Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 9.1 - 9.9专题 9.1 至 9.9BC
A curve is defined by $x(t) = t^{3}$ and $y(t) = t^{2} + 5t$. Find $\dfrac{dy}{dx}$ at $t = 1$.曲线由 $x(t) = t^{3}$ 与 $y(t) = t^{2} + 5t$ 定义。求 $t = 1$ 时的 $\dfrac{dy}{dx}$。
$\dfrac{dx}{dt} = 3t^{2}$ and $\dfrac{dy}{dt} = 2t+5$, so $\dfrac{dy}{dx} = \dfrac{2t+5}{3t^{2}}$. (M1)$\dfrac{dx}{dt} = 3t^{2}$，$\dfrac{dy}{dt} = 2t+5$，故 $\dfrac{dy}{dx} = \dfrac{2t+5}{3t^{2}}$。(M1)
At $t=1$: $\dfrac{dy}{dx} = \dfrac{2(1)+5}{3(1)^{2}} = \dfrac{7}{3}$. (A1)当 $t=1$ 时：$\dfrac{dy}{dx} = \dfrac{2(1)+5}{3(1)^{2}} = \dfrac{7}{3}$。(A1)
$x(t) = t^{2}-4t$, $y(t)=t^{3}-3t$. At which value(s) of $t$ does the curve have a horizontal tangent?$x(t) = t^{2}-4t$，$y(t)=t^{3}-3t$。曲线在哪个（些）$t$ 值处有水平切线？
$\dfrac{dy}{dt} = 3t^{2}-3 = 0 \Rightarrow t^{2}=1 \Rightarrow t=\pm 1$. (M1)$\dfrac{dy}{dt} = 3t^{2}-3 = 0 \Rightarrow t^{2}=1 \Rightarrow t=\pm 1$。(M1)
Check $\dfrac{dx}{dt}=2t-4$ at each: at $t=1$, $\dfrac{dx}{dt}=-2\ne 0$; at $t=-1$, $\dfrac{dx}{dt}=-6\ne 0$. Both give valid horizontal tangents, so $t=-1,1$. (A1)检验 $\dfrac{dx}{dt}=2t-4$：在 $t=1$ 处，$\dfrac{dx}{dt}=-2\ne 0$；在 $t=-1$ 处，$\dfrac{dx}{dt}=-6\ne 0$。两者均为有效水平切线，故 $t=-1,1$。(A1)
$x(t) = t^{3}$, $y(t) = t^{2}$. Find $\dfrac{d^{2}y}{dx^{2}}$ at $t=1$.$x(t) = t^{3}$，$y(t) = t^{2}$。求 $t=1$ 时的 $\dfrac{d^{2}y}{dx^{2}}$。
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{3t^{2}} = \dfrac{2}{3t}$. Differentiating w.r.t. $t$: $\dfrac{d}{dt}\left[\dfrac{2}{3t}\right] = -\dfrac{2}{3t^{2}}$. (M1)$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{3t^{2}} = \dfrac{2}{3t}$。对 $t$ 求导：$\dfrac{d}{dt}\left[\dfrac{2}{3t}\right] = -\dfrac{2}{3t^{2}}$。(M1)
Divide by $\dfrac{dx}{dt}=3t^{2}$: $\dfrac{d^{2}y}{dx^{2}} = \dfrac{-2/(3t^{2})}{3t^{2}} = -\dfrac{2}{9t^{4}}$. At $t=1$: $-\dfrac{2}{9}$. (A1)再除以 $\dfrac{dx}{dt}=3t^{2}$：$\dfrac{d^{2}y}{dx^{2}} = \dfrac{-2/(3t^{2})}{3t^{2}} = -\dfrac{2}{9t^{4}}$。当 $t=1$ 时：$-\dfrac{2}{9}$。(A1)
$x(t) = t^{2}+1$, $y(t) = t^{3}-12t$. Determine the concavity of the curve at $t=2$.$x(t) = t^{2}+1$，$y(t) = t^{3}-12t$。判断曲线在 $t=2$ 处的凹凸性。
$\dfrac{dy}{dx} = \dfrac{3t^{2}-12}{2t}$. (M1)$\dfrac{dy}{dx} = \dfrac{3t^{2}-12}{2t}$。(M1)
Simplify: $\dfrac{3t^{2}-12}{2t} = \dfrac{3t}{2}-\dfrac{6}{t}$, so $\dfrac{d}{dt}\left[\dfrac{3t}{2}-\dfrac{6}{t}\right] = \dfrac{3}{2}+\dfrac{6}{t^{2}}$. (M1)化简：$\dfrac{3t^{2}-12}{2t} = \dfrac{3t}{2}-\dfrac{6}{t}$，故 $\dfrac{d}{dt}\left[\dfrac{3t}{2}-\dfrac{6}{t}\right] = \dfrac{3}{2}+\dfrac{6}{t^{2}}$。(M1)
At $t=2$: numerator $=\dfrac{3}{2}+\dfrac{6}{4}=3$. Divide by $\dfrac{dx}{dt}=2t=4$: $\dfrac{d^{2}y}{dx^{2}}=\dfrac{3}{4}>0$, so the curve is concave up. (A1)当 $t=2$ 时：分子 $=\dfrac{3}{2}+\dfrac{6}{4}=3$。除以 $\dfrac{dx}{dt}=2t=4$：$\dfrac{d^{2}y}{dx^{2}}=\dfrac{3}{4}>0$，故曲线凹向上。(A1)
Which integral gives the arc length of $x(t) = t^{2}$, $y(t) = t^{3}$ for $0\le t\le 1$?下列哪个积分表示曲线 $x(t) = t^{2}$、$y(t) = t^{3}$ 在 $0\le t\le 1$ 上的弧长？
$\dfrac{dx}{dt}=2t$, $\dfrac{dy}{dt}=3t^{2}$, so the arc-length integrand is $\sqrt{(2t)^{2}+(3t^{2})^{2}} = \sqrt{4t^{2}+9t^{4}}$. (M1)$\dfrac{dx}{dt}=2t$，$\dfrac{dy}{dt}=3t^{2}$，故弧长被积函数为 $\sqrt{(2t)^{2}+(3t^{2})^{2}} = \sqrt{4t^{2}+9t^{4}}$。(M1)
This matches option (A). (A1)与选项 (A) 一致。(A1)
Find the length of $x(t) = 3\cos t$, $y(t) = 3\sin t$ for $0\le t\le \dfrac{\pi}{2}$.求曲线 $x(t) = 3\cos t$、$y(t) = 3\sin t$ 在 $0\le t\le \dfrac{\pi}{2}$ 上的弧长。
$\dfrac{dx}{dt}=-3\sin t$, $\dfrac{dy}{dt}=3\cos t$, so the integrand is $\sqrt{9\sin^{2}t+9\cos^{2}t} = \sqrt{9}=3$ (since $\sin^{2}t+\cos^{2}t=1$). (M1)$\dfrac{dx}{dt}=-3\sin t$，$\dfrac{dy}{dt}=3\cos t$，故被积函数为 $\sqrt{9\sin^{2}t+9\cos^{2}t} = \sqrt{9}=3$（因 $\sin^{2}t+\cos^{2}t=1$）。(M1)
$L=\displaystyle\int_{0}^{\pi/2}3\,dt = 3\cdot\dfrac{\pi}{2}=\dfrac{3\pi}{2}$. (A1)$L=\displaystyle\int_{0}^{\pi/2}3\,dt = 3\cdot\dfrac{\pi}{2}=\dfrac{3\pi}{2}$。(A1)
$\mathbf{r}(t) = \langle t^{2}, \ln t\rangle$. Find $\mathbf{r}'(t)$.$\mathbf{r}(t) = \langle t^{2}, \ln t\rangle$。求 $\mathbf{r}'(t)$。
$\dfrac{d}{dt}\left[t^{2}\right]=2t$. (M1)$\dfrac{d}{dt}\left[t^{2}\right]=2t$。(M1)
$\dfrac{d}{dt}\left[\ln t\right]=\dfrac{1}{t}$, so $\mathbf{r}'(t) = \left\langle 2t, \dfrac{1}{t}\right\rangle$. (A1)$\dfrac{d}{dt}\left[\ln t\right]=\dfrac{1}{t}$，故 $\mathbf{r}'(t) = \left\langle 2t, \dfrac{1}{t}\right\rangle$。(A1)
$\mathbf{r}'(2) = \langle 0, 5\rangle$. What does this tell you about the curve at $t=2$?$\mathbf{r}'(2) = \langle 0, 5\rangle$。这说明曲线在 $t=2$ 处具有怎样的性质？
The tangent vector's $x$-component is $0$ while its $y$-component is $5\ne 0$, so the direction of motion at $t=2$ is purely vertical (no horizontal component). (M1)切向量的 $x$ 分量为 $0$，而 $y$ 分量为 $5\ne 0$，故 $t=2$ 处运动方向纯为竖直（无水平分量）。(M1)
This matches the vertical-tangent criterion $\frac{dx}{dt}=0$ with $\frac{dy}{dt}\ne 0$, so the curve has a vertical tangent line at $t=2$. (A1)这正是竖直切线的判据 $\frac{dx}{dt}=0$ 且 $\frac{dy}{dt}\ne 0$，故曲线在 $t=2$ 处有竖直切线。(A1)
$\mathbf{r}'(t) = \langle 4t, 2e^{2t}\rangle$, $\mathbf{r}(0) = \langle -1, 3\rangle$. Find $x(1)$.$\mathbf{r}'(t) = \langle 4t, 2e^{2t}\rangle$，$\mathbf{r}(0) = \langle -1, 3\rangle$。求 $x(1)$。
$\int 4t\,dt = 2t^{2}+C_{1}$, so $x(t) = 2t^{2}+C_{1}$; applying $x(0)=0+C_{1}=-1$ gives $C_{1}=-1$, so $x(t)=2t^{2}-1$. (M1)$\int 4t\,dt = 2t^{2}+C_{1}$，故 $x(t) = 2t^{2}+C_{1}$；代入 $x(0)=0+C_{1}=-1$ 得 $C_{1}=-1$，故 $x(t)=2t^{2}-1$。(M1)
$x(1) = 2(1)^{2}-1=1$. (A1)$x(1) = 2(1)^{2}-1=1$。(A1)
$\displaystyle\int \langle 3t^{2}, \cos t\rangle\,dt =$
$\int 3t^{2}\,dt = t^{3}+C_{1}$. (M1)$\int 3t^{2}\,dt = t^{3}+C_{1}$。(M1)
$\int \cos t\,dt = \sin t + C_{2}$, so the integral is $\langle t^{3}, \sin t\rangle + \mathbf{C}$. (A1)$\int \cos t\,dt = \sin t + C_{2}$，故积分为 $\langle t^{3}, \sin t\rangle + \mathbf{C}$。(A1)
A particle moves with velocity $\mathbf{v}(t) = \langle 5, 12\rangle$. What is the particle's speed?质点以速度 $\mathbf{v}(t) = \langle 5, 12\rangle$ 运动，其速率是多少？
$|\mathbf{v}(t)| = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169}$. (M1)$|\mathbf{v}(t)| = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169}$。(M1)
$=13$. (A1)$=13$。(A1)
$\mathbf{v}(t) = \langle 2, 3t^{2}-3\rangle$, $\mathbf{r}(0) = \langle 0,0\rangle$. Find the displacement vector on $[0,2]$.$\mathbf{v}(t) = \langle 2, 3t^{2}-3\rangle$，$\mathbf{r}(0) = \langle 0,0\rangle$。求质点在 $[0,2]$ 上的位移向量。
$\displaystyle\int_{0}^{2} 2\,dt = 4$ and $\displaystyle\int_{0}^{2} (3t^{2}-3)\,dt = \left[t^{3}-3t\right]_{0}^{2} = (8-6)-0 = 2$. (M1)$\displaystyle\int_{0}^{2} 2\,dt = 4$，$\displaystyle\int_{0}^{2} (3t^{2}-3)\,dt = \left[t^{3}-3t\right]_{0}^{2} = (8-6)-0 = 2$。(M1)
Displacement $= \langle 4, 2\rangle$. (A1)位移 $= \langle 4, 2\rangle$。(A1)
$\mathbf{v}(t) = \langle 3, t^{2}-4\rangle$. Is the particle's speed increasing or decreasing at $t=1$?$\mathbf{v}(t) = \langle 3, t^{2}-4\rangle$。质点在 $t=1$ 时速率是增加还是减小？
$\mathbf{a}(t) = \mathbf{v}'(t) = \langle 0, 2t\rangle$. At $t=1$: $\mathbf{v}(1) = \langle 3, -3\rangle$, $\mathbf{a}(1) = \langle 0, 2\rangle$. (M1)$\mathbf{a}(t) = \mathbf{v}'(t) = \langle 0, 2t\rangle$。当 $t=1$ 时：$\mathbf{v}(1) = \langle 3, -3\rangle$，$\mathbf{a}(1) = \langle 0, 2\rangle$。(M1)
$\mathbf{v}(1)\cdot\mathbf{a}(1) = (3)(0)+(-3)(2) = -6 < 0$, so the speed is decreasing at $t=1$. (A1)$\mathbf{v}(1)\cdot\mathbf{a}(1) = (3)(0)+(-3)(2) = -6 < 0$，故 $t=1$ 时速率减小。(A1)
$r(\theta) = 2+2\cos\theta$. Find $\dfrac{dy}{dx}$ at $\theta=\dfrac{\pi}{2}$.$r(\theta) = 2+2\cos\theta$。求 $\theta=\dfrac{\pi}{2}$ 处的 $\dfrac{dy}{dx}$。
$f(\theta)=2+2\cos\theta$, $f'(\theta)=-2\sin\theta$. At $\theta=\frac{\pi}{2}$: $f=2$, $f'=-2$. Numerator $=f'\sin\theta+f\cos\theta = (-2)(1)+(2)(0)=-2$. Denominator $=f'\cos\theta-f\sin\theta=(-2)(0)-(2)(1)=-2$. (M1)$f(\theta)=2+2\cos\theta$，$f'(\theta)=-2\sin\theta$。当 $\theta=\frac{\pi}{2}$ 时：$f=2$，$f'=-2$。分子 $=f'\sin\theta+f\cos\theta = (-2)(1)+(2)(0)=-2$；分母 $=f'\cos\theta-f\sin\theta=(-2)(0)-(2)(1)=-2$。(M1)
$\dfrac{dy}{dx} = \dfrac{-2}{-2} = 1$. (A1)$\dfrac{dy}{dx} = \dfrac{-2}{-2} = 1$。(A1)
How many points on $r = 1+\cos\theta$, for $0\le\theta<2\pi$, have a vertical tangent line?心脏线 $r = 1+\cos\theta$（$0\le\theta<2\pi$）上有多少个点具有竖直切线？
$x=(1+\cos\theta)\cos\theta=\cos\theta+\cos^{2}\theta$, so $\dfrac{dx}{d\theta}=-\sin\theta-2\sin\theta\cos\theta=-\sin\theta(1+2\cos\theta)$. (M1)$x=(1+\cos\theta)\cos\theta=\cos\theta+\cos^{2}\theta$，故 $\dfrac{dx}{d\theta}=-\sin\theta-2\sin\theta\cos\theta=-\sin\theta(1+2\cos\theta)$。(M1)
Setting $\dfrac{dx}{d\theta}=0$: $\sin\theta=0 \Rightarrow \theta=0,\pi$; or $\cos\theta=-\tfrac12 \Rightarrow \theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$. (M1)令 $\dfrac{dx}{d\theta}=0$：$\sin\theta=0 \Rightarrow \theta=0,\pi$；或 $\cos\theta=-\tfrac12 \Rightarrow \theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$。(M1)
$y=(1+\cos\theta)\sin\theta$ gives $\dfrac{dy}{d\theta}=\cos\theta+\cos 2\theta$. At $\theta=0$: $1+1=2\ne 0$ (valid). At $\theta=\pi$: $-1+1=0$ also, this is the pole $r=0$, a degenerate cusp, not a genuine vertical tangent, excluded. At $\theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$: $-\tfrac12-\tfrac12=-1\ne 0$ (both valid). (M1)$y=(1+\cos\theta)\sin\theta$ 给出 $\dfrac{dy}{d\theta}=\cos\theta+\cos 2\theta$。在 $\theta=0$：$1+1=2\ne 0$（有效）。在 $\theta=\pi$：$-1+1=0$ 也为零，此为极点 $r=0$ 处的退化尖点，并非真正的竖直切线，予以排除。在 $\theta=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$：$-\tfrac12-\tfrac12=-1\ne 0$（均有效）。(M1)
Three genuine vertical tangent points remain: $\theta=0,\ \tfrac{2\pi}{3},\ \tfrac{4\pi}{3}$. (A1)剩下三个真正的竖直切线点：$\theta=0,\ \tfrac{2\pi}{3},\ \tfrac{4\pi}{3}$。(A1)
What is the area enclosed by $r=5$ for $0\le\theta\le 2\pi$?极坐标曲线 $r=5$ 在 $0\le\theta\le 2\pi$ 上所围的面积是多少？
$A=\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}(5)^{2}\,d\theta = \dfrac{1}{2}(25)(2\pi)$. (M1)$A=\dfrac{1}{2}\displaystyle\int_{0}^{2\pi}(5)^{2}\,d\theta = \dfrac{1}{2}(25)(2\pi)$。(M1)
$=25\pi$. (A1)$=25\pi$。(A1)
Find the area of one petal of $r = \sin(3\theta)$.求玫瑰线 $r = \sin(3\theta)$ 一个花瓣的面积。
One petal of $r=\sin(3\theta)$ spans $\theta\in\left[0,\tfrac{\pi}{3}\right]$ (where $r\ge 0$, the first petal). (M1)$r=\sin(3\theta)$ 的一个花瓣（第一瓣）跨越 $\theta\in\left[0,\tfrac{\pi}{3}\right]$（此区间 $r\ge 0$）。(M1)
$A=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/3}\sin^{2}(3\theta)\,d\theta = \dfrac{1}{4}\displaystyle\int_{0}^{\pi/3}\left[1-\cos(6\theta)\right]d\theta$, using $\sin^{2}u=\tfrac{1-\cos 2u}{2}$ with $u=3\theta$. (M1)$A=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/3}\sin^{2}(3\theta)\,d\theta = \dfrac{1}{4}\displaystyle\int_{0}^{\pi/3}\left[1-\cos(6\theta)\right]d\theta$，其中用到 $\sin^{2}u=\tfrac{1-\cos 2u}{2}$，$u=3\theta$。(M1)
$=\dfrac{1}{4}\left[\theta-\dfrac{\sin 6\theta}{6}\right]_{0}^{\pi/3} = \dfrac{1}{4}\left[\dfrac{\pi}{3}-0\right] = \dfrac{\pi}{12}$. (A1)$=\dfrac{1}{4}\left[\theta-\dfrac{\sin 6\theta}{6}\right]_{0}^{\pi/3} = \dfrac{1}{4}\left[\dfrac{\pi}{3}-0\right] = \dfrac{\pi}{12}$。(A1)
Which integral gives the area inside $r=3\sin\theta$ and outside $r=1+\sin\theta$?下列哪个积分表示位于 $r=3\sin\theta$ 内部且在 $r=1+\sin\theta$ 外部的区域面积？
Set $3\sin\theta=1+\sin\theta \Rightarrow 2\sin\theta=1 \Rightarrow \sin\theta=\tfrac12 \Rightarrow \theta=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ on $[0,\pi]$. (M1)令 $3\sin\theta=1+\sin\theta \Rightarrow 2\sin\theta=1 \Rightarrow \sin\theta=\tfrac12 \Rightarrow \theta=\tfrac{\pi}{6},\tfrac{5\pi}{6}$（在 $[0,\pi]$ 上）。(M1)
Testing $\theta=\tfrac{\pi}{2}$ (between the intersections): $3\sin\tfrac{\pi}{2}=3$ versus $1+\sin\tfrac{\pi}{2}=2$, so $r=3\sin\theta$ is the outer curve on $\left(\tfrac{\pi}{6},\tfrac{5\pi}{6}\right)$. (M1)检验 $\theta=\tfrac{\pi}{2}$（位于两交点之间）：$3\sin\tfrac{\pi}{2}=3$ 大于 $1+\sin\tfrac{\pi}{2}=2$，故在 $\left(\tfrac{\pi}{6},\tfrac{5\pi}{6}\right)$ 上 $r=3\sin\theta$ 为外曲线。(M1)
Subtracting inner$^2$ from outer$^2$ inside a single integral with the $\tfrac12$ factor outside gives option (A). (A1)在同一积分内以外曲线平方减内曲线平方，并在积分外保留 $\tfrac12$，即得选项 (A)。(A1)
A curve is defined by $x(t) = t^{3}-3t$ and $y(t) = t^{2}-4$.曲线由 $x(t) = t^{3}-3t$ 与 $y(t) = t^{2}-4$ 定义。
$\dfrac{dx}{dt}=3t^{2}-3$ and $\dfrac{dy}{dt}=2t$. (M1)$\dfrac{dx}{dt}=3t^{2}-3$，$\dfrac{dy}{dt}=2t$。(M1)
$\dfrac{dy}{dx}=\dfrac{2t}{3t^{2}-3}$. (A1)$\dfrac{dy}{dx}=\dfrac{2t}{3t^{2}-3}$。(A1)
At $t=2$: $x(2)=8-6=2$, $y(2)=4-4=0$, giving the point $(2,0)$; $\dfrac{dx}{dt}=9$, $\dfrac{dy}{dt}=4$, so the slope is $\dfrac{4}{9}$. (M1)当 $t=2$ 时：$x(2)=8-6=2$，$y(2)=4-4=0$，得点 $(2,0)$；$\dfrac{dx}{dt}=9$，$\dfrac{dy}{dt}=4$，故斜率为 $\dfrac{4}{9}$。(M1)
Tangent line: $y-0=\dfrac{4}{9}(x-2)$, i.e. $y=\dfrac{4}{9}(x-2)$. (A1)切线：$y-0=\dfrac{4}{9}(x-2)$，即 $y=\dfrac{4}{9}(x-2)$。(A1)
$\dfrac{d}{dt}\left[\dfrac{2t}{3t^{2}-3}\right]=\dfrac{-6(t^{2}+1)}{(3t^{2}-3)^{2}}$ by the quotient rule, so $\dfrac{d^{2}y}{dx^{2}}=\dfrac{-6(t^{2}+1)}{(3t^{2}-3)^{3}}$ after dividing by $\dfrac{dx}{dt}=3t^{2}-3$ once more. (M1)由商法则，$\dfrac{d}{dt}\left[\dfrac{2t}{3t^{2}-3}\right]=\dfrac{-6(t^{2}+1)}{(3t^{2}-3)^{2}}$，再除以一次 $\dfrac{dx}{dt}=3t^{2}-3$，得 $\dfrac{d^{2}y}{dx^{2}}=\dfrac{-6(t^{2}+1)}{(3t^{2}-3)^{3}}$。(M1)
At $t=2$: $3t^{2}-3=9$, so $\dfrac{d^{2}y}{dx^{2}}=\dfrac{-6(5)}{9^{3}}=\dfrac{-30}{729}=-\dfrac{10}{243}<0$, so the curve is concave down. (A1)当 $t=2$ 时：$3t^{2}-3=9$，故 $\dfrac{d^{2}y}{dx^{2}}=\dfrac{-6(5)}{9^{3}}=\dfrac{-30}{729}=-\dfrac{10}{243}<0$，曲线凹向下。(A1)
$L=\displaystyle\int_{0}^{2}\sqrt{(3t^{2}-3)^{2}+(2t)^{2}}\,dt$. (M1)$L=\displaystyle\int_{0}^{2}\sqrt{(3t^{2}-3)^{2}+(2t)^{2}}\,dt$。(M1)
A particle moves in the plane with velocity vector $\mathbf{v}(t) = \langle 2t,\, 3e^{-t}\rangle$ for $t \ge 0$. At time $t=0$, the particle is at the point $(1,4)$.质点在平面内运动，速度向量为 $\mathbf{v}(t) = \langle 2t,\, 3e^{-t}\rangle$，$t \ge 0$。$t=0$ 时质点位于点 $(1,4)$。
$\dfrac{d}{dt}(2t)=2$ and $\dfrac{d}{dt}(3e^{-t})=-3e^{-t}$. (M1)$\dfrac{d}{dt}(2t)=2$，$\dfrac{d}{dt}(3e^{-t})=-3e^{-t}$。(M1)
$\mathbf{a}(t)=\langle 2,-3e^{-t}\rangle$. (A1)$\mathbf{a}(t)=\langle 2,-3e^{-t}\rangle$。(A1)
$\int 2t\,dt=t^{2}+C_{1}$ and $\int 3e^{-t}\,dt=-3e^{-t}+C_{2}$, so $x(t)=t^{2}+C_{1}$ and $y(t)=-3e^{-t}+C_{2}$. (M1)$\int 2t\,dt=t^{2}+C_{1}$，$\int 3e^{-t}\,dt=-3e^{-t}+C_{2}$，故 $x(t)=t^{2}+C_{1}$，$y(t)=-3e^{-t}+C_{2}$。(M1)
Using $\mathbf{r}(0)=(1,4)$: $x(0)=C_{1}=1$ and $y(0)=-3+C_{2}=4 \Rightarrow C_{2}=7$. (M1)代入 $\mathbf{r}(0)=(1,4)$：$x(0)=C_{1}=1$，$y(0)=-3+C_{2}=4 \Rightarrow C_{2}=7$。(M1)
$\mathbf{r}(t)=\langle t^{2}+1,\,-3e^{-t}+7\rangle$. (A1)$\mathbf{r}(t)=\langle t^{2}+1,\,-3e^{-t}+7\rangle$。(A1)
$\mathbf{v}(1)=\langle 2,3e^{-1}\rangle \approx \langle 2, 1.104\rangle$. (M1)$\mathbf{v}(1)=\langle 2,3e^{-1}\rangle \approx \langle 2, 1.104\rangle$。(M1)
Speed $=\sqrt{2^{2}+1.104^{2}} \approx \sqrt{4+1.218}\approx\sqrt{5.218}\approx 2.284$. (A1)速率 $=\sqrt{2^{2}+1.104^{2}} \approx \sqrt{4+1.218}\approx\sqrt{5.218}\approx 2.284$。(A1)
$\mathbf{a}(1)=\langle 2,-3e^{-1}\rangle\approx\langle 2,-1.104\rangle$, so $\mathbf{v}(1)\cdot\mathbf{a}(1)=(2)(2)+\left(\tfrac{3}{e}\right)\left(-\tfrac{3}{e}\right)=4-\tfrac{9}{e^{2}}\approx 4-1.218=2.782$. (M1)$\mathbf{a}(1)=\langle 2,-3e^{-1}\rangle\approx\langle 2,-1.104\rangle$，故 $\mathbf{v}(1)\cdot\mathbf{a}(1)=(2)(2)+\left(\tfrac{3}{e}\right)\left(-\tfrac{3}{e}\right)=4-\tfrac{9}{e^{2}}\approx 4-1.218=2.782$。(M1)
Since $\mathbf{v}(1)\cdot\mathbf{a}(1)>0$, the particle's speed is increasing at $t=1$. (R1)因 $\mathbf{v}(1)\cdot\mathbf{a}(1)>0$，质点在 $t=1$ 时速率增加。(R1)
A particle moves along a curve with velocity vector $\mathbf{v}(t) = \langle \cos t,\, \sin(2t)\rangle$ for $0 \le t \le \pi$.质点沿曲线运动，速度向量为 $\mathbf{v}(t) = \langle \cos t,\, \sin(2t)\rangle$，$0 \le t \le \pi$。
$\mathbf{v}\left(\tfrac{\pi}{4}\right)=\left\langle \cos\tfrac{\pi}{4}, \sin\tfrac{\pi}{2}\right\rangle=\left\langle \tfrac{\sqrt2}{2}, 1\right\rangle$. (M1)$\mathbf{v}\left(\tfrac{\pi}{4}\right)=\left\langle \cos\tfrac{\pi}{4}, \sin\tfrac{\pi}{2}\right\rangle=\left\langle \tfrac{\sqrt2}{2}, 1\right\rangle$。(M1)
Speed $=\sqrt{\tfrac12+1}=\sqrt{\tfrac32}\approx 1.225$. (A1)速率 $=\sqrt{\tfrac12+1}=\sqrt{\tfrac32}\approx 1.225$。(A1)
$\displaystyle\int_{0}^{\pi}\cos t\,dt=[\sin t]_{0}^{\pi}=0-0=0$. (M1)$\displaystyle\int_{0}^{\pi}\cos t\,dt=[\sin t]_{0}^{\pi}=0-0=0$。(M1)
$\displaystyle\int_{0}^{\pi}\sin(2t)\,dt=\left[-\tfrac{\cos 2t}{2}\right]_{0}^{\pi}=\left(-\tfrac12\right)-\left(-\tfrac12\right)=0$. (M1)$\displaystyle\int_{0}^{\pi}\sin(2t)\,dt=\left[-\tfrac{\cos 2t}{2}\right]_{0}^{\pi}=\left(-\tfrac12\right)-\left(-\tfrac12\right)=0$。(M1)
Displacement $=\langle 0,0\rangle$: despite moving throughout $[0,\pi]$, the particle returns exactly to its starting point. (A1)位移 $=\langle 0,0\rangle$：尽管质点在 $[0,\pi]$ 上持续运动，最终却恰好回到出发点。(A1)
Total distance $=\displaystyle\int_{0}^{\pi}|\mathbf{v}(t)|\,dt$. (M1)总路程 $=\displaystyle\int_{0}^{\pi}|\mathbf{v}(t)|\,dt$。(M1)
$=\displaystyle\int_{0}^{\pi}\sqrt{\cos^{2}t+\sin^{2}(2t)}\,dt$. (A1)$=\displaystyle\int_{0}^{\pi}\sqrt{\cos^{2}t+\sin^{2}(2t)}\,dt$。(A1)
At $t=\tfrac{\pi}{4}$: $v_{x}\left(\tfrac{\pi}{4}\right)=\cos\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}>0$. (M1)当 $t=\tfrac{\pi}{4}$ 时：$v_{x}\left(\tfrac{\pi}{4}\right)=\cos\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}>0$。(M1)
A positive horizontal velocity component means the particle is moving in the positive $x$-direction (rightward) at this instant. (R1)水平速度分量为正，说明质点此刻沿 $x$ 正方向（向右）运动。(R1)
Consider the polar curve $r = 2 + 4\cos\theta$ for $0 \le \theta \le 2\pi$.考虑极坐标曲线 $r = 2 + 4\cos\theta$，$0 \le \theta \le 2\pi$。
$f(\theta)=2+4\cos\theta$, so $f'(\theta)=-4\sin\theta$. (M1)$f(\theta)=2+4\cos\theta$，故 $f'(\theta)=-4\sin\theta$。(M1)
$\dfrac{dy}{dx}=\dfrac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta}$. (M1)$\dfrac{dy}{dx}=\dfrac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta}$。(M1)
$=\dfrac{-4\sin^{2}\theta+(2+4\cos\theta)\cos\theta}{-4\sin\theta\cos\theta-(2+4\cos\theta)\sin\theta}$. (A1)$=\dfrac{-4\sin^{2}\theta+(2+4\cos\theta)\cos\theta}{-4\sin\theta\cos\theta-(2+4\cos\theta)\sin\theta}$。(A1)
At $\theta=\tfrac{\pi}{3}$: $\cos\theta=\tfrac12$, $\sin\theta=\tfrac{\sqrt3}{2}$, so $f\left(\tfrac{\pi}{3}\right)=2+4\left(\tfrac12\right)=4$ and $f'\left(\tfrac{\pi}{3}\right)=-4\left(\tfrac{\sqrt3}{2}\right)=-2\sqrt3$. (M1)当 $\theta=\tfrac{\pi}{3}$ 时：$\cos\theta=\tfrac12$，$\sin\theta=\tfrac{\sqrt3}{2}$，故 $f\left(\tfrac{\pi}{3}\right)=2+4\left(\tfrac12\right)=4$，$f'\left(\tfrac{\pi}{3}\right)=-4\left(\tfrac{\sqrt3}{2}\right)=-2\sqrt3$。(M1)
Numerator $=(-2\sqrt3)\left(\tfrac{\sqrt3}{2}\right)+(4)\left(\tfrac12\right)=-3+2=-1$. Denominator $=(-2\sqrt3)\left(\tfrac12\right)-(4)\left(\tfrac{\sqrt3}{2}\right)=-\sqrt3-2\sqrt3=-3\sqrt3$. (M1)分子 $=(-2\sqrt3)\left(\tfrac{\sqrt3}{2}\right)+(4)\left(\tfrac12\right)=-3+2=-1$。分母 $=(-2\sqrt3)\left(\tfrac12\right)-(4)\left(\tfrac{\sqrt3}{2}\right)=-\sqrt3-2\sqrt3=-3\sqrt3$。(M1)
$\dfrac{dy}{dx}=\dfrac{-1}{-3\sqrt3}=\dfrac{1}{3\sqrt3}=\dfrac{\sqrt3}{9}\approx 0.192$. (A1)$\dfrac{dy}{dx}=\dfrac{-1}{-3\sqrt3}=\dfrac{1}{3\sqrt3}=\dfrac{\sqrt3}{9}\approx 0.192$。(A1)
$2+4\cos\theta=0 \Rightarrow \cos\theta=-\tfrac12$. (M1)$2+4\cos\theta=0 \Rightarrow \cos\theta=-\tfrac12$。(M1)
On $[0,2\pi)$: $\theta=\tfrac{2\pi}{3}$ and $\theta=\tfrac{4\pi}{3}$. (A1)在 $[0,2\pi)$ 上：$\theta=\tfrac{2\pi}{3}$ 与 $\theta=\tfrac{4\pi}{3}$。(A1)
Let $r_{1} = 4\cos\theta$ and $r_{2} = 2$ for $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$.设 $r_{1} = 4\cos\theta$、$r_{2} = 2$，$-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$。
$4\cos\theta=2 \Rightarrow \cos\theta=\tfrac12$. (M1)$4\cos\theta=2 \Rightarrow \cos\theta=\tfrac12$。(M1)
On $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$: $\theta=-\tfrac{\pi}{3}$ and $\theta=\tfrac{\pi}{3}$. (A1)在 $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ 上：$\theta=-\tfrac{\pi}{3}$ 与 $\theta=\tfrac{\pi}{3}$。(A1)
For $\theta\in\left(-\tfrac{\pi}{3},\tfrac{\pi}{3}\right)$, testing $\theta=0$ gives $r_{1}=4>2=r_{2}$, so $r_{1}$ is outer there: Area $=\dfrac12\displaystyle\int_{-\pi/3}^{\pi/3}\left[(4\cos\theta)^{2}-(2)^{2}\right]d\theta$. (M1)对于 $\theta\in\left(-\tfrac{\pi}{3},\tfrac{\pi}{3}\right)$，检验 $\theta=0$ 得 $r_{1}=4>2=r_{2}$，故此区间 $r_{1}$ 为外曲线：面积 $=\dfrac12\displaystyle\int_{-\pi/3}^{\pi/3}\left[(4\cos\theta)^{2}-(2)^{2}\right]d\theta$。(M1)
$(4\cos\theta)^{2}-4=16\cos^{2}\theta-4=8(1+\cos2\theta)-4=4+8\cos2\theta$, using $\cos^{2}\theta=\tfrac{1+\cos2\theta}{2}$. (M1)$(4\cos\theta)^{2}-4=16\cos^{2}\theta-4=8(1+\cos2\theta)-4=4+8\cos2\theta$，其中用到 $\cos^{2}\theta=\tfrac{1+\cos2\theta}{2}$。(M1)
$\displaystyle\int(4+8\cos2\theta)\,d\theta=4\theta+4\sin2\theta$; evaluated from $-\tfrac{\pi}{3}$ to $\tfrac{\pi}{3}$: $\left[\tfrac{4\pi}{3}+2\sqrt3\right]-\left[-\tfrac{4\pi}{3}-2\sqrt3\right]=\tfrac{8\pi}{3}+4\sqrt3$. (M1)$\displaystyle\int(4+8\cos2\theta)\,d\theta=4\theta+4\sin2\theta$；从 $-\tfrac{\pi}{3}$ 到 $\tfrac{\pi}{3}$ 求值：$\left[\tfrac{4\pi}{3}+2\sqrt3\right]-\left[-\tfrac{4\pi}{3}-2\sqrt3\right]=\tfrac{8\pi}{3}+4\sqrt3$。(M1)
Area $=\dfrac12\left(\tfrac{8\pi}{3}+4\sqrt3\right)=\dfrac{4\pi}{3}+2\sqrt3\approx 7.653$. (A1)面积 $=\dfrac12\left(\tfrac{8\pi}{3}+4\sqrt3\right)=\dfrac{4\pi}{3}+2\sqrt3\approx 7.653$。(A1)