Companion to the University-Style Practice Set配套大学风格练习题集
Sections 1 to 7: parametric curves, parametric calculus, parametric arc length, polar coordinates, polar area, polar arc length, conic sections第 1 至 7 节：参数曲线、参数微积分、参数弧长、极坐标、极坐标面积、极坐标弧长、圆锥曲线CALC II
Eliminate the parameter and identify each curve: (a) $x=2t-1$, $y=t^{2}+3$; (b) $x=3\cos\theta$, $y=5\sin\theta$; (c) $x=e^{t}$, $y=e^{2t}-4$.消去参数并判断各曲线类型：(a) $x=2t-1$，$y=t^{2}+3$；(b) $x=3\cos\theta$，$y=5\sin\theta$；(c) $x=e^{t}$，$y=e^{2t}-4$。
From $x=2t-1$, isolate $t=\dfrac{x+1}{2}$. (M1) Substitute into $y$:由 $x=2t-1$，解出 $t=\dfrac{x+1}{2}$。(M1) 代入 $y$：
$$ y = \left(\frac{x+1}{2}\right)^{2}+3 = \frac{(x+1)^{2}}{4}+3. $$This is an upward-opening parabola with vertex $(-1,\,3)$. Since $t\in\mathbb{R}$, there is no restriction on $x$. (A1)这是顶点为 $(-1,\,3)$ 的开口向上抛物线。由于 $t\in\mathbb{R}$，$x$ 没有限制。(A1)
From $\cos\theta=x/3$ and $\sin\theta=y/5$, apply $\cos^{2}\theta+\sin^{2}\theta=1$: (M1)由 $\cos\theta=x/3$ 和 $\sin\theta=y/5$，应用 $\cos^{2}\theta+\sin^{2}\theta=1$：(M1)
$$ \frac{x^{2}}{9}+\frac{y^{2}}{25}=1. $$This is an ellipse with semi-axes $a=5$ (vertical) and $b=3$ (horizontal), centred at the origin. (A1)这是一个椭圆，半轴 $a=5$（竖直）和 $b=3$（水平），以原点为中心。(A1)
Since $x=e^{t}$ and $e^{t}>0$ for all $t$, we have $x>0$ and $e^{t}=x$. (M1) Then $e^{2t}=(e^{t})^{2}=x^{2}$, so $y=x^{2}-4$. The curve is the right branch ($x>0$) of the parabola $y=x^{2}-4$. (A1)由于对所有 $t$ 均有 $x=e^{t}$ 且 $e^{t}>0$，故 $x>0$ 且 $e^{t}=x$。(M1) 则 $e^{2t}=(e^{t})^{2}=x^{2}$，从而 $y=x^{2}-4$。曲线是抛物线 $y=x^{2}-4$ 的右支（$x>0$）。(A1)
For $x=t^{3}-3t$, $y=t^{2}-1$: (a) find $dy/dx$; (b) the tangent at $t=2$; (c) all horizontal and vertical tangents with their Cartesian points.对 $x=t^{3}-3t$，$y=t^{2}-1$：(a) 求 $dy/dx$；(b) $t=2$ 处的切线；(c) 所有水平切线与竖直切线及其直角坐标点。
Differentiate: $dx/dt=3t^{2}-3=3(t^{2}-1)$ and $dy/dt=2t$. (M1) Therefore求导：$dx/dt=3t^{2}-3=3(t^{2}-1)$，$dy/dt=2t$。(M1) 因此
$$ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t}{3(t^{2}-1)}, \quad t\ne\pm 1. $$(A1)
At $t=2$: $x=8-6=2$, $y=4-1=3$. The slope is $m=\dfrac{2(2)}{3(4-1)}=\dfrac{4}{9}$. (M1)在 $t=2$ 处：$x=8-6=2$，$y=4-1=3$。斜率为 $m=\dfrac{2(2)}{3(4-1)}=\dfrac{4}{9}$。(M1)
Point-slope form: $y-3=\dfrac{4}{9}(x-2)$. (A1) Rearranging:点斜式：$y-3=\dfrac{4}{9}(x-2)$。(A1) 整理得：
$$ y=\frac{4}{9}x-\frac{8}{9}+3=\frac{4}{9}x+\frac{19}{9}. $$(A1)
Horizontal tangent: $dy/dt=0$ and $dx/dt\ne 0$. Setting $2t=0$ gives $t=0$. At $t=0$: $dx/dt=-3\ne 0$, so this is a genuine horizontal tangent. (M1) The Cartesian point is $(0-0,\,0-1)=(0,-1)$. (A1)水平切线：$dy/dt=0$ 且 $dx/dt\ne 0$。令 $2t=0$ 得 $t=0$。在 $t=0$ 处：$dx/dt=-3\ne 0$，故这是一个真正的水平切线。(M1) 直角坐标点为 $(0-0,\,0-1)=(0,-1)$。(A1)
Vertical tangent: $dx/dt=0$ and $dy/dt\ne 0$. Setting $3(t^{2}-1)=0$ gives $t=\pm 1$. At $t=1$: $dy/dt=2\ne 0$ and at $t=-1$: $dy/dt=-2\ne 0$. (M1)竖直切线：$dx/dt=0$ 且 $dy/dt\ne 0$。令 $3(t^{2}-1)=0$ 得 $t=\pm 1$。在 $t=1$ 处：$dy/dt=2\ne 0$；在 $t=-1$ 处：$dy/dt=-2\ne 0$。(M1)
At $t=1$: $(x,y)=(1-3,\,1-1)=(-2,\,0)$. At $t=-1$: $(x,y)=(-1+3,\,1-1)=(2,\,0)$. (A1)在 $t=1$ 处：$(x,y)=(1-3,\,1-1)=(-2,\,0)$。在 $t=-1$ 处：$(x,y)=(-1+3,\,1-1)=(2,\,0)$。(A1)
For $x=\ln t$, $y=t^{2}+t^{-1}$, $t>0$: (a) compute $dy/dx$ and $d^{2}y/dx^{2}$; (b) determine concavity at $t=1$ and give the Cartesian coordinates of that point.对 $x=\ln t$，$y=t^{2}+t^{-1}$，$t>0$：(a) 计算 $dy/dx$ 和 $d^{2}y/dx^{2}$；(b) 判断 $t=1$ 处的凹凸性，并给出该点的直角坐标。
Differentiate: $dx/dt=1/t$ and $dy/dt=2t-t^{-2}$. (M1) Then求导：$dx/dt=1/t$，$dy/dt=2t-t^{-2}$。(M1) 则
$$ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t-t^{-2}}{1/t}=t\bigl(2t-t^{-2}\bigr)=2t^{2}-t^{-1}. $$(A1)
For the second derivative, differentiate $dy/dx$ with respect to $t$ and divide by $dx/dt$: (M1)对于二阶导数，将 $dy/dx$ 对 $t$ 求导后除以 $dx/dt$：(M1)
$$ \frac{d}{dt}\!\left(\frac{dy}{dx}\right)=\frac{d}{dt}\!\left(2t^{2}-t^{-1}\right)=4t+t^{-2}. $$ $$ \frac{d^{2}y}{dx^{2}}=\frac{4t+t^{-2}}{dx/dt}=\frac{4t+t^{-2}}{1/t}=t\bigl(4t+t^{-2}\bigr)=4t^{2}+t^{-1}. $$(A1)
At $t=1$: $d^{2}y/dx^{2}=4(1)^{2}+(1)^{-1}=4+1=5>0$. Since the second derivative is positive, the curve is concave up at this point. (A1)在 $t=1$ 处：$d^{2}y/dx^{2}=4(1)^{2}+(1)^{-1}=4+1=5>0$。由于二阶导数为正，曲线在该点处上凸。(A1)
The Cartesian coordinates are $x=\ln 1=0$ and $y=1^{2}+1^{-1}=2$, giving the point $(0,\,2)$. (A1)直角坐标为 $x=\ln 1=0$，$y=1^{2}+1^{-1}=2$，即点 $(0,\,2)$。(A1)
(a) Convert $(-\sqrt{3},1)$ to polar; (b) convert $r=4\sin\theta$ to Cartesian and identify; (c) identify the type of each: (i) $r=2+2\cos\theta$; (ii) $r=\sin 3\theta$.(a) 将 $(-\sqrt{3},1)$ 转为极坐标；(b) 将 $r=4\sin\theta$ 转为直角坐标并判断类型；(c) 判断各曲线类型：(i) $r=2+2\cos\theta$；(ii) $r=\sin 3\theta$。
Compute $r=\sqrt{(-\sqrt{3})^{2}+1^{2}}=\sqrt{3+1}=2$. (M1) For the angle, $\tan\theta=y/x=1/(-\sqrt{3})=-1/\sqrt{3}$. Since $x<0$ and $y>0$ the point is in the second quadrant, so $\theta=\pi-\pi/6=5\pi/6$. (A1) Polar form: $(2,\,5\pi/6)$.计算 $r=\sqrt{(-\sqrt{3})^{2}+1^{2}}=\sqrt{3+1}=2$。(M1) 对于角度，$\tan\theta=y/x=1/(-\sqrt{3})=-1/\sqrt{3}$。由于 $x<0$ 且 $y>0$，点在第二象限，故 $\theta=\pi-\pi/6=5\pi/6$。(A1) 极坐标形式：$(2,\,5\pi/6)$。
Multiply both sides by $r$: $r^{2}=4r\sin\theta$. (M1) Substitute $r^{2}=x^{2}+y^{2}$ and $r\sin\theta=y$:两边乘以 $r$：$r^{2}=4r\sin\theta$。(M1) 代入 $r^{2}=x^{2}+y^{2}$ 和 $r\sin\theta=y$：
$$ x^{2}+y^{2}=4y \implies x^{2}+y^{2}-4y=0 \implies x^{2}+(y-2)^{2}=4. $$Circle with centre $(0,2)$ and radius $2$. (A1)圆心为 $(0,2)$、半径为 $2$ 的圆。(A1)
(i) $r=2+2\cos\theta$ has the form $r=a+b\cos\theta$ with $a=b=2$. When $|a|=|b|$ the curve is a cardioid; distinguishing feature: the curve passes through the origin once ($r=0$ at $\theta=\pi$), producing a single cusp. (A1)(i) $r=2+2\cos\theta$ 具有 $r=a+b\cos\theta$ 的形式，其中 $a=b=2$。当 $|a|=|b|$ 时曲线为心形线；判别特征：曲线过原点一次（$\theta=\pi$ 时 $r=0$），产生一个尖点。(A1)
(ii) $r=\sin 3\theta$ is a rose with $n=3$ (odd), giving exactly $3$ petals. Distinguishing feature: for $r=\sin(n\theta)$ with $n$ odd, there are exactly $n$ petals, each spanning a $\pi/n$-wide interval of $\theta$. (A1)(ii) $r=\sin 3\theta$ 是 $n=3$（奇数）的玫瑰线，恰好有 $3$ 片花瓣。判别特征：对于 $n$ 为奇数的 $r=\sin(n\theta)$，恰好有 $n$ 片花瓣，每片花瓣跨越宽度为 $\pi/n$ 的 $\theta$ 区间。(A1)
Divide numerator and denominator of $r=\dfrac{6}{2+\cos\theta}$ by $2$: (M1)将 $r=\dfrac{6}{2+\cos\theta}$ 的分子分母同除以 $2$：(M1)
$$ r = \frac{3}{1 + \tfrac{1}{2}\cos\theta}. $$Matching $r=\dfrac{ed}{1+e\cos\theta}$ gives $e=\tfrac{1}{2}$ and $ed=3$, so $d=6$. (A1) Since $e=\tfrac{1}{2}<1$, the conic is an ellipse with one focus at the origin and directrix $x=6$ to the right.对照 $r=\dfrac{ed}{1+e\cos\theta}$ 得 $e=\tfrac{1}{2}$ 且 $ed=3$，故 $d=6$。(A1) 由于 $e=\tfrac{1}{2}<1$，该圆锥曲线是以原点为焦点、右侧准线 $x=6$ 的椭圆。
Derive $A=\tfrac{1}{2}\int_{\alpha}^{\beta}r^{2}\,d\theta$ by partitioning $[\alpha,\beta]$ into $n$ sectors, applying the sector area formula, forming a Riemann sum, and passing to the limit.将 $[\alpha,\beta]$ 分成 $n$ 个扇形，应用扇形面积公式，构造黎曼和后取极限，推导出 $A=\tfrac{1}{2}\int_{\alpha}^{\beta}r^{2}\,d\theta$。
A circular sector of radius $\rho$ and central angle $\Delta\theta$ (in radians) has area $\tfrac{1}{2}\rho^{2}\Delta\theta$. This follows from the proportion $\Delta\theta/(2\pi)$ of the full disk of area $\pi\rho^{2}$. (A1)半径为 $\rho$、圆心角为 $\Delta\theta$（弧度）的扇形面积为 $\tfrac{1}{2}\rho^{2}\Delta\theta$。这由面积 $\pi\rho^{2}$ 的完整圆盘中占比 $\Delta\theta/(2\pi)$ 得出。(A1)
Partition $[\alpha,\beta]$ into $n$ equal subintervals $[\theta_{i-1},\theta_{i}]$ of width $\Delta\theta=(\beta-\alpha)/n$, and choose a sample point $\theta_{i}^{*}\in[\theta_{i-1},\theta_{i}]$. (M1) The swept-area element near $\theta_{i}^{*}$ is approximated by a sector of radius $f(\theta_{i}^{*})$ and angle $\Delta\theta$, giving area $\tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta$. The total area is approximated by将 $[\alpha,\beta]$ 等分为 $n$ 个宽度为 $\Delta\theta=(\beta-\alpha)/n$ 的子区间 $[\theta_{i-1},\theta_{i}]$，取样本点 $\theta_{i}^{*}\in[\theta_{i-1},\theta_{i}]$。(M1) $\theta_{i}^{*}$ 附近的扫掠面积微元由半径为 $f(\theta_{i}^{*})$、圆心角为 $\Delta\theta$ 的扇形近似，面积为 $\tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta$。总面积近似为
$$ A \approx \sum_{i=1}^{n} \tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta. $$(A1) As $n\to\infty$ (equivalently $\Delta\theta\to 0$), each sector becomes an infinitesimally thin wedge, and the sector approximation becomes exact because the arc length of the boundary $\Delta\theta\cdot f(\theta_{i}^{*})$ is negligible compared to the sector radius. (R1)(A1) 当 $n\to\infty$（等价地 $\Delta\theta\to 0$）时，每个扇形变为无限薄的楔形，扇形近似趋于精确，因为边界弧长 $\Delta\theta\cdot f(\theta_{i}^{*})$ 相对于扇形半径可忽略不计。(R1)
The Riemann sum $\sum_{i=1}^{n}\tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta$ has the exact form required by the definition of the definite integral of the function $\theta\mapsto\tfrac{1}{2}[f(\theta)]^{2}$ over $[\alpha,\beta]$. (M1) Since $f$ is continuous, the integral exists, and黎曼和 $\sum_{i=1}^{n}\tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta$ 恰好具有函数 $\theta\mapsto\tfrac{1}{2}[f(\theta)]^{2}$ 在 $[\alpha,\beta]$ 上的定积分定义所要求的形式。(M1) 由于 $f$ 连续，积分存在，且
$$ \lim_{n\to\infty}\sum_{i=1}^{n}\tfrac{1}{2}[f(\theta_{i}^{*})]^{2}\Delta\theta = \frac{1}{2}\int_{\alpha}^{\beta}[f(\theta)]^{2}\,d\theta. $$(A1) This completes the derivation of the polar area formula.(A1) 至此完成了极坐标面积公式的推导。
The formula correctly measures the swept area only when the curve is traced exactly once as $\theta$ ranges from $\alpha$ to $\beta$; if any region is swept multiple times, its area is counted multiple times. (A1) Example: $r=\cos\theta$ on $[0,2\pi]$ traces the circle $x^{2}+(y-\tfrac{1}{2})^{2}=\tfrac{1}{4}$ twice, so $\tfrac{1}{2}\int_{0}^{2\pi}\cos^{2}\theta\,d\theta=\tfrac{\pi}{2}$ gives double the actual area $\tfrac{\pi}{4}$. The correct interval is $[0,\pi]$. (R1)只有当 $\theta$ 从 $\alpha$ 到 $\beta$ 变化时曲线恰好被描绘一次，该公式才能正确量度扫掠面积；若任何区域被多次扫掠，其面积将被重复计算。(A1) 例如：$r=\cos\theta$ 在 $[0,2\pi]$ 上将圆 $x^{2}+(y-\tfrac{1}{2})^{2}=\tfrac{1}{4}$ 描绘了两次，故 $\tfrac{1}{2}\int_{0}^{2\pi}\cos^{2}\theta\,d\theta=\tfrac{\pi}{2}$ 给出了实际面积 $\tfrac{\pi}{4}$ 的两倍。正确的区间为 $[0,\pi]$。(R1)
Derive $L=\int_{a}^{b}\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}\,dt$ by partitioning, applying the MVT, forming a Riemann sum, and reducing to the ordinary arc-length formula.通过划分区间、应用中值定理、构造黎曼和，推导出 $L=\int_{a}^{b}\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}\,dt$，并化简至普通弧长公式。
Partition $[a,b]$ into $n$ equal subintervals $[t_{i-1},t_{i}]$ of width $\Delta t=(b-a)/n$. The endpoints of the $i$-th chord are $(f(t_{i-1}),g(t_{i-1}))$ and $(f(t_{i}),g(t_{i}))$. (M1) By the distance formula the chord length is将 $[a,b]$ 等分为 $n$ 个宽度为 $\Delta t=(b-a)/n$ 的子区间 $[t_{i-1},t_{i}]$。第 $i$ 段弦的端点为 $(f(t_{i-1}),g(t_{i-1}))$ 和 $(f(t_{i}),g(t_{i}))$。(M1) 由距离公式，弦长为
$$ \ell_{i} = \sqrt{[f(t_{i})-f(t_{i-1})]^{2}+[g(t_{i})-g(t_{i-1})]^{2}} = \sqrt{(\Delta x_{i})^{2}+(\Delta y_{i})^{2}}. $$(A1)
Since $f$ and $g$ are differentiable, the MVT guarantees $t_{i}^{*},t_{i}^{**}\in(t_{i-1},t_{i})$ such that $\Delta x_{i}=f(t_{i})-f(t_{i-1})=f'(t_{i}^{*})\Delta t$ and $\Delta y_{i}=g'(t_{i}^{**})\Delta t$. (M1) Substituting:由于 $f$ 和 $g$ 可微，中值定理保证存在 $t_{i}^{*},t_{i}^{**}\in(t_{i-1},t_{i})$ 使得 $\Delta x_{i}=f(t_{i})-f(t_{i-1})=f'(t_{i}^{*})\Delta t$ 且 $\Delta y_{i}=g'(t_{i}^{**})\Delta t$。(M1) 代入：
$$ \ell_{i} = \sqrt{[f'(t_{i}^{*})]^{2}+[g'(t_{i}^{**})]^{2}}\,\Delta t. $$(A1)
The total length approximation is $\sum_{i=1}^{n}\sqrt{[f'(t_{i}^{*})]^{2}+[g'(t_{i}^{**})]^{2}}\,\Delta t$. In general, a Riemann sum requires a single sample point per subinterval, but here two different MVT points $t_{i}^{*}\ne t_{i}^{**}$ appear. (M1) Because $f'$ and $g'$ are continuous on $[a,b]$, they are uniformly continuous, so the maximum difference $|f'(t_{i}^{*})-f'(t_{i}^{**})|\to 0$ as $\Delta t\to 0$. This means replacing both by a single point introduces error that vanishes as $n\to\infty$. Consequently the sum converges to总长度近似为 $\sum_{i=1}^{n}\sqrt{[f'(t_{i}^{*})]^{2}+[g'(t_{i}^{**})]^{2}}\,\Delta t$。一般而言，黎曼和在每个子区间只需一个样本点，但此处出现了两个不同的中值定理点 $t_{i}^{*}\ne t_{i}^{**}$。(M1) 由于 $f'$ 和 $g'$ 在 $[a,b]$ 上连续，它们一致连续，故最大差值 $|f'(t_{i}^{*})-f'(t_{i}^{**})|\to 0$（当 $\Delta t\to 0$）。这意味着将两者替换为单个点所引入的误差在 $n\to\infty$ 时消失。因此和式收敛到
$$ L = \int_{a}^{b}\sqrt{[f'(t)]^{2}+[g'(t)]^{2}}\,dt. $$(A1)
For the parametrisation $x=t$, $y=F(t)$, we have $f'(t)=1$ and $g'(t)=F'(t)$. (M1) Substituting into the parametric formula:对参数化 $x=t$，$y=F(t)$，有 $f'(t)=1$ 且 $g'(t)=F'(t)$。(M1) 代入参数公式：
$$ L = \int_{a}^{b}\sqrt{1^{2}+[F'(t)]^{2}}\,dt = \int_{a}^{b}\sqrt{1+[F'(t)]^{2}}\,dt, $$which is exactly the standard arc-length integral for $y=F(x)$. (A1)这恰好是 $y=F(x)$ 的标准弧长积分。(A1)
Derive the polar slope formula, apply it to the cardioid $r=1+\cos\theta$ to find all horizontal tangents, and evaluate the slope at $\theta=\pi/3$.推导极坐标斜率公式，将其应用于心形线 $r=1+\cos\theta$ 求所有水平切线，并计算 $\theta=\pi/3$ 处的斜率。
Treating $r=f(\theta)$ as a function of $\theta$ and applying the product rule: (M1)将 $r=f(\theta)$ 视为 $\theta$ 的函数并应用乘积法则：(M1)
$$ \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta. $$(A1)
The slope of the polar curve in the $xy$-plane is $dy/dx=(dy/d\theta)/(dx/d\theta)$. A horizontal tangent occurs when the numerator $dy/d\theta=0$ and the denominator $dx/d\theta\ne 0$. (A1) A vertical tangent occurs when the denominator $dx/d\theta=0$ and the numerator $dy/d\theta\ne 0$. (A1) When both are simultaneously zero the analysis requires a separate limit.极坐标曲线在 $xy$ 平面中的斜率为 $dy/dx=(dy/d\theta)/(dx/d\theta)$。当分子 $dy/d\theta=0$ 且分母 $dx/d\theta\ne 0$ 时出现水平切线。(A1) 当分母 $dx/d\theta=0$ 且分子 $dy/d\theta\ne 0$ 时出现竖直切线。(A1) 当两者同时为零时，需要单独分析极限。
For $r=1+\cos\theta$, $dr/d\theta=-\sin\theta$. (M1) Compute the numerator:对 $r=1+\cos\theta$，$dr/d\theta=-\sin\theta$。(M1) 计算分子：
$$ \frac{dy}{d\theta}=(-\sin\theta)\sin\theta+(1+\cos\theta)\cos\theta=-\sin^{2}\theta+\cos\theta+\cos^{2}\theta. $$Using $\cos^{2}\theta-\sin^{2}\theta=\cos(2\theta)=2\cos^{2}\theta-1$:利用 $\cos^{2}\theta-\sin^{2}\theta=\cos(2\theta)=2\cos^{2}\theta-1$：
$$ \frac{dy}{d\theta}=\cos(2\theta)+\cos\theta=2\cos^{2}\theta-1+\cos\theta=(2\cos\theta-1)(\cos\theta+1). $$(M1) Set $dy/d\theta=0$: either $\cos\theta=\tfrac{1}{2}$ (giving $\theta=\pi/3$ or $\theta=5\pi/3$) or $\cos\theta=-1$ (giving $\theta=\pi$). (A1)(M1) 令 $dy/d\theta=0$：或 $\cos\theta=\tfrac{1}{2}$（得 $\theta=\pi/3$ 或 $\theta=5\pi/3$），或 $\cos\theta=-1$（得 $\theta=\pi$）。(A1)
Check $\theta=\pi$: the denominator $dx/d\theta=-\sin\pi\cos\pi-(1+\cos\pi)\sin\pi=0$. Both numerator and denominator vanish at $\theta=\pi$, which is the cusp at the pole ($r=0$ there); this is not a genuine horizontal tangent. The valid horizontal tangents are at $\theta=\pi/3$ and $\theta=5\pi/3$. (A1)检验 $\theta=\pi$：分母 $dx/d\theta=-\sin\pi\cos\pi-(1+\cos\pi)\sin\pi=0$。分子和分母在 $\theta=\pi$ 处同时为零，这是极点处的尖点（该处 $r=0$），不是真正的水平切线。有效的水平切线在 $\theta=\pi/3$ 和 $\theta=5\pi/3$ 处。(A1)
At $\theta=\pi/3$: $r=1+\cos(\pi/3)=1+\tfrac{1}{2}=\tfrac{3}{2}$ and $dr/d\theta=-\sin(\pi/3)=-\tfrac{\sqrt{3}}{2}$. (M1)在 $\theta=\pi/3$ 处：$r=1+\cos(\pi/3)=1+\tfrac{1}{2}=\tfrac{3}{2}$，$dr/d\theta=-\sin(\pi/3)=-\tfrac{\sqrt{3}}{2}$。(M1)
Numerator: $dy/d\theta=(2\cos(\pi/3)-1)(\cos(\pi/3)+1)=(2\cdot\tfrac{1}{2}-1)(\tfrac{1}{2}+1)=0\cdot\tfrac{3}{2}=0$.分子：$dy/d\theta=(2\cos(\pi/3)-1)(\cos(\pi/3)+1)=(2\cdot\tfrac{1}{2}-1)(\tfrac{1}{2}+1)=0\cdot\tfrac{3}{2}=0$。
Denominator: $dx/d\theta=(-\tfrac{\sqrt{3}}{2})\cos(\pi/3)-(1+\tfrac{1}{2})\sin(\pi/3)=-\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2}-\tfrac{3}{2}\cdot\tfrac{\sqrt{3}}{2}=-\tfrac{\sqrt{3}}{4}-\tfrac{3\sqrt{3}}{4}=-\sqrt{3}\ne 0$.分母：$dx/d\theta=(-\tfrac{\sqrt{3}}{2})\cos(\pi/3)-(1+\tfrac{1}{2})\sin(\pi/3)=-\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2}-\tfrac{3}{2}\cdot\tfrac{\sqrt{3}}{2}=-\tfrac{\sqrt{3}}{4}-\tfrac{3\sqrt{3}}{4}=-\sqrt{3}\ne 0$。
Therefore $dy/dx=0/(-\sqrt{3})=0$. (A1) This confirms $\theta=\pi/3$ is a horizontal tangent point, consistent with part (c).因此 $dy/dx=0/(-\sqrt{3})=0$。(A1) 这确认 $\theta=\pi/3$ 是水平切线点，与第(c)部分一致。
For the involute $x=\cos t+t\sin t$, $y=\sin t - t\cos t$, $0\le t\le\pi$: (a) compute $dx/dt$, $dy/dt$; (b) simplify the speed squared; (c) find the arc length; (d) set up and evaluate the surface-area integral for revolution about the $x$-axis over $[0,\pi/2]$.对渐开线 $x=\cos t+t\sin t$，$y=\sin t - t\cos t$，$0\le t\le\pi$：(a) 计算 $dx/dt$、$dy/dt$；(b) 化简速度的平方；(c) 求弧长；(d) 建立并计算绕 $x$ 轴旋转在 $[0,\pi/2]$ 上的表面积积分。
Differentiate $x=\cos t+t\sin t$: (M1)对 $x=\cos t+t\sin t$ 求导：(M1)
$$ \frac{dx}{dt}=-\sin t+\sin t + t\cos t = t\cos t. $$Differentiate $y=\sin t - t\cos t$:对 $y=\sin t - t\cos t$ 求导：
$$ \frac{dy}{dt}=\cos t - \cos t + t\sin t = t\sin t. $$(A1) (Note the elegant cancellations from the product rule.)(A1)（注意乘积法则产生的优雅消去。）
Compute (M1)计算 (M1)
$$ \left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}=(t\cos t)^{2}+(t\sin t)^{2}=t^{2}\cos^{2}t+t^{2}\sin^{2}t=t^{2}(\cos^{2}t+\sin^{2}t)=t^{2}. $$The identity used is $\cos^{2}t+\sin^{2}t=1$. (A1)所用恒等式为 $\cos^{2}t+\sin^{2}t=1$。(A1)
The arc length formula gives (M1)弧长公式给出 (M1)
$$ L=\int_{0}^{\pi}\sqrt{t^{2}}\,dt=\int_{0}^{\pi}|t|\,dt=\int_{0}^{\pi}t\,dt=\left[\frac{t^{2}}{2}\right]_{0}^{\pi}=\frac{\pi^{2}}{2}. $$(Here $t\ge 0$ on $[0,\pi]$, so $|t|=t$.) (A1)（这里 $t\ge 0$ 在 $[0,\pi]$ 上，故 $|t|=t$。）(A1)
The surface area generated by revolving about the $x$-axis is (M1)绕 $x$ 轴旋转所得旋转体的表面积为 (M1)
$$ S=2\pi\int_{0}^{\pi/2}y\,\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt = 2\pi\int_{0}^{\pi/2}(\sin t-t\cos t)\cdot t\,dt = 2\pi\int_{0}^{\pi/2}(t\sin t - t^{2}\cos t)\,dt. $$Integrate each term by parts. For the first: $\int t\sin t\,dt = -t\cos t+\sin t+C$. For the second: $\int t^{2}\cos t\,dt = t^{2}\sin t+2t\cos t-2\sin t+C$ (two applications of IBP). Therefore对每一项分部积分。第一项：$\int t\sin t\,dt = -t\cos t+\sin t+C$。第二项：$\int t^{2}\cos t\,dt = t^{2}\sin t+2t\cos t-2\sin t+C$（两次分部积分）。因此
$$ \int(t\sin t-t^{2}\cos t)\,dt = (-t\cos t+\sin t)-(t^{2}\sin t+2t\cos t-2\sin t)+C = 3\sin t-3t\cos t-t^{2}\sin t+C. $$Evaluating from $0$ to $\pi/2$: at $t=\pi/2$, $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$, giving $3(1)-3(\pi/2)(0)-(\pi/2)^{2}(1)=3-\pi^{2}/4$. At $t=0$ the expression equals $0$. Therefore (A1)从 $0$ 到 $\pi/2$ 求值：在 $t=\pi/2$ 处，$\sin(\pi/2)=1$，$\cos(\pi/2)=0$，得 $3(1)-3(\pi/2)(0)-(\pi/2)^{2}(1)=3-\pi^{2}/4$。在 $t=0$ 处表达式等于 $0$。因此 (A1)
$$ S = 2\pi\!\left(3-\frac{\pi^{2}}{4}\right) = 6\pi - \frac{\pi^{3}}{2}. $$For $r=3\cos\theta$ and $r=1+\cos\theta$: (a) find all intersection points, including the pole; (b) identify the outer curve on $[-\pi/3,\pi/3]$; (c) compute the area inside $r=3\cos\theta$ and outside $r=1+\cos\theta$.对 $r=3\cos\theta$ 和 $r=1+\cos\theta$：(a) 求所有交点，包括极点；(b) 判断 $[-\pi/3,\pi/3]$ 上的外边界曲线；(c) 计算位于 $r=3\cos\theta$ 内且位于 $r=1+\cos\theta$ 外的面积。
Set $3\cos\theta=1+\cos\theta$: $2\cos\theta=1$, so $\cos\theta=\tfrac{1}{2}$. (M1) Solutions in $[0,2\pi)$: $\theta=\pi/3$ and $\theta=5\pi/3$ (equivalently $\theta=-\pi/3$). At both, $r=3\cdot\tfrac{1}{2}=\tfrac{3}{2}$. The intersection points in Cartesian form are (A1)令 $3\cos\theta=1+\cos\theta$：$2\cos\theta=1$，故 $\cos\theta=\tfrac{1}{2}$。(M1) 在 $[0,2\pi)$ 中的解为 $\theta=\pi/3$ 和 $\theta=5\pi/3$（等价地 $\theta=-\pi/3$）。在这两处，$r=3\cdot\tfrac{1}{2}=\tfrac{3}{2}$。直角坐标形式的交点为 (A1)
$$ \left(\frac{3}{2}\cos\frac{\pi}{3},\,\frac{3}{2}\sin\frac{\pi}{3}\right)=\left(\frac{3}{4},\,\frac{3\sqrt{3}}{4}\right) \quad\text{and}\quad \left(\frac{3}{4},\,-\frac{3\sqrt{3}}{4}\right). $$The curves also meet at the pole. For $r=3\cos\theta=0$, we need $\theta=\pi/2$; for $r=1+\cos\theta=0$, we need $\cos\theta=-1$, i.e. $\theta=\pi$. (A1) These are different parameter values, so the pole does not appear as a solution to the simultaneous equation $3\cos\theta=1+\cos\theta$: each curve passes through the origin but at a different value of $\theta$, which is why the algebraic method misses it. (R1)两曲线也在极点相交。对 $r=3\cos\theta=0$，需要 $\theta=\pi/2$；对 $r=1+\cos\theta=0$，需要 $\cos\theta=-1$，即 $\theta=\pi$。(A1) 这些是不同的参数值，故极点不出现在联立方程 $3\cos\theta=1+\cos\theta$ 的解中：每条曲线都过原点，但对应不同的 $\theta$ 值，这就是代数方法漏掉它的原因。(R1)
At $\theta=0$: $r=3\cos 0=3$ for the circle and $r=1+\cos 0=2$ for the cardioid. (M1) Since $3>2$, the circle $r=3\cos\theta$ is the outer (larger) curve on the interval $[-\pi/3,\pi/3]$, and the cardioid is the inner curve. (A1)在 $\theta=0$ 处：圆的 $r=3\cos 0=3$，心形线的 $r=1+\cos 0=2$。(M1) 由于 $3>2$，圆 $r=3\cos\theta$ 在区间 $[-\pi/3,\pi/3]$ 上是外（较大的）曲线，心形线是内曲线。(A1)
The area between two polar curves is $A = \tfrac{1}{2}\int_{\alpha}^{\beta}(r_{\text{outer}}^{2} - r_{\text{inner}}^{2})\,d\theta$ over the interval $[-\pi/3,\pi/3]$ where the circle is the outer curve: (M1)两极坐标曲线之间的面积为 $A = \tfrac{1}{2}\int_{\alpha}^{\beta}(r_{\text{outer}}^{2} - r_{\text{inner}}^{2})\,d\theta$，在圆为外曲线的区间 $[-\pi/3,\pi/3]$ 上：(M1)
$$ A = \frac{1}{2}\int_{-\pi/3}^{\pi/3}\bigl[(3\cos\theta)^{2}-(1+\cos\theta)^{2}\bigr]\,d\theta = \frac{1}{2}\int_{-\pi/3}^{\pi/3}\bigl[9\cos^{2}\theta - 1 - 2\cos\theta - \cos^{2}\theta\bigr]\,d\theta. $$Simplify the integrand: $9\cos^{2}\theta - 1 - 2\cos\theta - \cos^{2}\theta = 8\cos^{2}\theta - 2\cos\theta - 1$. Using $\cos^{2}\theta = \tfrac{1}{2}(1+\cos 2\theta)$: (M1)化简被积函数：$9\cos^{2}\theta - 1 - 2\cos\theta - \cos^{2}\theta = 8\cos^{2}\theta - 2\cos\theta - 1$。利用 $\cos^{2}\theta = \tfrac{1}{2}(1+\cos 2\theta)$：(M1)
$$ 8\cdot\tfrac{1}{2}(1+\cos 2\theta) - 2\cos\theta - 1 = 4 + 4\cos 2\theta - 2\cos\theta - 1 = 3 + 4\cos 2\theta - 2\cos\theta. $$The integrand is even in $\theta$, so the full integral with the $\tfrac{1}{2}$ polar factor becomes: (A1)被积函数关于 $\theta$ 为偶函数，故带有极坐标因子 $\tfrac{1}{2}$ 的完整积分变为：(A1)
$$ A = \frac{1}{2}\int_{-\pi/3}^{\pi/3}(3 + 4\cos 2\theta - 2\cos\theta)\,d\theta = \frac{1}{2}\cdot 2\int_{0}^{\pi/3}(3 + 4\cos 2\theta - 2\cos\theta)\,d\theta. $$ $$ = \Bigl[3\theta + 2\sin 2\theta - 2\sin\theta\Bigr]_{0}^{\pi/3}. $$Evaluate at $\theta = \pi/3$: $3\cdot\dfrac{\pi}{3} + 2\sin\dfrac{2\pi}{3} - 2\sin\dfrac{\pi}{3} = \pi + 2\cdot\dfrac{\sqrt{3}}{2} - 2\cdot\dfrac{\sqrt{3}}{2} = \pi + \sqrt{3} - \sqrt{3} = \pi$. (A1)在 $\theta = \pi/3$ 处求值：$3\cdot\dfrac{\pi}{3} + 2\sin\dfrac{2\pi}{3} - 2\sin\dfrac{\pi}{3} = \pi + 2\cdot\dfrac{\sqrt{3}}{2} - 2\cdot\dfrac{\sqrt{3}}{2} = \pi + \sqrt{3} - \sqrt{3} = \pi$。(A1)
At $\theta = 0$ the bracket equals $0$. Therefore $A = \pi - 0 = \pi$. (A1)在 $\theta = 0$ 处括号内等于 $0$。因此 $A = \pi - 0 = \pi$。(A1)
For the rose $r=2\cos(2\theta)$: (a) number of petals and the $\theta$-interval for one petal; (b) area of one petal; (c) arc-length integrand for one petal, simplified to $2\sqrt{1+3\sin^{2}(2\theta)}$.对玫瑰线 $r=2\cos(2\theta)$：(a) 花瓣总数及一片花瓣的 $\theta$ 区间；(b) 一片花瓣的面积；(c) 一片花瓣弧长的被积函数，化简为 $2\sqrt{1+3\sin^{2}(2\theta)}$。
The rose $r=2\cos(n\theta)$ with $n=2$ (even) has $2n=4$ petals. (A1) The petal along the positive $x$-axis occurs where $r\ge 0$, i.e. $\cos(2\theta)\ge 0$. The primary interval satisfying this near $\theta=0$ is $2\theta\in[-\pi/2,\pi/2]$, giving $\theta\in[-\pi/4,\pi/4]$. (A1)玫瑰线 $r=2\cos(n\theta)$，$n=2$（偶数），有 $2n=4$ 片花瓣。(A1) 沿正 $x$ 轴方向的花瓣出现在 $r\ge 0$ 处，即 $\cos(2\theta)\ge 0$。在 $\theta=0$ 附近满足此条件的主区间为 $2\theta\in[-\pi/2,\pi/2]$，即 $\theta\in[-\pi/4,\pi/4]$。(A1)
Apply the polar area formula to the interval $[-\pi/4,\pi/4]$: (M1)将极坐标面积公式应用于区间 $[-\pi/4,\pi/4]$：(M1)
$$ A=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\bigl[2\cos(2\theta)\bigr]^{2}\,d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}4\cos^{2}(2\theta)\,d\theta=2\int_{-\pi/4}^{\pi/4}\cos^{2}(2\theta)\,d\theta. $$Use the identity $\cos^{2}u=\tfrac{1}{2}(1+\cos 2u)$ with $u=2\theta$: (M1)用恒等式 $\cos^{2}u=\tfrac{1}{2}(1+\cos 2u)$，令 $u=2\theta$：(M1)
$$ A=2\int_{-\pi/4}^{\pi/4}\frac{1+\cos 4\theta}{2}\,d\theta=\int_{-\pi/4}^{\pi/4}(1+\cos 4\theta)\,d\theta. $$Since the integrand is even in $\theta$: (A1)由于被积函数关于 $\theta$ 为偶函数：(A1)
$$ A=2\int_{0}^{\pi/4}(1+\cos 4\theta)\,d\theta=2\left[\theta+\frac{\sin 4\theta}{4}\right]_{0}^{\pi/4}=2\!\left(\frac{\pi}{4}+\frac{\sin\pi}{4}\right)=2\cdot\frac{\pi}{4}=\frac{\pi}{2}. $$(A1) The area of one petal is $\pi/2$.(A1) 一片花瓣的面积为 $\pi/2$。
Compute $dr/d\theta=-4\sin(2\theta)$. (M1) The polar arc-length formula gives integrand计算 $dr/d\theta=-4\sin(2\theta)$。(M1) 极坐标弧长公式给出被积函数
$$ \sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}=\sqrt{[2\cos(2\theta)]^{2}+[-4\sin(2\theta)]^{2}}=\sqrt{4\cos^{2}(2\theta)+16\sin^{2}(2\theta)}. $$(M1) Factor:(M1) 提取公因子：
$$ \sqrt{4\cos^{2}(2\theta)+16\sin^{2}(2\theta)}=\sqrt{4\cos^{2}(2\theta)+4\sin^{2}(2\theta)+12\sin^{2}(2\theta)}=\sqrt{4\bigl(\cos^{2}(2\theta)+\sin^{2}(2\theta)\bigr)+12\sin^{2}(2\theta)}. $$$=\sqrt{4+12\sin^{2}(2\theta)}=2\sqrt{1+3\sin^{2}(2\theta)}$. (A1) The arc length of one petal is therefore$=\sqrt{4+12\sin^{2}(2\theta)}=2\sqrt{1+3\sin^{2}(2\theta)}$。(A1) 因此一片花瓣的弧长为
$$ L=\int_{-\pi/4}^{\pi/4}2\sqrt{1+3\sin^{2}(2\theta)}\,d\theta. $$(A1) This is an elliptic integral and cannot be expressed in closed form using elementary functions.(A1) 这是一个椭圆积分，无法用初等函数表达为闭合形式。