2024 AP AP Calculus AB Practice Paper with Answers

(a)
$$B'(7) \approx \frac{B(10) - B(3)}{10 - 3} = \frac{122 - 185}{7} = \frac{-63}{7} = -9\text{ degrees Celsius per minute}$$

(b)
$$\int_0^{16} B(t)\,dt \approx (3 - 0) \cdot B(3) + (10 - 3) \cdot B(10) + (16 - 10) \cdot B(16)$$
$$= 3 \cdot 185 + 7 \cdot 122 + 6 \cdot 80 = 555 + 854 + 480 = 1889$$

$\frac{1}{16}\int_0^{16} B(t)\,dt$ represents the average temperature of the ceramic block (in degrees Celsius) over the time interval from $t = 0$ to $t = 16$ minutes.

(c)
$$B(30) = B(16) + \int_{16}^{30} B'(t)\,dt$$
$$B(30) = 80 + \int_{16}^{30} \frac{-36.8e^{0.02t}}{t}\,dt$$
$$= 80 - 27.647808 = 52.352192$$

The temperature of the ceramic block at time $t = 30$ is approximately $52.352$ degrees Celsius.

(d)
Because $t - 50 < 0$ for all $t$ in the interval $16 < t < 30$, while $0.736e^{0.02t} > 0$ and $t^2 > 0$, it follows that $B''(t) < 0$ on the interval $16 < t < 30$.

Because $B''(t) < 0$ on $16 < t < 30$, the rate of change of the temperature, $B'(t)$, is decreasing on this interval. Therefore, the temperature of the ceramic block is changing at a decreasing rate.

Part (a): 2 points
- 1 point for the estimate with supporting difference quotient $\frac{122 - 185}{10 - 3}$ or equivalent unsimplified form.
- 1 point for correct units of measure (degrees Celsius per minute or $^\circ\text{C}/\text{min}$).

Scoring notes for (a):
- The first point requires an explicit quotient and difference using values from the table.
- Units point can be earned even if the estimate arithmetic is slightly miscalculated, provided units are attached to a numerical rate.

Part (b): 3 points
- 1 point for form of right Riemann sum: $(3)(185) + (7)(122) + (6)(80)$ or equivalent.
- 1 point for value of the estimate ($1889$).
- 1 point for interpretation: must mention "average temperature" (or average value of temperature) and the specific interval $t = 0$ to $t = 16$ minutes.

Part (c): 3 points
- 1 point for the definite integral $\int_{16}^{30} B'(t)\,dt$.
- 1 point for using the initial condition $B(16) = 80$ in a valid setup: $80 + \int_{16}^{30} B'(t)\,dt$.
- 1 point for the answer ($52.352$).

Scoring notes for (c):
- An answer of $52.352$ with no supporting setup earns $0$ out of $3$ points.

Part (d): 1 point
- 1 point for answer "decreasing rate" with reason referencing the sign of the second derivative ($B''(t) < 0$) on the interval $16 < t < 30$.

(a) The particle changes direction when its velocity changes sign. Setting \(v(t) = 0\) on the interval \(0 < t < 4\):
\[ 3\sin(0.4t^2) - 0.5t - 1 = 0 \implies t \approx 1.18209 \text{ and } t \approx 2.40646 \]
On the interval \((0, 1.182)\), \(v(t) < 0\).
On the interval \((1.182, 2.406)\), \(v(t) > 0\).
On the interval \((2.406, 4)\), \(v(t) < 0\).
Because \(v(t)\) changes sign at \(t = 1.182\) (from negative to positive) and at \(t = 2.406\) (from positive to negative), the particle changes direction at \(t = 1.182\) and \(t = 2.406\).

(b) The acceleration of the particle is given by \(a(t) = v'(t)\).
At time \(t = 2\):
\[ a(2) = v'(2) = -0.64016 \approx -0.640 \]
Evaluating the velocity at time \(t = 2\):
\[ v(2) = 0.99871 \approx 0.999 > 0 \]
Because \(v(2)\) and \(a(2)\) have opposite signs (\(v(2) > 0\) and \(a(2) < 0\)), the speed of the particle is decreasing at time \(t = 2\).

(c) Using the Fundamental Theorem of Calculus:
\[ x(3) = x(0) + \int_0^3 v(t)\,dt \]
\[ x(3) = 5 + \int_0^3 \left(3\sin(0.4t^2) - 0.5t - 1\right) dt \]
\[ x(3) = 5 - 1.118747 = 3.881253 \]
The position of the particle at time \(t = 3\) is \(3.881\) (or \(3.882\)).

(d) The total distance traveled by the particle over the interval \(0 \le t \le 3\) is given by the integral of speed:
\[ \text{Total distance} = \int_0^3 |v(t)|\,dt = \int_0^3 |3\sin(0.4t^2) - 0.5t - 1|\,dt \approx 3.635 \]

Part (a): 2 points
- 1 point for setting \(v(t) = 0\) and identifying \(t = 1.182\) and \(t = 2.406\)
- 1 point for justification based on the sign change of \(v(t)\)

Part (b): 2 points
- 1 point for acceleration setup and value \(a(2) = v'(2) = -0.640\)
- 1 point for concluding speed is decreasing with an explanation comparing the signs of \(v(2)\) and \(a(2)\)

Part (c): 3 points
- 1 point for the definite integral \(\int_0^3 v(t)\,dt\)
- 1 point for using the initial condition \(x(0) = 5\)
- 1 point for the final answer \(x(3) = 3.881\) (or \(3.882\))

Part (d): 2 points
- 1 point for the integral setup \(\int_0^3 |v(t)|\,dt\)
- 1 point for the answer \(3.635\)

(a) Evaluate the slope of the tangent line at \((2, 7)\):
\[\left.\frac{dy}{dx}\right|_{(2,7)} = \frac{2(7 - 5)}{2} = 2\]
An equation of the tangent line is:
\[y - 7 = 2(x - 2) \implies y = 7 + 2(x - 2)\]
Using the tangent line at \(x = 2.1\):
\[f(2.1) \approx 7 + 2(2.1 - 2) = 7 + 2(0.1) = 7.2\]

(b) Differentiating \(\frac{dy}{dx} = \frac{1}{2}x(y - 5)\) with respect to \(x\) using the product rule:
\[\frac{d^2y}{dx^2} = \frac{1}{2}(y - 5) + \frac{1}{2}x\frac{dy}{dx} = \frac{1}{2}(y - 5) + \frac{1}{2}x\left(\frac{x(y - 5)}{2}\right) = \frac{1}{2}(y - 5) + \frac{x^2(y - 5)}{4}\]
At the point \((2, 7)\):
\[\left.\frac{d^2y}{dx^2}\right|_{(2,7)} = \frac{1}{2}(7 - 5) + \frac{1}{2}(2)(2) = 1 + 2 = 3 > 0\]
Since \(\frac{d^2y}{dx^2} > 0\) on an interval containing \(x = 2\), the graph of \(f\) is concave up near \(x = 2\). Therefore, the tangent line lies below the graph of \(f\), and the approximation \(7.2\) is an underestimate.

(c) Separate the variables in \(\frac{dy}{dx} = \frac{x(y - 5)}{2}\):
\[\frac{1}{y - 5}\,dy = \frac{x}{2}\,dx\]
Integrate both sides:
\[\int \frac{1}{y - 5}\,dy = \int \frac{x}{2}\,dx\]
\[\ln |y - 5| = \frac{x^2}{4} + C\]
Apply the initial condition \(f(2) = 7\):
\[\ln |7 - 5| = \frac{2^2}{4} + C \implies \ln 2 = 1 + C \implies C = \ln 2 - 1\]
Substitute \(C\) back into the equation:
\[\ln |y - 5| = \frac{x^2}{4} + \ln 2 - 1\]
Since \(y(2) = 7 > 5\), \(|y - 5| = y - 5\):
\[y - 5 = e^{\frac{x^2}{4} - 1 + \ln 2} = e^{\ln 2} \cdot e^{\frac{x^2 - 4}{4}} = 2e^{\frac{x^2 - 4}{4}}\]
\[y = 5 + 2e^{\frac{x^2 - 4}{4}}\]

Part (a): 2 points
- 1 point: Tangent line equation (or slope at \(x = 2\))
- 1 point: Approximation of \(f(2.1)\)

Part (b): 2 points
- 1 point: \(\frac{d^2y}{dx^2}\) expression or evaluated at \((2, 7)\)
- 1 point: Underestimate with justification based on concavity (\(\frac{d^2y}{dx^2} > 0\))

Part (c): 5 points
- 1 point: Separation of variables
- 2 points: Antiderivatives (1 point for \(\ln|y-5|\), 1 point for \(\frac{x^2}{4}\))
- 1 point: Constant of integration and uses initial condition \(f(2) = 7\)
- 1 point: Solves for \(y\) explicitly in terms of \(x\)

Scoring Notes:
- A response with no separation of variables earns 0 out of 5 points in part (c).
- In part (c), a response without a constant of integration earns at most 3 of the 5 points.

(a)
\( g(-4) = \int_{0}^{-4} f(t)\, dt = -\int_{-4}^{0} f(t)\, dt \)
On the interval \([-4, 0]\), the graph of \( f \) is a line with \( t \)-intercept at \( t = -2 \).
\( \int_{-4}^{-2} f(t)\, dt = -\frac{1}{2}(2)(3) = -3 \)
\( \int_{-2}^{0} f(t)\, dt = \frac{1}{2}(2)(3) = 3 \)
Thus, \( \int_{-4}^{0} f(t)\, dt = -3 + 3 = 0 \implies g(-4) = -0 = 0 \).

\( g(6) = \int_{0}^{6} f(t)\, dt = \int_{0}^{2} f(t)\, dt + \int_{2}^{6} f(t)\, dt \)
On \([0, 2]\), the area under the line segment is a triangle of base \( 2 \) and height \( 3 \):
\( \int_{0}^{2} f(t)\, dt = \frac{1}{2}(2)(3) = 3 \)
On \([2, 6]\), the region under the semicircle of radius \( r = 2 \) has area:
\( \int_{2}^{6} f(t)\, dt = \frac{1}{2}\pi(2)^2 = 2\pi \)
Therefore, \( g(6) = 3 + 2\pi \).

(b)
By the Fundamental Theorem of Calculus, \( g'(x) = f(x) \).
The critical points of \( g \) on \((-4, 6)\) occur where \( g'(x) = f(x) = 0 \), which are \( x = -2 \) and \( x = 2 \).
Evaluating \( g(x) \) at the critical points and endpoints:
• \( g(-4) = 0 \)
• \( g(-2) = \int_{0}^{-2} f(t)\, dt = -\int_{-2}^{0} f(t)\, dt = -3 \)
• \( g(2) = \int_{0}^{2} f(t)\, dt = 3 \)
• \( g(6) = 3 + 2\pi \)

Comparing these values, the absolute minimum value of \( g \) on \([-4, 6]\) is \( -3 \) (occurring at \( x = -2 \)).

(c)
Using the chain rule and the Fundamental Theorem of Calculus:
\( p'(x) = 2g(x)g'(x) - 3f'(x) = 2g(x)f(x) - 3f'(x) \)
At \( x = 1 \):
• \( g(1) = \int_{0}^{1} f(t)\, dt = \int_{0}^{1} \left(-\frac{3}{2}t + 3\right) dt = \left[ -\frac{3}{4}t^2 + 3t \right]_{0}^{1} = -\frac{3}{4} + 3 = \frac{9}{4} \)
• \( f(1) = -\frac{3}{2}(1) + 3 = \frac{3}{2} \)
• Since \( f \) is linear on \((0, 2)\), \( f'(1) = -\frac{3}{2} \)

\( p'(1) = 2\left(\frac{9}{4}\right)\left(\frac{3}{2}\right) - 3\left(-\frac{3}{2}\right) = \frac{27}{4} + \frac{9}{2} = \frac{45}{4} \).

(d)
Since \( g(0) = 0 \), we have:
\( \lim_{x \to 0} (g(x) - 3x) = 0 - 0 = 0 \) and \( \lim_{x \to 0} x^2 = 0 \).
Because this produces the indeterminate form \( \frac{0}{0} \), L'Hospital's Rule can be applied:
\( \lim_{x \to 0} \frac{g(x) - 3x}{x^2} = \lim_{x \to 0} \frac{g'(x) - 3}{2x} = \lim_{x \to 0} \frac{f(x) - 3}{2x} \)

Evaluating the one-sided limits:
• From the left (\( x < 0 \)), \( f(x) = \frac{3}{2}x + 3 \):
\( \lim_{x \to 0^-} \frac{\left(\frac{3}{2}x + 3\right) - 3}{2x} = \lim_{x \to 0^-} \frac{\frac{3}{2}x}{2x} = \frac{3}{4} \)
• From the right (\( x > 0 \)), \( f(x) = -\frac{3}{2}x + 3 \):
\( \lim_{x \to 0^+} \frac{\left(-\frac{3}{2}x + 3\right) - 3}{2x} = \lim_{x \to 0^+} \frac{-\frac{3}{2}x}{2x} = -\frac{3}{4} \)

Since \( \lim_{x \to 0^-} \frac{f(x) - 3}{2x} \ne \lim_{x \to 0^+} \frac{f(x) - 3}{2x} \), \( \lim_{x \to 0} \frac{f(x) - 3}{2x} \) does not exist.
Therefore, \( \lim_{x \to 0} \frac{g(x) - 3x}{x^2} \) does not exist.

Part (a): 2 points
• 1 point for \( g(-4) = 0 \) with supporting geometric work
• 1 point for \( g(6) = 3 + 2\pi \) with supporting geometric work

Part (b): 3 points
• 1 point for setting \( g'(x) = f(x) = 0 \) to identify critical point candidates (\( x = -2 \) and \( x = 2 \))
• 1 point for correctly evaluating \( g(x) \) at candidates and endpoints (or using a sign chart / analysis of \( g' \))
• 1 point for answer of \( -3 \) with complete justification (Candidates Test or equivalent reasoning)

Part (c): 2 points
• 1 point for applying the product/chain rule to express \( p'(x) = 2g(x)f(x) - 3f'(x) \)
• 1 point for the correct numerical value \( p'(1) = \frac{45}{4} \)

Part (d): 2 points
• 1 point for applying L'Hospital's Rule to obtain \( \lim_{x \to 0} \frac{f(x) - 3}{2x} \)
• 1 point for finding unequal one-sided limits (\( \frac{3}{4} \) and \( -\frac{3}{4} \)) and concluding that the limit does not exist

(a) Evaluate the derivative at \((3, 2)\):
\[\left.\frac{dy}{dx}\right|_{(3,2)} = -\frac{4(3) + 2}{3 + 2(2)} = -\frac{14}{7} = -2\]
The equation of the tangent line at \((3, 2)\) is \(y - 2 = -2(x - 3)\), or \(y = 2 - 2(x - 3)\).
For \(x = 3.2\):
\[y \approx 2 - 2(3.2 - 3) = 2 - 2(0.2) = 1.6\]

(b) A horizontal tangent occurs when \(\frac{dy}{dx} = 0\), which requires \(-(4x + y) = 0 \implies y = -4x\) (with \(x + 2y \neq 0\)).
Substitute \(y = -4x\) into the original equation \(2x^2 + xy + y^2 = 28\):
\[2x^2 + x(-4x) + (-4x)^2 = 28\]
\[2x^2 - 4x^2 + 16x^2 = 28\]
\[14x^2 = 28 \implies x^2 = 2 \implies x = \pm\sqrt{2}\]
For \(x = \pm\sqrt{2}\), the denominator is \(x + 2y = x + 2(-4x) = -7x = \mp 7\sqrt{2} \neq 0\).
Thus, the \(x\)-coordinates of the points with horizontal tangent lines are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

(c) A vertical tangent occurs when the denominator of \(\frac{dy}{dx}\) is zero while the numerator is nonzero:
\[x + 2y = 0 \implies x = -2y\]
Substitute \(x = -2y\) into the original equation:
\[2(-2y)^2 + (-2y)y + y^2 = 28\]
\[8y^2 - 2y^2 + y^2 = 28\]
\[7y^2 = 28 \implies y^2 = 4 \implies y = \pm 2\]
For \(y = 2\), \(x = -4\) and the numerator is \(-(4(-4) + 2) = 14 \neq 0\).
For \(y = -2\), \(x = 4\) and the numerator is \(-(4(4) + (-2)) = -14 \neq 0\).
Thus, the \(y\)-coordinates of the points with vertical tangent lines are \(y = 2\) and \(y = -2\).

(d) Differentiating \(x^3 + 3xy^2 = 28\) implicitly with respect to \(t\):
\[3x^2\frac{dx}{dt} + 3y^2\frac{dx}{dt} + 6xy\frac{dy}{dt} = 0\]
Substitute \(x = 1\), \(y = 3\), and \(\frac{dx}{dt} = 4\):
\[3(1)^2(4) + 3(3)^2(4) + 6(1)(3)\frac{dy}{dt} = 0\]
\[12 + 108 + 18\frac{dy}{dt} = 0\]
\[120 + 18\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{120}{18} = -\frac{20}{3}\]
The rate of change of the \(y\)-coordinate with respect to time is \(-\frac{20}{3}\) units per second.

Part (a): 2 points
- 1 point for the slope of the tangent line, \(\left.\frac{dy}{dx}\right|_{(3,2)} = -2\)
- 1 point for the tangent line approximation, \(y \approx 1.6\) (or \(2 - 2(3.2 - 3)\))

Part (b): 2 points
- 1 point for setting \(4x + y = 0\) (or \(y = -4x\))
- 1 point for the answer, \(x = \pm\sqrt{2}\) (both roots required)

Part (c): 1 point
- 1 point for setting \(x + 2y = 0\) and finding \(y = \pm 2\)

Part (d): 4 points
- 1 point for attempting implicit differentiation with respect to \(t\) (product/chain rule)
- 1 point for the correct differentiated equation, \(3x^2\frac{dx}{dt} + 3y^2\frac{dx}{dt} + 6xy\frac{dy}{dt} = 0\)
- 1 point for correctly substituting \(x = 1\), \(y = 3\), and \(\frac{dx}{dt} = 4\)
- 1 point for the final answer, \(\frac{dy}{dt} = -\frac{20}{3}\)

(a) To find the points of intersection of \( f \) and \( g \):
\[ 3 + 2x - x^2 = 3 - x \implies 3x - x^2 = 0 \implies x(3 - x) = 0 \]
The curves intersect at \( x = 0 \) and \( x = 3 \). On the interval \([0, 3]\), \( f(x) \ge g(x) \).
\[ \text{Area} = \int_{0}^{3} (f(x) - g(x))\,dx = \int_{0}^{3} (3x - x^2)\,dx = \left[ \frac{3}{2}x^2 - \frac{1}{3}x^3 \right]_{0}^{3} = \left( \frac{27}{2} - 9 \right) - 0 = \frac{9}{2} \]

(b) The side length of each square cross section is \( s(x) = f(x) - g(x) = 3x - x^2 \).
The area of each cross section is \( A(x) = (s(x))^2 = (3x - x^2)^2 = 9x^2 - 6x^3 + x^4 \).
The volume of the solid is:
\[ \text{Volume} = \int_{0}^{3} A(x)\,dx = \int_{0}^{3} (9x^2 - 6x^3 + x^4)\,dx \]
\[ = \left[ 3x^3 - \frac{3}{2}x^4 + \frac{1}{5}x^5 \right]_{0}^{3} = \left( 3(27) - \frac{3}{2}(81) + \frac{1}{5}(243) \right) - 0 \]
\[ = 81 - \frac{243}{2} + \frac{243}{5} = \frac{810 - 1215 + 486}{10} = \frac{81}{10} \]

(c) The solid is formed by revolving region \( R \) around \( y = -2 \).
The outer radius is \( R(x) = f(x) - (-2) = (3 + 2x - x^2) + 2 = 5 + 2x - x^2 \).
The inner radius is \( r(x) = g(x) - (-2) = (3 - x) + 2 = 5 - x \).
The volume of the solid is given by:
\[ \text{Volume} = \pi \int_{0}^{3} \left( [R(x)]^2 - [r(x)]^2 \right) dx = \pi \int_{0}^{3} \left[ (5 + 2x - x^2)^2 - (5 - x)^2 \right] dx \]

Part (a): 3 points
- 1 point for integrand \( f(x) - g(x) \) (or \( 3x - x^2 \))
- 1 point for a correct antiderivative
- 1 point for the correct answer \( \frac{9}{2} \) (simplification not required, e.g., \( \frac{27}{2} - 9 \) earns the point)

Part (b): 3 points
- 1 point for integrand \( (f(x) - g(x))^2 \) (or \( (3x - x^2)^2 \) or \( 9x^2 - 6x^3 + x^4 \))
- 1 point for a correct antiderivative
- 1 point for the correct answer \( \frac{81}{10} \) (or unsimplified equivalent \( 81 - \frac{243}{2} + \frac{243}{5} \))

Part (c): 3 points
- 1 point for the limits \( [0, 3] \) and the constant \( \pi \)
- 1 point for outer radius \( f(x) + 2 \) and inner radius \( g(x) + 2 \)
- 1 point for the correct form of the integrand \( [R(x)]^2 - [r(x)]^2 \)