Learn Extracted exam questions AP Calculus AB 2021 Free Response
2021 Free Response
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The density of a bacteria population in a circular petri dish at a distance $r$ centimeters from the center of the dish is given by an increasing, differentiable function $f$, where $f(r)$ is measured in milligrams per square centimeter. Values of $f(r)$ for selected values of $r$ are given in the table above.
| $r$ (centimeters) | 0 | 1 | 2 | 2.5 | 4 |
|---|---|---|---|---|---|
| $f(r)$ (milligrams per square centimeter) | 1 | 2 | 6 | 10 | 18 |
Use the data in the table to estimate $f'(2.25)$. Using correct units, interpret the meaning of your answer in the context of this problem.
The total mass, in milligrams, of bacteria in the petri dish is given by the integral expression $2\pi \displaystyle\int_0^4 r f(r)\, dr$. Approximate the value of $2\pi \displaystyle\int_0^4 r f(r)\, dr$ using a right Riemann sum with the four subintervals indicated by the data in the table.
Is the approximation found in part (b) an overestimate or underestimate of the total mass of bacteria in the petri dish? Explain your reasoning.
The density of bacteria in the petri dish, for $1 \le r \le 4$, is modeled by the function $g$ defined by
A particle, $P$, is moving along the $x$-axis. The velocity of particle $P$ at time $t$ is given by $v_P(t) = \sin\!\left(t^{1.5}\right)$ for $0 \le t \le \pi$. At time $t = 0$, particle $P$ is at position $x = 5$.
A second particle, $Q$, also moves along the $x$-axis. The velocity of particle $Q$ at time $t$ is given by $v_Q(t) = (t - 1.8) \cdot 1.25^{t}$ for $0 \le t \le \pi$. At time $t = 0$, particle $Q$ is at position $x = 10$.
Find the positions of particles $P$ and $Q$ at time $t = 1$.
Are particles $P$ and $Q$ moving toward each other or away from each other at time $t = 1$? Explain your reasoning.
Find the acceleration of particle $Q$ at time $t = 1$. Is the speed of particle $Q$ increasing or decreasing at time $t = 1$? Explain your reasoning.
Find the total distance traveled by particle $P$ over the time interval $0 \le t \le \pi$.
[Graph showing a curve in the first quadrant starting at the origin $O$, rising to a rounded peak, then curving back down to meet the $x$-axis; $x$-axis and $y$-axis labeled, no numeric scale marks shown.]
A company designs spinning toys using the family of functions $y = cx\sqrt{4 - x^2}$, where $c$ is a positive constant. The figure above shows the region in the first quadrant bounded by the $x$-axis and the graph of $y = cx\sqrt{4 - x^2}$, for some $c$. Each spinning toy is in the shape of the solid generated when such a region is revolved about the $x$-axis. Both $x$ and $y$ are measured in inches.
Find the area of the region in the first quadrant bounded by the $x$-axis and the graph of $y = cx\sqrt{4 - x^2}$ for $c = 6$.
It is known that, for $y = cx\sqrt{4 - x^2}$, $\dfrac{dy}{dx} = \dfrac{c(4 - 2x^2)}{\sqrt{4 - x^2}}$. For a particular spinning toy, the radius of the largest cross-sectional circular slice is $1.2$ inches. What is the value of $c$ for this spinning toy?
For another spinning toy, the volume is $2\pi$ cubic inches. What is the value of $c$ for this spinning toy?
[Graph of $f$ on a grid, $x$-axis from $-4$ to $6$, $y$-axis from $-4$ to $6$, gridlines every 2 units on the $y$-axis and every 2 units on the $x$-axis; the graph consists of four connected line segments: from $(-4, 0)$ up to $(-2, 6)$, then down to $(0, 4)$, then down to $(2, -4)$, then up to $(6, 0)$. Labeled "Graph of $f$".]
Let $f$ be a continuous function defined on the closed interval $-4 \le x \le 6$. The graph of $f$, consisting of four line segments, is shown above. Let $G$ be the function defined by $G(x) = \displaystyle\int_0^x f(t)\, dt$.
On what open intervals is the graph of $G$ concave up? Give a reason for your answer.
Let $P$ be the function defined by $P(x) = G(x) \cdot f(x)$. Find $P'(3)$.
Find $\displaystyle\lim_{x \to 2} \dfrac{G(x)}{x^2 - 2x}$.
Find the average rate of change of $G$ on the interval $[-4, 2]$. Does the Mean Value Theorem guarantee a value $c$, $-4 < c < 2$, for which $G'(c)$ is equal to this average rate of change? Justify your answer.
Consider the function $y = f(x)$ whose curve is given by the equation $2y^2 - 6 = y\sin x$ for $y > 0$.
Show that $\dfrac{dy}{dx} = \dfrac{y\cos x}{4y - \sin x}$.
Write an equation for the line tangent to the curve at the point $(0, \sqrt{3})$.
For $0 \le x \le \pi$ and $y > 0$, find the coordinates of the point where the line tangent to the curve is horizontal.
Determine whether $f$ has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.
A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time $t$ hours is modeled by a function $y = A(t)$ that satisfies the differential equation $\dfrac{dy}{dt} = \dfrac{12 - y}{3}$. At time $t = 0$ hours, there are $0$ milligrams of the medication in the patient.
A portion of the slope field for the differential equation $\dfrac{dy}{dt} = \dfrac{12 - y}{3}$ is given below. Sketch the solution curve through the point $(0, 0)$.
[Slope field diagram, $t$-axis horizontal from 0 rightward, $y$-axis vertical; short line segments drawn at grid points showing slope direction: segments near the top of the grid slope downward (negative slope, since $y > 12$ there), a horizontal row of flat dashes at $y = 12$ (zero slope), and segments below slope upward (positive slope, since $y < 12$), with segments closer to $y = 0$ sloping steeply upward and flattening as $y$ approaches 12 across each column.]
Using correct units, interpret the statement $\displaystyle\lim_{t \to \infty} A(t) = 12$ in the context of this problem.
Use separation of variables to find $y = A(t)$, the particular solution to the differential equation $\dfrac{dy}{dt} = \dfrac{12 - y}{3}$ with initial condition $A(0) = 0$.
A different procedure is used to administer the medication to a second patient. The amount, in milligrams, of the medication in the second patient at time $t$ hours is modeled by a function $y = B(t)$ that satisfies the differential equation $\dfrac{dy}{dt} = 3 - \dfrac{y}{t+2}$. At time $t = 1$ hour, there are $2.5$ milligrams of the medication in the second patient. Is the rate of change of the amount of medication in the second patient increasing or decreasing at time $t = 1$? Give a reason for your answer.