Learn Extracted exam questions AP Calculus BC 2014 Free Response
2014 Free Response
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Grass clippings are placed in a bin, where they decompose. For $0 \le t \le 30$, the amount of grass clippings remaining in the bin is modeled by $A(t) = 6.687(0.931)^t$, where $A(t)$ is measured in pounds and $t$ is measured in days.
Find the average rate of change of $A(t)$ over the interval $0 \le t \le 30$. Indicate units of measure.
Find the value of $A'(15)$. Using correct units, interpret the meaning of the value in the context of the problem.
Find the time $t$ for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval $0 \le t \le 30$.
For $t > 30$, $L(t)$, the linear approximation to $A$ at $t = 30$, is a better model for the amount of grass clippings remaining in the bin. Use $L(t)$ to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.
[Graph showing two polar curves for $0 \le \theta \le \pi$: the circle $r = 3$ and the curve $r = 3 - 2\sin(2\theta)$, plotted on an $xy$-plane with axis labeled $1$ near the origin $O$. The shaded region $R$, labeled near the point $(1,1)$ roughly, is the region inside both curves — it is bounded above by the circle $r=3$ arc and has a scalloped lower/right boundary from the curve $r = 3-2\sin(2\theta)$, meeting the $x$-axis near $x \approx 3$.]
The graphs of the polar curves $r = 3$ and $r = 3 - 2\sin(2\theta)$ are shown in the figure above for $0 \le \theta \le \pi$.
Let $R$ be the shaded region that is inside the graph of $r = 3$ and inside the graph of $r = 3 - 2\sin(2\theta)$. Find the area of $R$.
For the curve $r = 3 - 2\sin(2\theta)$, find the value of $\dfrac{dx}{d\theta}$ at $\theta = \dfrac{\pi}{6}$.
The distance between the two curves changes for $0 < \theta < \dfrac{\pi}{2}$. Find the rate at which the distance between the two curves is changing with respect to $\theta$ when $\theta = \dfrac{\pi}{3}$.
A particle is moving along the curve $r = 3 - 2\sin(2\theta)$ so that $\dfrac{d\theta}{dt} = 3$ for all times $t \ge 0$. Find the value of $\dfrac{dr}{dt}$ at $\theta = \dfrac{\pi}{6}$.
[Graph of $f$: a piecewise-linear function on $[-5,4]$ consisting of three line segments. An open/solid point at $(-5,2)$ connects down to a point at approximately $(-3,0)$ on the $x$-axis; from there the graph rises linearly to a peak above the $y$-axis near $x=0$ (peak value not numerically labeled, appears to be above 2); from the peak the graph falls linearly, crossing the $x$-axis near $x=1$, continuing down to the point $(4,-4)$. Gridlines/labels shown: $1$ marked on both axes, origin labeled $O$.]
The function $f$ is defined on the closed interval $[-5, 4]$. The graph of $f$ consists of three line segments and is shown in the figure above. Let $g$ be the function defined by $g(x) = \displaystyle\int_{-3}^{x} f(t)\,dt$.
Find $g(3)$.
On what open intervals contained in $-5 < x < 4$ is the graph of $g$ both increasing and concave down? Give a reason for your answer.
The function $h$ is defined by $h(x) = \dfrac{g(x)}{5x}$. Find $h'(3)$.
The function $p$ is defined by $p(x) = f(x^2 - x)$. Find the slope of the line tangent to the graph of $p$ at the point where $x = -1$.
Train $A$ runs back and forth on an east-west section of railroad track. Train $A$'s velocity, measured in meters per minute, is given by a differentiable function $v_A(t)$, where time $t$ is measured in minutes. Selected values for $v_A(t)$ are given in the table above.
| $t$ (minutes) | 0 | 2 | 5 | 8 | 12 |
|---|---|---|---|---|---|
| $v_A(t)$ (meters/minute) | 0 | 100 | 40 | $-120$ | $-150$ |
Find the average acceleration of train $A$ over the interval $2 \le t \le 8$.
Do the data in the table support the conclusion that train $A$'s velocity is $-100$ meters per minute at some time $t$ with $5 < t < 8$? Give a reason for your answer.
At time $t = 2$, train $A$'s position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train $A$, in meters from the Origin Station, at time $t = 12$. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time $t = 12$.
A second train, train $B$, travels north from the Origin Station. At time $t$ the velocity of train $B$ is given by $v_B(t) = -5t^2 + 60t + 25$, and at time $t = 2$ the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train $A$ and train $B$ is changing at time $t = 2$.
[Graph showing the shaded region $R$ in the first quadrant (and dipping slightly below the $x$-axis) bounded by the curve $y = xe^{x^2}$ (rising steeply and concave up from the origin $O$ through $x=1$), the line $y = -2x$ (a straight line from the origin through the fourth quadrant), and the vertical line $x = 1$; axis labels $O$ at the origin, $1$ marked on the $x$-axis, $x$ and $y$ axis arrows shown; region $R$ labeled inside the shaded area.]
Let $R$ be the shaded region bounded by the graph of $y = xe^{x^2}$, the line $y = -2x$, and the vertical line $x = 1$, as shown in the figure above.
Find the area of $R$.
Write, but do not evaluate, an integral expression that gives the volume of the solid generated when $R$ is rotated about the horizontal line $y = -2$.
Write, but do not evaluate, an expression involving one or more integrals that gives the perimeter of $R$.
The Taylor series for a function $f$ about $x = 1$ is given by $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{2}{n}(x-1)^n$ and converges to $f(x)$ for $|x - 1| < R$, where $R$ is the radius of convergence of the Taylor series.
Find the value of $R$.
Find the first three nonzero terms and the general term of the Taylor series for $f'$, the derivative of $f$, about $x = 1$.
The Taylor series for $f'$ about $x = 1$, found in part (b), is a geometric series. Find the function $f'$ to which the series converges for $|x - 1| < R$. Use this function to determine $f$ for $|x - 1| < R$.