = 0\)
- 1 point for interior critical point \(t = 3.998\) (or \(3.997\))
- 1 point for justification and answer

(a) The speed of the particle at time \(t = 2\) is given by:
\[ \text{Speed} = \sqrt{(x'(2))^2 + (y'(2))^2} \]
\[ x'(2) = (2+1)\cos\left(\frac{2^2}{4}\right) = 3\cos(1) \approx 1.620907 \]
\[ y'(2) = 3\sqrt{2} - e^{0.4} \approx 2.750849 \]
\[ \text{Speed} = \sqrt{(1.620907)^2 + (2.750849)^2} \approx 3.193 \]

(b) The acceleration vector is \(\vec{a}(t) = \langle x''(t), y''(t) \rangle\).
Using numerical differentiation at \(t = 2\):
\[ x''(2) = \left.\frac{d}{dt}\left[(t+1)\cos\left(\frac{t^2}{4}\right)\right]\right|_{t=2} = \cos(1) - 3\sin(1) \approx -4.509 \]
\[ y''(2) = \left.\frac{d}{dt}\left[3\sqrt{t} - e^{0.2t}\right]\right|_{t=2} = \frac{3}{2\sqrt{2}} - 0.2e^{0.4} \approx 0.762 \]
\[ \vec{a}(2) = \langle -4.509, 0.762 \rangle \]

(c) The total distance traveled over \(0 \le t \le 3\) is given by the integral of speed:
\[ \text{Total distance} = \int_0^3 \sqrt{(x'(t))^2 + (y'(t))^2}\,dt \]
\[ = \int_0^3 \sqrt{\left((t+1)\cos\left(\frac{t^2}{4}\right)\right)^2 + \left(3\sqrt{t} - e^{0.2t}\right)^2} dt \approx 10.457 \]

(d) By the Fundamental Theorem of Calculus:
\[ y(0) = y(2) + \int_2^0 y'(t)\,dt = 5 + \int_2^0 \left(3\sqrt{t} - e^{0.2t}\right)dt \]
\[ = 5 - \int_0^2 \left(3\sqrt{t} - e^{0.2t}\right)dt = 5 - 3.208226 = 1.792 \]

(e) The slope of the tangent line at \(t = 2\) is:
\[ \frac{dy}{dx} = \frac{y'(2)}{x'(2)} = \frac{2.750849}{1.620907} \approx 1.697 \]
Since the particle is at \((3, 5)\) at \(t = 2\), the equation of the tangent line is:
\[ y - 5 = 1.697(x - 3) \]

(a) 2 points:
- 1 point for speed setup \(\sqrt{(x'(2))^2 + (y'(2))^2}\)
- 1 point for answer \(3.193\)

(b) 1 point:
- 1 point for acceleration vector \(\langle -4.509, 0.762 \rangle\)

(c) 2 points:
- 1 point for integral setup \(\int_0^3 \sqrt{(x'(t))^2 + (y'(t))^2}\,dt\)
- 1 point for answer \(10.457\)

(d) 2 points:
- 1 point for definite integral setup \(\int_2^0 y'(t)\,dt\) or \(\int_0^2 y'(t)\,dt\)
- 1 point for using initial condition \(y(2)=5\) and finding \(y(0) = 1.792\)

(e) 2 points:
- 1 point for slope \(\frac{dy}{dx} = \frac{y'(2)}{x'(2)} \approx 1.697\)
- 1 point for equation of tangent line

(a) For \(f(0) = 7 > 4\), since solutions do not cross the horizontal asymptote \(y = 4\), \(y - 4 > 0\) for the domain considered.
Setting \(\frac{dy}{dt} = 0\):
\[\frac{1}{3}(y - 4)\cos\left(\frac{t}{3}\right) = 0 \implies \cos\left(\frac{t}{3}\right) = 0\]
For \(0 < t < 2\pi\), \(0 < \frac{t}{3} < \frac{2\pi}{3}\), so \(\frac{t}{3} = \frac{\pi}{2} \implies t = \frac{3\pi}{2}\).

To determine the type of extremum:
- For \(0 < t < \frac{3\pi}{2}\), \(0 < \frac{t}{3} < \frac{\pi}{2}\), so \(\cos\left(\frac{t}{3}\right) > 0\) and \(\frac{dy}{dt} > 0\).
- For \(\frac{3\pi}{2} < t < 2\pi\), \(\frac{\pi}{2} < \frac{t}{3} < \frac{2\pi}{3}\), so \(\cos\left(\frac{t}{3}\right) < 0\) and \(\frac{dy}{dt} < 0\).

Since \(\frac{dy}{dt}\) changes from positive to negative at \(t = \frac{3\pi}{2}\), \(f\) has a relative maximum at \(t = \frac{3\pi}{2}\).

(b) At \(t = 0\), \(y = 7\):
\[\left.\frac{dy}{dt}\right|_{(0,7)} = \frac{1}{3}(7 - 4)\cos(0) = \frac{1}{3}(3)(1) = 1\]
The tangent line is \(y - 7 = 1(t - 0) \implies y = t + 7\).
\[f(0.3) \approx 0.3 + 7 = 7.3\]

(c) Differentiating \(\frac{dy}{dt} = \frac{1}{3}(y - 4)\cos\left(\frac{t}{3}\right)\) with respect to \(t\) using the product rule and chain rule:
\[\frac{d^2y}{dt^2} = \frac{1}{3}\frac{dy}{dt}\cos\left(\frac{t}{3}\right) + \frac{1}{3}(y - 4)\left(-\frac{1}{3}\sin\left(\frac{t}{3}\right)\right)\]
Substitute \(\frac{dy}{dt} = \frac{1}{3}(y - 4)\cos\left(\frac{t}{3}\right)\):
\[\frac{d^2y}{dt^2} = \frac{1}{9}(y - 4)\cos^2\left(\frac{t}{3}\right) - \frac{1}{9}(y - 4)\sin\left(\frac{t}{3}\right)\]

(d) Separating variables:
\[\frac{1}{y - 4}\,dy = \frac{1}{3}\cos\left(\frac{t}{3}\right)\,dt\]
Integrating both sides:
\[\int \frac{1}{y - 4}\,dy = \int \frac{1}{3}\cos\left(\frac{t}{3}\right)\,dt\]
\[\ln|y - 4| = \sin\left(\frac{t}{3}\right) + C\]
Using the initial condition \(f(0) = 7\):
\[\ln|7 - 4| = \sin(0) + C \implies \ln 3 = C\]
Because \(y(0) = 7 > 4\), \(|y - 4| = y - 4\):
\[\ln(y - 4) = \sin\left(\frac{t}{3}\right) + \ln 3\]
\[y - 4 = e^{\sin(t/3) + \ln 3} = 3e^{\sin(t/3)}\]
\[y = f(t) = 4 + 3e^{\sin(t/3)}\]

Part (a): 2 points
- 1 point: Identifies \(t = \frac{3\pi}{2}\)
- 1 point: Relative maximum with justification (sign change of \(\frac{dy}{dt}\))

Part (b): 2 points
- 1 point: Tangent line equation
- 1 point: Approximation \(7.3\)

Part (c): 1 point
- 1 point: \(\frac{d^2y}{dt^2}\) in terms of \(y\) and \(t\)

Part (d): 4 points
- 1 point: Separation of variables
- 1 point: Antiderivatives
- 1 point: Constant of integration and uses initial condition \((0,7)\)
- 1 point: Solves for \(y\) explicitly

(a)
\[g(-4) = \int_{-2}^{-4} f(t)\,dt = -\int_{-4}^{-2} f(t)\,dt\]
The region under \(f\) from \(x = -4\) to \(x = -2\) is a triangle with base \(2\) and height \(3\):
\[\text{Area} = \frac{1}{2}(2)(3) = 3 \implies g(-4) = -3\]

\[g(2) = \int_{-2}^2 f(t)\,dt = \int_{-2}^0 f(t)\,dt + \int_0^2 f(t)\,dt\]
From \(x = -2\) to \(x = 0\), the area is a trapezoid with bases \(3\) and \(2\) and height \(2\):
\[\text{Area} = \frac{3 + 2}{2} \cdot 2 = 5\]
From \(x = 0\) to \(x = 2\), the region is a quarter-circle of radius \(2\):
\[\text{Area} = \frac{1}{4}\pi(2^2) = \pi\]
Thus, \(g(2) = 5 + \pi\).

(b) By the Fundamental Theorem of Calculus, \(g'(x) = f(x)\).
Setting \(g'(x) = f(x) = 0\) on \([-4, 6]\):
\(f(x) = 0\) at \(x = -4\) and \(x = 2\).

Candidates for absolute extrema on \([-4, 6]\) are the critical points and endpoints: \(x = -4, 2, 6\).
- \(g(-4) = -3\)
- \(g(2) = 5 + \pi\)
- \(g(6) = g(2) + \int_2^6 f(t)\,dt\)
From \(x = 2\) to \(x = 6\), the region is a triangle below the \(x\)-axis with base \(4\) and height \(-4\):
\[\int_2^6 f(t)\,dt = -\frac{1}{2}(4)(4) = -8 \implies g(6) = 5 + \pi - 8 = \pi - 3 \approx 0.142\]
Comparing the values:
\(g(-4) = -3\), \(g(2) = 5 + \pi\), \(g(6) = \pi - 3\).
The absolute maximum value of \(g\) on \([-4, 6]\) is \(5 + \pi\) (at \(x = 2\)).

(c) The graph of \(g\) has a point of inflection where \(g'(x) = f(x)\) changes from increasing to decreasing or from decreasing to increasing.
- At \(x = -2\), \(f\) changes from increasing (slope \(1.5\)) to decreasing (slope \(-0.5\)).
- At \(x = 0\), \(f\) changes from decreasing linearly to decreasing along the circular arc, but let's look at slope: from \(x = -2\) to \(0\), \(f'(x) = -0.5\); for \(0 < x < 2\), \(f'(x) = -\frac{x}{\sqrt{4-x^2}}\), which is decreasing (slope becomes more negative from \(0\) to \(-\infty\)). Thus \(f\) continues decreasing across \(x = 0\).
Wait, let's re-verify the slope behavior on \([0,2]\): \(f(x) = \sqrt{4-x^2}\), \(f'(x) = \frac{-x}{\sqrt{4-x^2}}\). \(f''(x) = -\frac{4}{(4-x^2)^{3/2}} < 0\), so \(f'\) is decreasing on \((0,2)\).
On \([-2, 0]\), \(f'(x) = -\frac{1}{2}\).
On \([2, 6]\), the line has slope \(\frac{-4 - 0}{6 - 2} = -1\). So at \(x = 2\), \(f'\) changes from \(-\infty\) to \(-1\) (increasing).
Thus \(f\) changes direction of increase/decrease only at \(x = -2\) (increasing to decreasing).

(d)
\[h'(x) = \frac{d}{dx}[x \cdot g(x)] = 1 \cdot g(x) + x \cdot g'(x) = g(x) + x f(x)\]
At \(x = 2\):
\[h'(2) = g(2) + 2 f(2) = (5 + \pi) + 2(0) = 5 + \pi\]

Part (a): 2 points
- 1 point: \(g(-4) = -3\)
- 1 point: \(g(2) = 5 + \pi\)

Part (b): 3 points
- 1 point: Identifies \(x = 2\) as critical point (sets \(g'(x) = f(x) = 0\))
- 1 point: Considers values at critical points and endpoints
- 1 point: Answer \(5 + \pi\) with justification

Part (c): 2 points
- 1 point: Identifies \(x = -2\)
- 1 point: Reason mentioning \(g'(x) = f(x)\) changes from increasing to decreasing

Part (d): 2 points
- 1 point: Product rule for \(h'(x)\)
- 1 point: Value \(h'(2) = 5 + \pi\)

(a) Step size \(\Delta x = \frac{5 - 1}{2} = 2\).

Step 1:
\[x_0 = 1, \quad y_0 = f(1) = 2\]
\[f(3) \approx y_1 = y_0 + f'(1)\Delta x = 2 + (4)(2) = 10\]

Step 2:
\[x_1 = 3, \quad y_1 = 10\]
\[f(5) \approx y_2 = y_1 + f'(3)\Delta x = 10 + (-2)(2) = 10 - 4 = 6\]

(b) By the Fundamental Theorem of Calculus:
\[k'(x) = \sqrt{1 + [f'(x)]^2}\]
Evaluating at \(x = 3\):
\[k'(3) = \sqrt{1 + [f'(3)]^2} = \sqrt{1 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\]

(c) The expression \(\int_1^5 \sqrt{1 + [f'(x)]^2}\,dx\) represents the arc length (or length) of the graph of \(y = f(x)\) over the interval \([1, 5]\).

(d) Use integration by parts:
Let \(u = 2t + 1 \implies du = 2\,dt\).
Let \(dv = e^{-3t}\,dt \implies v = -\frac{1}{3}e^{-3t}\).

Using \(\int u\,dv = uv - \int v\,du\):
\[\int (2t + 1)e^{-3t}\,dt = (2t + 1)\left(-\frac{1}{3}e^{-3t}\right) - \int \left(-\frac{1}{3}e^{-3t}\right)(2\,dt)\]
\[= -\frac{1}{3}(2t + 1)e^{-3t} + \frac{2}{3}\int e^{-3t}\,dt\]
\[= -\frac{1}{3}(2t + 1)e^{-3t} - \frac{2}{9}e^{-3t} + C\]
\[\left(\text{or } -\left(\frac{2}{3}t + \frac{5}{9}\right)e^{-3t} + C\right)\]

Part (a): 2 points
- 1 point: Euler's method setup with 2 steps
- 1 point: Approximation \(f(5) \approx 6\)

Part (b): 2 points
- 1 point: Fundamental Theorem of Calculus applied to find \(k'(x)\)
- 1 point: Value \(\sqrt{5}\)

Part (c): 2 points
- 1 point: Arc length of the graph of \(f\)
- 1 point: Specifies interval \([1, 5]\)

Part (d): 3 points
- 1 point: Identifies \(u\) and \(dv\) correctly
- 1 point: Correct application of integration by parts formula
- 1 point: Correct antiderivative including constant of integration \(+ C\)

(a) Apply the Ratio Test:
Let \(a_n = \frac{(n+2)x^n}{n \cdot 4^n}\).
\[\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+3)x^{n+1}}{(n+1)4^{n+1}} \cdot \frac{n \cdot 4^n}{(n+2)x^n}\right| = \left(\frac{n(n+3)}{(n+1)(n+2)}\right) \frac{|x|}{4}\]
Evaluating the limit as \(n \to \infty\):
\[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left(\frac{n^2 + 3n}{n^2 + 3n + 2}\right) \frac{|x|}{4} = (1) \frac{|x|}{4} = \frac{|x|}{4}\]
By the Ratio Test, the series converges when \(\frac{|x|}{4} < 1 \implies |x| < 4\).
Thus, the radius of convergence is \(R = 4\).

(b) At \(x = 4\), the series is:
\[\sum_{n=1}^\infty \frac{(n+2)4^n}{n \cdot 4^n} = \sum_{n=1}^\infty \frac{n+2}{n}\]
Since \(\lim_{n \to \infty} \frac{n+2}{n} = 1 \ne 0\), the series diverges by the \(n\)th-term Divergence Test.
(Alternatively: \(\frac{n+2}{n} > \frac{1}{n}\) for all \(n \ge 1\), and \(\sum \frac{1}{n}\) is the divergent harmonic series, so it diverges by the direct comparison test).

(c) At \(x = -4\), the series is:
\[\sum_{n=1}^\infty \frac{(n+2)(-4)^n}{n \cdot 4^n} = \sum_{n=1}^\infty (-1)^n \frac{n+2}{n}\]
Consider the limit of the terms:
\[\lim_{n \to \infty} \left|(-1)^n \frac{n+2}{n}\right| = \lim_{n \to \infty} \frac{n+2}{n} = 1 \ne 0\]
Since \(\lim_{n \to \infty} (-1)^n \frac{n+2}{n} \ne 0\), the series diverges at \(x = -4\) by the \(n\)th-term Divergence Test.

(d) Differentiating term-by-term:
General term:
\[\frac{d}{dx}\left[\frac{(n+2)x^n}{n \cdot 4^n}\right] = \frac{n(n+2)x^{n-1}}{n \cdot 4^n} = \frac{(n+2)x^{n-1}}{4^n}\]
First three nonzero terms (for \(n = 1, 2, 3\)):
- \(n = 1\): \(\frac{(1+2)x^0}{4^1} = \frac{3}{4}\)
- \(n = 2\): \(\frac{(2+2)x^1}{4^2} = \frac{4x}{16} = \frac{x}{4}\) (or \(\frac{4x}{16}\))
- \(n = 3\): \(\frac{(3+2)x^2}{4^3} = \frac{5x^2}{64}\)

Thus, \(f'(x) = \frac{3}{4} + \frac{x}{4} + \frac{5x^2}{64} + \dots + \frac{(n+2)x^{n-1}}{4^n} + \dots\)

Part (a): 3 points
- 1 point: Sets up ratio \(\left|\frac{a_{n+1}}{a_n}\right|\)
- 1 point: Evaluates limit \(\frac{|x|}{4}\)
- 1 point: Radius of convergence \(R = 4\)

Part (b): 2 points
- 1 point: Considers series at \(x = 4\)
- 1 point: Conclusion of divergence with valid reason (e.g., \(n\)th-term test)

Part (c): 2 points
- 1 point: Considers series at \(x = -4\)
- 1 point: Conclusion of divergence with reason (\(\lim_{n \to \infty} a_n \ne 0\))

Part (d): 2 points
- 1 point: First three nonzero terms
- 1 point: General term \(\frac{(n+2)x^{n-1}}{4^n}\)