Learn Extracted exam questions AP Calculus BC 2024 Free Response
2024 Free Response
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The temperature of coffee in a cup at time $t$ minutes is modeled by a decreasing differentiable function $C$, where $C(t)$ is measured in degrees Celsius. For $0 \le t \le 12$, selected values of $C(t)$ are given in the table shown.
| $t$ (minutes) | 0 | 3 | 7 | 12 |
|---|---|---|---|---|
| $C(t)$ (degrees Celsius) | 100 | 85 | 69 | 55 |
Approximate $C'(5)$ using the average rate of change of $C$ over the interval $3 \le t \le 7$. Show the work that leads to your answer and include units of measure.
Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the value of $\displaystyle\int_0^{12} C(t)\,dt$. Interpret the meaning of $\dfrac{1}{12}\displaystyle\int_0^{12} C(t)\,dt$ in the context of the problem.
For $12 \le t \le 20$, the rate of change of the temperature of the coffee is modeled by $C'(t) = \dfrac{-24.55e^{0.01t}}{t}$, where $C'(t)$ is measured in degrees Celsius per minute. Find the temperature of the coffee at time $t = 20$. Show the setup for your calculations.
For the model defined in part (c), it can be shown that $C''(t) = \dfrac{0.2455e^{0.01t}(100 - t)}{t^2}$. For $12 < t < 20$, determine whether the temperature of the coffee is changing at a decreasing rate or at an increasing rate. Give a reason for your answer.
A particle moving along a curve in the $xy$-plane has position $(x(t), y(t))$ at time $t$ seconds, where $x(t)$ and $y(t)$ are measured in centimeters. It is known that $x'(t) = 8t - t^2$ and $y'(t) = -t + \sqrt{t^{1.2} + 20}$. At time $t = 2$ seconds, the particle is at the point $(3, 6)$.
Find the speed of the particle at time $t = 2$ seconds. Show the setup for your calculations.
Find the total distance traveled by the particle over the time interval $0 \le t \le 2$. Show the setup for your calculations.
Find the $y$-coordinate of the position of the particle at the time $t = 0$. Show the setup for your calculations.
For $2 \le t \le 8$, the particle remains in the first quadrant. Find all times $t$ in the interval $2 \le t \le 8$ when the particle is moving toward the $x$-axis. Give a reason for your answer.
The depth of seawater at a location can be modeled by the function $H$ that satisfies the differential equation $\dfrac{dH}{dt} = \dfrac{1}{2}(H - 1)\cos\!\left(\dfrac{t}{2}\right)$, where $H(t)$ is measured in feet and $t$ is measured in hours after noon ($t = 0$). It is known that $H(0) = 4$.
A portion of the slope field for the differential equation is provided. Sketch the solution curve, $y = H(t)$, through the point $(0, 4)$.
[Slope field graph; $t$-axis (horizontal) from 0 to 5 with gridlines at 1, 2, 3, 4, 5; $y$-axis (vertical) with gridlines at 5 and 10. A dot marks the point $(0, 4)$ on the $y$-axis. The slope field shows short line segments at grid points across the region $0 \le t \le 5$: for small $y$ (near the $t$-axis) the segments are nearly horizontal (flat dashes); moving up toward $y = 5$–$10$, the segments tilt, transitioning from forward slashes ($/$, positive slope) on the left side of the grid (small $t$) through nearly vertical/horizontal ($-$ or $\sim$) in the middle, to backslashes ($\backslash$, negative slope) on the right side (larger $t$), reflecting the $\cos(t/2)$ factor changing sign.]
For $0 < t < 5$, it can be shown that $H(t) > 1$. Find the value of $t$, for $0 < t < 5$, at which $H$ has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.
Use separation of variables to find $y = H(t)$, the particular solution to the differential equation $\dfrac{dH}{dt} = \dfrac{1}{2}(H - 1)\cos\!\left(\dfrac{t}{2}\right)$ with initial condition $H(0) = 4$.
[Graph of $f$; $x$-axis from $-6$ to 7 with gridlines at every integer, $y$-axis from $-2$ to 3 with gridlines at $-2, -1, 1, 2, 3$. The labeled point $(-6, 0.5)$ marks the left endpoint of the curve on the $y$-axis side. From $x = -6$ to $x = 4$, the curve rises smoothly from about $(-6, 0.5)$ to a rounded peak near $(-2, 2.3)$ (horizontal tangent at $x = -2$), then descends, crossing the $x$-axis at $x = 4$, continuing as a straight line down to the labeled point $(6, -1)$ and beyond to $x = 7$. The shaded region $R$ lies in the second quadrant, bounded above by the curve, below by the $x$-axis, and on the left by the vertical line $x = -6$, between $x = -6$ and $x = 0$.]
The graph of the differentiable function $f$, shown for $-6 \le x \le 7$, has a horizontal tangent at $x = -2$ and is linear for $0 \le x \le 7$. Let $R$ be the region in the second quadrant bounded by the graph of $f$, the vertical line $x = -6$, and the $x$- and $y$-axes. Region $R$ has area 12.
The function $g$ is defined by $g(x) = \displaystyle\int_0^{x} f(t)\,dt$. Find the values of $g(-6)$, $g(4)$, and $g(6)$.
For the function $g$ defined in part (a), find all values of $x$ in the interval $0 \le x \le 6$ at which the graph of $g$ has a critical point. Give a reason for your answer.
The function $h$ is defined by $h(x) = \displaystyle\int_{-6}^{x} f'(t)\,dt$. Find the values of $h(6)$, $h'(6)$, and $h''(6)$. Show the work that leads to your answers.
| $x$ | 0 | $\pi$ | $2\pi$ |
|---|---|---|---|
| $f'(x)$ | 5 | 6 | 0 |
The function $f$ is twice differentiable for all $x$ with $f(0) = 0$. Values of $f'$, the derivative of $f$, are given in the table for selected values of $x$.
For $x \ge 0$, the function $h$ is defined by $h(x) = \displaystyle\int_0^{x} \sqrt{1 + (f'(t))^2}\,dt$. Find the value of $h'(\pi)$. Show the work that leads to your answer.
What information does $\displaystyle\int_0^{\pi} \sqrt{1 + (f'(x))^2}\,dx$ provide about the graph of $f$?
Use Euler's method, starting at $x = 0$ with two steps of equal size, to approximate $f(2\pi)$. Show the computations that lead to your answer.
Find $\displaystyle\int (t + 5)\cos\!\left(\dfrac{t}{4}\right)\,dt$. Show the work that leads to your answer.
The Maclaurin series for a function $f$ is given by $\displaystyle\sum_{n=1}^{\infty} \dfrac{(n+1)x^n}{n^2 6^n}$ and converges to $f(x)$ for all $x$ in the interval of convergence. It can be shown that the Maclaurin series for $f$ has a radius of convergence of 6.
Determine whether the Maclaurin series for $f$ converges or diverges at $x = 6$. Give a reason for your answer.
It can be shown that $f(-3) = \displaystyle\sum_{n=1}^{\infty} \dfrac{(n+1)(-3)^n}{n^2 6^n} = \displaystyle\sum_{n=1}^{\infty} \dfrac{n+1}{n^2}\left(-\dfrac{1}{2}\right)^n$ and that the first three terms of this series sum to $S_3 = -\dfrac{125}{144}$. Show that $\left| f(-3) - S_3 \right| < \dfrac{1}{50}$.
Find the general term of the Maclaurin series for $f'$, the derivative of $f$. Find the radius of convergence of the Maclaurin series for $f'$.
Let $g(x) = \displaystyle\sum_{n=1}^{\infty} \dfrac{(n+1)x^{2n}}{n^2 3^n}$. Use the ratio test to determine the radius of convergence of the Maclaurin series for $g$.