Learn Extracted exam questions AP Calculus AB 2023 Free Response
2023 Free Response
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A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function $f$, where $f(t)$ is measured in gallons per second and $t$ is measured in seconds since pumping began. Selected values of $f(t)$ are given in the table.
| $t$ (seconds) | 0 | 60 | 90 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|
| $f(t)$ (gallons per second) | 0 | 0.1 | 0.15 | 0.1 | 0.05 | 0 |
Using correct units, interpret the meaning of $\displaystyle\int_{60}^{135} f(t)\,dt$ in the context of the problem. Use a right Riemann sum with the three subintervals $[60, 90]$, $[90, 120]$, and $[120, 135]$ to approximate the value of $\displaystyle\int_{60}^{135} f(t)\,dt$.
Must there exist a value of $c$, for $60 < c < 120$, such that $f'(c) = 0$? Justify your answer.
The rate of flow of gasoline, in gallons per second, can also be modeled by $g(t) = \left(\dfrac{t}{500}\right)\cos\!\left[\left(\dfrac{t}{120}\right)^{2}\right]$ for $0 \le t \le 150$. Using this model, find the average rate of flow of gasoline over the time interval $0 \le t \le 150$.
Show the setup for your calculations.
Using the model $g$ defined in part (c), find the value of $g'(140)$. Interpret the meaning of your answer in the context of the problem.
Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen's velocity is modeled by $v(t) = 2.38e^{-0.02t}\sin\!\left(\dfrac{\pi}{56}t\right)$, where $t$ is measured in seconds and $v(t)$ is measured in meters per second.
Find all times $t$ in the interval $0 < t < 90$ at which Stephen changes direction. Give a reason for your answer.
Find Stephen's acceleration at time $t = 60$ seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time $t = 60$ seconds? Give a reason for your answer.
Find the distance between Stephen's position at time $t = 20$ seconds and his position at time $t = 80$ seconds. Show the setup for your calculations.
Find the total distance Stephen swims over the time interval $0 \le t \le 90$ seconds. Show the setup for your calculations.
A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function $M$ models the temperature of the milk at time $t$, where $M(t)$ is measured in degrees Celsius ($^\circ\text{C}$) and $t$ is the number of minutes since the bottle was placed in the pan. $M$ satisfies the differential equation $\dfrac{dM}{dt} = \dfrac{1}{4}(40 - M)$. At time $t = 0$, the temperature of the milk is $5^\circ\text{C}$. It can be shown that $M(t) < 40$ for all values of $t$.
A slope field for the differential equation $\dfrac{dM}{dt} = \dfrac{1}{4}(40 - M)$ is shown. Sketch the solution curve through the point $(0, 5)$.
[Slope field diagram: axes labeled $M(t)$ (vertical, with gridlines at $5$ and $70$ marked) and $t$ (horizontal, with a gridline at $15$ marked), origin $O$. Short line segments are drawn at grid points across the region $0 \le t \le 15$-ish, $0 \le M(t) \le 70$-ish: segments are nearly horizontal (flat, slightly downward-sloping) near $M(t) = 70$, becoming steeper negative slope moving down toward $M(t) \approx 40$ (segments are horizontal/flat right at $M(t)=40$), then the slopes become positive and increasingly steep as $M(t)$ decreases from $40$ down toward $M(t) = 5$, with the steepest (near-vertical, slope $\approx 1$) segments at the bottom of the field near $M(t) = 5$. The point $(0, 5)$ is marked with a star on the vertical axis.]
Use the line tangent to the graph of $M$ at $t = 0$ to approximate $M(2)$, the temperature of the milk at time $t = 2$ minutes.
Write an expression for $\dfrac{d^2M}{dt^2}$ in terms of $M$. Use $\dfrac{d^2M}{dt^2}$ to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of $M(2)$. Give a reason for your answer.
Use separation of variables to find an expression for $M(t)$, the particular solution to the differential equation $\dfrac{dM}{dt} = \dfrac{1}{4}(40 - M)$ with initial condition $M(0) = 5$.
The function $f$ is defined on the closed interval $[-2, 8]$ and satisfies $f(2) = 1$. The graph of $f'$, the derivative of $f$, consists of two line segments and a semicircle, as shown in the figure.
[Graph of $f'$: x-axis from $-2$ to $8$ with gridlines at every integer, y-axis from $-2$ to $2$ with gridlines at every integer. The graph consists of: a line segment from $(-2, 2)$ down to $(0, -2)$; a line segment from $(0, -2)$ up to $(4, 2)$; and a semicircle (bulging downward, below the x-axis) connecting $(4, 2)$ down through the x-axis and back up to $(8, 2)$, dipping to a minimum of about $0$ at $x=6$. Solid dots are marked at $(-2, 2)$, $(0,-2)$, $(4,2)$, and $(8,2)$.]
Does $f$ have a relative minimum, a relative maximum, or neither at $x = 6$? Give a reason for your answer.
On what open intervals, if any, is the graph of $f$ concave down? Give a reason for your answer.
Find the value of $\displaystyle\lim_{x \to 2} \dfrac{6f(x) - 3x}{x^2 - 5x + 6}$, or show that it does not exist. Justify your answer.
Find the absolute minimum value of $f$ on the closed interval $[-2, 8]$. Justify your answer.
The functions $f$ and $g$ are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of $x$.
| $x$ | 0 | 2 | 4 | 7 |
|---|---|---|---|---|
| $f(x)$ | 10 | 7 | 4 | 5 |
| $f'(x)$ | $\dfrac{3}{2}$ | $-8$ | 3 | 6 |
| $g(x)$ | 1 | 2 | $-3$ | 0 |
| $g'(x)$ | 5 | 4 | 2 | 8 |
Let $h$ be the function defined by $h(x) = f(g(x))$. Find $h'(7)$. Show the work that leads to your answer.
Let $k$ be a differentiable function such that $k'(x) = (f(x))^2 \cdot g(x)$. Is the graph of $k$ concave up or concave down at the point where $x = 4$? Give a reason for your answer.
Let $m$ be the function defined by $m(x) = 5x^3 + \displaystyle\int_0^x f'(t)\,dt$. Find $m(2)$. Show the work that leads to your answer.
Is the function $m$ defined in part (c) increasing, decreasing, or neither at $x = 2$? Justify your answer.
Consider the curve given by the equation $6xy = 2 + y^3$.
Show that $\dfrac{dy}{dx} = \dfrac{2y}{y^2 - 2x}$.
Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
A particle is moving along the curve. At the instant when the particle is at the point $\left(\dfrac{1}{2}, -2\right)$, its horizontal position is increasing at a rate of $\dfrac{dx}{dt} = \dfrac{2}{3}$ unit per second. What is the value of $\dfrac{dy}{dt}$, the rate of change of the particle's vertical position, at that instant?