Learn Extracted exam questions AP Calculus BC 2019 Free Response
2019 Free Response
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Fish enter a lake at a rate modeled by the function $E$ given by $E(t) = 20 + 15\sin\!\left(\dfrac{\pi t}{6}\right)$. Fish leave the lake at a rate modeled by the function $L$ given by $L(t) = 4 + 2^{0.1t^2}$. Both $E(t)$ and $L(t)$ are measured in fish per hour, and $t$ is measured in hours since midnight ($t = 0$).
How many fish enter the lake over the 5-hour period from midnight ($t = 0$) to 5 A.M. ($t = 5$)? Give your answer to the nearest whole number.
What is the average number of fish that leave the lake per hour over the 5-hour period from midnight ($t = 0$) to 5 A.M. ($t = 5$)?
At what time $t$, for $0 \le t \le 8$, is the greatest number of fish in the lake? Justify your answer.
Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. ($t = 5$)? Explain your reasoning.
Let $S$ be the region bounded by the graph of the polar curve $r(\theta) = 3\sqrt{\theta}\sin\!\left(\theta^2\right)$ for $0 \le \theta \le \sqrt{\pi}$, as shown in the figure above.
[Figure: A polar-coordinate graph with axes $x$ (horizontal) and $y$ (vertical), origin labeled $O$. The shaded region $S$ is a teardrop/loop shape starting at the origin, bulging out to the right and extending upward, bounded by the curve $r(\theta) = 3\sqrt{\theta}\sin(\theta^2)$ for $0 \le \theta \le \sqrt{\pi}$.]
Find the area of $S$.
What is the average distance from the origin to a point on the polar curve $r(\theta) = 3\sqrt{\theta}\sin\!\left(\theta^2\right)$ for $0 \le \theta \le \sqrt{\pi}$?
There is a line through the origin with positive slope $m$ that divides the region $S$ into two regions with equal areas. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of $m$.
For $k > 0$, let $A(k)$ be the area of the portion of region $S$ that is also inside the circle $r = k\cos\theta$. Find $\displaystyle\lim_{k\to\infty} A(k)$.
[Figure: Graph of $f$ on a grid with $x$-axis from $-2$ to $5$ and $y$-axis from $-1$ to $3$. The graph consists of two line segments and a curve: a line segment from $(-2, 1)$ down to $(-1, -1)$; a line segment from $(-1,-1)$ up to $(2, 3)$; then a curve (quarter circle centered at $(5,3)$) from $(2,3)$ down to $(5,0)$, passing through the point $(3, 3-\sqrt{5})$.]
The continuous function $f$ is defined on the closed interval $-6 \le x \le 5$. The figure above shows a portion of the graph of $f$, consisting of two line segments and a quarter of a circle centered at the point $(5, 3)$. It is known that the point $\left(3,\, 3 - \sqrt{5}\right)$ is on the graph of $f$.
If $\displaystyle\int_{-6}^{5} f(x)\,dx = 7$, find the value of $\displaystyle\int_{-6}^{-2} f(x)\,dx$. Show the work that leads to your answer.
Evaluate $\displaystyle\int_{3}^{5} \left(2f'(x) + 4\right) dx$.
The function $g$ is given by $g(x) = \displaystyle\int_{-2}^{x} f(t)\,dt$. Find the absolute maximum value of $g$ on the interval $-2 \le x \le 5$. Justify your answer.
Find $\displaystyle\lim_{x\to 1} \dfrac{10^{x} - 3f'(x)}{f(x) - \arctan x}$.
[Figure: A vertical cylindrical barrel with diameter labeled "2 ft" across the top. The barrel is shaded (filled with water) from the bottom up to a level marked with a bracket labeled "$h$ ft", with an unshaded empty region above the water level up to the rim.]
A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height $h$ of the water in the barrel with respect to time $t$ is modeled by $\dfrac{dh}{dt} = -\dfrac{1}{10}\sqrt{h}$, where $h$ is measured in feet and $t$ is measured in seconds. (The volume $V$ of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$.)
Find the rate of change of the volume of water in the barrel with respect to time when the height of the water is 4 feet. Indicate units of measure.
When the height of the water is 3 feet, is the rate of change of the height of the water with respect to time increasing or decreasing? Explain your reasoning.
At time $t = 0$ seconds, the height of the water is 5 feet. Use separation of variables to find an expression for $h$ in terms of $t$.
Consider the family of functions $f(x) = \dfrac{1}{x^2 - 2x + k}$, where $k$ is a constant.
Find the value of $k$, for $k > 0$, such that the slope of the line tangent to the graph of $f$ at $x = 0$ equals $6$.
For $k = -8$, find the value of $\displaystyle\int_{0}^{1} f(x)\,dx$.
For $k = 1$, find the value of $\displaystyle\int_{0}^{2} f(x)\,dx$ or show that it diverges.
[Figure: A graph showing a portion of the graph of $f$ (a curve resembling a parabola opening upward, passing through $(0,3)$ with a minimum near $(1, 1.8)$, rising steeply for $x>1$ and for $x<-1$) together with the line tangent to the graph of $f$ at $x=0$ (a straight line through $(0,3)$ sloping downward to the right, passing near $(1.5, 0)$), on a grid with $x$-axis from $-2$ to $2$ and $y$-axis from $0$ to $5$ (approximately).]
| $n$ | $f^{(n)}(0)$ |
|---|---|
| 2 | 3 |
| 3 | $-\dfrac{23}{2}$ |
| 4 | 54 |
A function $f$ has derivatives of all orders for all real numbers $x$. A portion of the graph of $f$ is shown above, along with the line tangent to the graph of $f$ at $x = 0$. Selected derivatives of $f$ at $x = 0$ are given in the table above.
Write the third-degree Taylor polynomial for $f$ about $x = 0$.
Write the first three nonzero terms of the Maclaurin series for $e^x$. Write the second-degree Taylor polynomial for $e^x f(x)$ about $x = 0$.
Let $h$ be the function defined by $h(x) = \displaystyle\int_{0}^{x} f(t)\,dt$. Use the Taylor polynomial found in part (a) to find an approximation for $h(1)$.
It is known that the Maclaurin series for $h$ converges to $h(x)$ for all real numbers $x$. It is also known that the individual terms of the series for $h(1)$ alternate in sign and decrease in absolute value to $0$. Use the alternating series error bound to show that the approximation found in part (c) differs from $h(1)$ by at most $0.45$.