Show all supporting work on scratch paper. On the AP Exam, Section I is split into a no-calculator and a calculator-allowed part, each question below is labeled accordingly.请将所有辅助步骤写在草稿纸上。AP 考试第一部分分为禁用计算器和允许使用计算器两类，每道题均已标注。
A population grows at a rate proportional to the current population. Which differential equation models this situation?某种群的增长速率与当前种群数量成正比。哪个微分方程描述了这一情形？
A cup of coffee cools at a rate proportional to the difference between its temperature $T$ and a room temperature of $70$°F. Which equation models this?一杯咖啡的冷却速率与其温度 $T$ 和室温 $70$°F 之差成正比。哪个方程描述了这一情形？
Let $y$ be the solution to $\dfrac{dy}{dx}=y^{2}$ with $y(0)=1$. The largest open interval containing $0$ on which the solution is defined is设 $y$ 是满足 $\dfrac{dy}{dx}=y^{2}$，且 $y(0)=1$ 的解，该解定义在包含 $0$ 的最大开区间是
A radioactive substance decays so that $\dfrac{dN}{dt}=-0.04\,N$, where $t$ is in years. To the nearest year, the half-life is某放射性物质的衰变满足 $\dfrac{dN}{dt}=-0.04\,N$，其中 $t$ 以年为单位。精确到最近的整年，其半衰期为
Let $y=f(x)$ be the solution to $\dfrac{dy}{dx}=x+y$ with $f(1)=2$. The tangent line to $y=f(x)$ at $x=1$ is used to approximate $f(1.2)$. The approximation is设 $y=f(x)$ 是满足 $\dfrac{dy}{dx}=x+y$，且 $f(1)=2$ 的解，利用 $y=f(x)$ 在 $x=1$ 处的切线近似 $f(1.2)$，近似值为
A bacterial culture has $1000$ bacteria. After $4$ hours, the culture has $4000$ bacteria. If the growth rate is proportional to the current population, how many bacteria are present after $6$ hours?某细菌培养液初始含有 $1000$ 个细菌，$4$ 小时后增至 $4000$ 个。若增长速率与当前数量成正比，$6$ 小时后共有多少个细菌？
Consider $\dfrac{dy}{dt}=y(2-y)$. The equilibrium solutions are考虑 $\dfrac{dy}{dt}=y(2-y)$，其平衡解为
(A)$y=0$ only仅 $y=0$
(B)$y=2$ only仅 $y=2$
(C)$y=0$ and $y=2$$y=0$ 和 $y=2$
(D)none exist不存在
PART IIShow All Work展示全部步骤
Free-Response Questions自由作答题
Free-response answers require complete algebraic work: separation of variables, antiderivatives, $+C$, use of initial conditions, and correct solving for $y$. On the AP Exam, skipping any of these steps will cost points.自由作答需展示完整代数步骤：分离变量、求不定积分、写 $+C$、代入初始条件，以及正确解出 $y$。AP 考试中，省略任何步骤均会扣分。
A tank initially contains $500$ gallons of water. Water leaks out at a rate proportional to the amount remaining: $\dfrac{dW}{dt}=k\,W$, where $t$ is in minutes. After $20$ minutes, the tank contains $400$ gallons.水箱初始含有 $500$ 加仑水，水以与剩余水量成正比的速率漏出，满足 $\dfrac{dW}{dt}=k\,W$，$t$ 以分钟为单位。$20$ 分钟后，水箱中剩余 $400$ 加仑。
(a)Write an expression for $W(t)$, the amount of water at time $t$, in terms of $k$.写出以 $k$ 表示的 $W(t)$ 表达式，即 $t$ 时刻水箱中的水量。
(b)Find the value of $k$. Round to four decimal places.求 $k$ 的值，精确到小数点后四位。
(c)How much water, to the nearest gallon, remains in the tank after $60$ minutes?$60$ 分钟后水箱中剩余多少加仑水（精确到最近的整加仑）？
(d)At what rate, in gallons per minute, is water leaking from the tank at $t=60$? Indicate units.$t=60$ 时水漏出的速率是多少（以加仑/分钟为单位）？请标明单位。
Consider the differential equation $\dfrac{dy}{dx}=\dfrac{2y}{x+1}$.考虑微分方程 $\dfrac{dy}{dx}=\dfrac{2y}{x+1}$。
(a)Verify that $y=C(x+1)^{2}$ is a solution to the DE for any constant $C$.验证对任意常数 $C$，$y=C(x+1)^{2}$ 均是该微分方程的解。
(b)Find the particular solution $y=f(x)$ with initial condition $f(0)=3$.求满足初始条件 $f(0)=3$ 的特解 $y=f(x)$。
(c)Find the largest open interval containing $x=0$ on which the solution from part (b) is defined, and justify your answer.求包含 $x=0$ 的最大开区间，使得 (b) 中的解在该区间上有定义，并说明理由。
(d)Find the value of $f''(0)$ for the particular solution from part (b).求 (b) 中特解的 $f''(0)$ 的值。
A biologist studies a fish population $P(t)$ in a lake, where $t$ is in years. The population is modeled by the differential equation $\dfrac{dP}{dt}=0.1\,P\left(1-\dfrac{P}{1000}\right)$. (You are not required to solve this differential equation.)某生物学家研究湖中鱼类种群 $P(t)$，$t$ 以年为单位，种群由微分方程 $\dfrac{dP}{dt}=0.1\,P\left(1-\dfrac{P}{1000}\right)$ 建模。（不要求求解该微分方程。）
(a)Find $\dfrac{dP}{dt}$ when $P=400$. Interpret the value in context, with correct units.求 $P=400$ 时的 $\dfrac{dP}{dt}$，并在语境中用正确单位解释该值。
(b)For what value(s) of $P$ is $\dfrac{dP}{dt}=0$? What do these values represent?对哪些 $P$ 值有 $\dfrac{dP}{dt}=0$？这些值代表什么？
(c)Suppose $P(0)=400$. Use the tangent line to the graph of $P$ at $t=0$ to approximate $P(2)$.设 $P(0)=400$，利用 $P$ 在 $t=0$ 处的切线近似 $P(2)$。
(d)Use implicit differentiation to find $\dfrac{d^{2}P}{dt^{2}}$ in terms of $P$. Use this to determine whether the tangent-line approximation in part (c) is an over- or underestimate. Justify.利用隐式微分求以 $P$ 表示的 $\dfrac{d^{2}P}{dt^{2}}$，并据此判断 (c) 中切线近似是高估还是低估，并说明理由。
PART III: (BC) EXTENSIONSTopics 7.5, 7.9 - BC ONLY主题 7.5, 7.9 - 仅 BC
BC-Only Practice仅 BC 练习
BC ONLY.The following items cover Euler's method (Topic 7.5) and logistic models (Topic 7.9). AB students may skip this section. BC students should be fluent with: (i) Euler iteration $y_{n+1}=y_n+h\cdot f(x_n,y_n)$; (ii) the logistic equation $\dfrac{dy}{dt}=ky\!\left(1-\dfrac{y}{K}\right)$ with carrying capacity $K$ and inflection at $y=K/2$.以下题目涵盖欧拉法（主题 7.5）和逻辑斯谛模型（主题 7.9）。AB 学生可跳过本节。BC 学生应熟练掌握：(i) 欧拉迭代 $y_{n+1}=y_n+h\cdot f(x_n,y_n)$；(ii) 逻辑斯谛方程 $\dfrac{dy}{dt}=ky\!\left(1-\dfrac{y}{K}\right)$，其中 $K$ 为承载容量，拐点在 $y=K/2$。
Let $y=f(x)$ be the solution to $\dfrac{dy}{dx}=x+y$ with $f(0)=1$. Use Euler's method with two steps of equal size $h=0.5$ starting at $x=0$ to approximate $f(1)$.设 $y=f(x)$ 是满足 $\dfrac{dy}{dx}=x+y$，且 $f(0)=1$ 的解，从 $x=0$ 出发，用步长 $h=0.5$ 的两步欧拉法近似 $f(1)$。
Suppose the solution $y=f(x)$ to a differential equation is concave up on the interval $[a,b]$. An Euler's-method approximation of $f(b)$ starting from $f(a)$ with positive step size $h$ will be设某微分方程的解 $y=f(x)$ 在区间 $[a,b]$ 上是上凸的（即凹形）。从 $f(a)$ 出发用正步长 $h$ 的欧拉法近似 $f(b)$，结果将
(A)always greater than the true value $f(b)$.始终大于真实值 $f(b)$。
(B)always less than the true value $f(b)$.始终小于真实值 $f(b)$。
(C)equal to the true value $f(b)$ regardless of $h$.无论 $h$ 取何值，均等于真实值 $f(b)$。
(D)indeterminate without knowing $f''$.在不知道 $f''$ 的情况下无法确定。
A population $P(t)$ satisfies $\dfrac{dP}{dt}=0.04\,P\!\left(1-\dfrac{P}{500}\right)$. The carrying capacity and the population value at which $P$ is increasing fastest are, respectively,种群 $P(t)$ 满足 $\dfrac{dP}{dt}=0.04\,P\!\left(1-\dfrac{P}{500}\right)$，承载容量以及 $P$ 增长最快时的种群值分别为
Consider the differential equation $\dfrac{dy}{dx}=x-y$ with initial condition $y(0)=2$.考虑微分方程 $\dfrac{dy}{dx}=x-y$，初始条件为 $y(0)=2$。
(a)Use Euler's method with two steps of size $h=0.5$, starting at $x=0$, to approximate $y(1)$. Show all computations.从 $x=0$ 出发，用步长 $h=0.5$ 的两步欧拉法近似 $y(1)$，展示全部计算过程。
(b)Find $\dfrac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$. Use this to determine whether the Euler approximation in part (a) is an over- or underestimate of $y(1)$. Justify.用 $x$ 和 $y$ 表示 $\dfrac{d^{2}y}{dx^{2}}$，并据此判断 (a) 中的欧拉近似是 $y(1)$ 的高估还是低估，说明理由。
(c)A second student uses Euler's method with four steps of size $h=0.25$ and obtains a different value. Without computing it, predict whether this second approximation will be closer to or farther from the true value of $y(1)$, and explain why.另一位同学用步长 $h=0.25$ 的四步欧拉法得到不同的结果。无需计算，预测该近似是更接近还是更偏离 $y(1)$ 的真实值，并解释原因。
A wildlife biologist models a deer population $P(t)$, where $t$ is in years, by $\dfrac{dP}{dt}=0.2\,P\!\left(1-\dfrac{P}{800}\right)$, with $P(0)=100$.某野生生物学家以 $\dfrac{dP}{dt}=0.2\,P\!\left(1-\dfrac{P}{800}\right)$ 建模鹿群数量 $P(t)$，$t$ 以年为单位，初始条件 $P(0)=100$。
(a)State the carrying capacity. Find $\displaystyle\lim_{t\to\infty}P(t)$ and justify.写出承载容量。求 $\displaystyle\lim_{t\to\infty}P(t)$ 并说明理由。
(b)At what value of $P$ is the population growing fastest? Find $\dfrac{dP}{dt}$ at that value, with units.$P$ 取何值时种群增长最快？求该值处的 $\dfrac{dP}{dt}$，并标明单位。
(c)Find $\dfrac{d^{2}P}{dt^{2}}$ in terms of $P$ alone. Use it to identify the value of $P$ at which the graph of $P(t)$ has an inflection point. Justify your answer using a sign analysis of $\dfrac{d^{2}P}{dt^{2}}$.仅以 $P$ 表示 $\dfrac{d^{2}P}{dt^{2}}$，并据此确定 $P(t)$ 图像拐点处的 $P$ 值，用 $\dfrac{d^{2}P}{dt^{2}}$ 的符号分析加以说明。
(d)Sketch a qualitative graph of $P(t)$ for $t\ge 0$. Label the inflection value and the horizontal asymptote.对 $t\ge 0$，画出 $P(t)$ 的示意图，标出拐点处的值和水平渐近线。