Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 6: functions of several variables, domain and level curves, limits and continuity in two variables, partial derivatives, Clairaut's theorem, the multivariable chain rule, and the gradientCALC III第 1 至 6 节：多元函数、定义域与水平曲线、二元极限与连续性、偏导数、克莱罗定理、多元链式法则及梯度CALC III
Let $f(x,y) = \ln(4 - x^2 - y^2)$. (a) State the natural domain. (b) Find level curves $f = k$ and their range. (c) Describe how the level curves change as $k$ increases toward $\ln 4$.设 $f(x,y) = \ln(4 - x^2 - y^2)$。(a) 给出自然定义域。(b) 求水平曲线 $f = k$ 及其取值范围。(c) 描述当 $k$ 趋向 $\ln 4$ 时水平曲线的变化。
The logarithm requires a strictly positive argument. (M1)对数要求真数严格为正。(M1)
$$ 4 - x^2 - y^2 > 0 \iff x^2 + y^2 < 4. $$The domain is the open disk of radius $2$ centred at the origin. The boundary circle $x^2 + y^2 = 4$ makes the argument zero, so $\ln 0$ is undefined; the boundary is excluded. (A1, R1)定义域是以原点为圆心、半径为 $2$ 的开圆盘。边界圆 $x^2 + y^2 = 4$ 使真数为零，$\ln 0$ 无意义，故边界不包含在内。(A1, R1)
Set $f(x,y) = k$: (M1)令 $f(x,y) = k$：(M1)
$$ \ln(4 - x^2 - y^2) = k \implies 4 - x^2 - y^2 = e^k \implies x^2 + y^2 = 4 - e^k. $$For this circle to exist in the domain we need $4 - e^k > 0$, i.e. $e^k < 4$, i.e. $k < \ln 4$. There is no upper bound below $\ln 4$: any $k < \ln 4$ works. The range of $f$ is $(-\infty, \ln 4)$. (A1, A1)要使该圆在定义域内存在，需 $4 - e^k > 0$，即 $e^k < 4$，即 $k < \ln 4$。在 $\ln 4$ 以下没有下界，任意 $k < \ln 4$ 均可。$f$ 的值域为 $(-\infty, \ln 4)$。(A1, A1)
As $k$ increases toward $\ln 4$, the radius $\sqrt{4 - e^k} \to 0$, so the level circles shrink toward the centre $(0,0)$. (M1) Level curves crowding together near the centre signals that the surface $z = \ln(4-x^2-y^2)$ is dropping steeply: the function falls to $-\infty$ as you approach the boundary from inside, so near the centre (high $k$) the contours compress, indicating steepness. (A1)当 $k$ 趋向 $\ln 4$ 时，半径 $\sqrt{4 - e^k} \to 0$，水平圆向圆心 $(0,0)$ 收缩。(M1) 水平曲线在圆心附近聚集，表明曲面 $z = \ln(4-x^2-y^2)$ 急剧下降：从内部趋向边界时，函数趋向 $-\infty$，因此在圆心附近（$k$ 较大处）等值线密集，反映出曲面的陡峭程度。(A1)
Let $f(x,y) = x^3 y^2 - 4xy + e^{x+2y}$. Compute all first and second partial derivatives; verify $f_{xy} = f_{yx}$.设 $f(x,y) = x^3 y^2 - 4xy + e^{x+2y}$。计算所有一阶和二阶偏导数，验证 $f_{xy} = f_{yx}$。
Differentiate with respect to $x$, treating $y$ as a constant: (M1)对 $x$ 求偏导，将 $y$ 视为常数：(M1)
$$ f_x = 3x^2 y^2 - 4y + e^{x+2y}. $$(A1) Differentiate with respect to $y$, treating $x$ as a constant:(A1) 对 $y$ 求偏导，将 $x$ 视为常数：
$$ f_y = 2x^3 y - 4x + 2e^{x+2y}. $$(A1) Note that $\frac{\partial}{\partial y}e^{x+2y} = 2e^{x+2y}$ by the chain rule with inner derivative $2$.(A1) 注意由链式法则，内层导数为 $2$，故 $\frac{\partial}{\partial y}e^{x+2y} = 2e^{x+2y}$。
Differentiate $f_x$ with respect to $x$: (M1)对 $f_x$ 关于 $x$ 求导：(M1)
$$ f_{xx} = 6xy^2 + e^{x+2y}. $$Differentiate $f_y$ with respect to $y$:对 $f_y$ 关于 $y$ 求导：
$$ f_{yy} = 2x^3 + 4e^{x+2y}. $$Mixed partial, $f_{xy} = \frac{\partial}{\partial y}(f_x)$: differentiate $f_x = 3x^2y^2 - 4y + e^{x+2y}$ with respect to $y$: (A1)混合偏导数 $f_{xy} = \frac{\partial}{\partial y}(f_x)$：对 $f_x = 3x^2y^2 - 4y + e^{x+2y}$ 关于 $y$ 求导：(A1)
$$ f_{xy} = 6x^2 y - 4 + 2e^{x+2y}. $$Mixed partial, $f_{yx} = \frac{\partial}{\partial x}(f_y)$: differentiate $f_y = 2x^3 y - 4x + 2e^{x+2y}$ with respect to $x$: (A1)混合偏导数 $f_{yx} = \frac{\partial}{\partial x}(f_y)$：对 $f_y = 2x^3 y - 4x + 2e^{x+2y}$ 关于 $x$ 求导：(A1)
$$ f_{yx} = 6x^2 y - 4 + 2e^{x+2y}. $$(A1) The two expressions are identical.(A1) 两个表达式完全相同。
$f_{xy} = f_{yx} = 6x^2y - 4 + 2e^{x+2y}$, confirming Clairaut's theorem, which states that if $f_{xy}$ and $f_{yx}$ are both continuous on an open disk, then they are equal there. Since $f$ is a polynomial plus an exponential, all its partials are continuous everywhere, so the equality is guaranteed. (R1)$f_{xy} = f_{yx} = 6x^2y - 4 + 2e^{x+2y}$，验证了克莱罗定理：若 $f_{xy}$ 和 $f_{yx}$ 在某开圆盘上均连续，则二者在该圆盘上相等。由于 $f$ 是多项式与指数函数之和，其所有偏导数处处连续，等式因此成立。(R1)
Let $w = x^2 y - y^2 z$ with $x = t^2$, $y = \sin t$, $z = e^t$. Draw a dependency tree and find $dw/dt$.设 $w = x^2 y - y^2 z$，其中 $x = t^2$，$y = \sin t$，$z = e^t$。画出依赖树并求 $dw/dt$。
The tree has $w$ at the top, three intermediate variables $x$, $y$, $z$, and $t$ at the bottom. Each intermediate links to $t$ with one edge. (M1)树以 $w$ 为顶，$x$、$y$、$z$ 为中间变量，$t$ 在底层。每个中间变量通过一条边连接至 $t$。(M1)
The six elementary derivatives are:六个初等导数为：
$$ w_x = 2xy, \quad w_y = x^2 - 2yz, \quad w_z = -y^2, $$ $$ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = \cos t, \quad \frac{dz}{dt} = e^t. $$(A1 for the $w$-partials, A1 for the $t$-derivatives)（$w$ 的偏导数得 A1，$t$ 的导数得 A1）
Sum one term per path from $t$ up to $w$: (M1)对从 $t$ 到 $w$ 的每条路径求和一项：(M1)
$$ \frac{dw}{dt} = w_x \frac{dx}{dt} + w_y \frac{dy}{dt} + w_z \frac{dz}{dt}. $$ $$ = (2xy)(2t) + (x^2 - 2yz)(\cos t) + (-y^2)(e^t). $$Substitute $x = t^2$, $y = \sin t$, $z = e^t$: (A1)代入 $x = t^2$，$y = \sin t$，$z = e^t$：(A1)
$$ = 2(t^2)(\sin t)(2t) + \bigl((t^2)^2 - 2(\sin t)(e^t)\bigr)\cos t + (-\sin^2 t)(e^t). $$ $$ = 4t^3 \sin t + (t^4 - 2e^t \sin t)\cos t - e^t \sin^2 t. $$ $$ = 4t^3 \sin t + t^4 \cos t - 2e^t \sin t \cos t - e^t \sin^2 t. $$(A1)
Let $f(x,y) = x^2 - xy + 2y^2$. (a) $\nabla f$ at $(2,-1)$; (b) unit steepest-ascent direction and max rate; (c) level curve equation and perpendicularity check.设 $f(x,y) = x^2 - xy + 2y^2$。(a) 求点 $(2,-1)$ 处的 $\nabla f$；(b) 最速上升方向的单位向量及最大增长率；(c) 水平曲线方程及垂直性验证。
Compute the first partials: (M1)计算一阶偏导数：(M1)
$$ f_x = 2x - y, \quad f_y = -x + 4y. $$Evaluate at $(2,-1)$: $f_x(2,-1) = 4-(-1) = 5$ and $f_y(2,-1) = -2 + 4(-1) = -6$. So在点 $(2,-1)$ 处求值：$f_x(2,-1) = 4-(-1) = 5$，$f_y(2,-1) = -2 + 4(-1) = -6$。故
$$ \nabla f(2,-1) = \langle 5, -6\rangle. \quad \text{(A1)} $$The direction of steepest increase is $\nabla f / |\nabla f|$. (M1) The magnitude is最速上升方向为 $\nabla f / |\nabla f|$。(M1) 模长为
$$ |\nabla f(2,-1)| = \sqrt{25 + 36} = \sqrt{61}. $$The unit direction is $\mathbf{u} = \dfrac{1}{\sqrt{61}}\langle 5,-6\rangle$ and the maximum rate of increase is $\sqrt{61}$. (A1)单位方向向量为 $\mathbf{u} = \dfrac{1}{\sqrt{61}}\langle 5,-6\rangle$，最大增长率为 $\sqrt{61}$。(A1)
Evaluate $f(2,-1) = 4 - (2)(-1) + 2(1) = 4 + 2 + 2 = 8$. The level curve through $(2,-1)$ is (M1)计算 $f(2,-1) = 4 - (2)(-1) + 2(1) = 4 + 2 + 2 = 8$。过点 $(2,-1)$ 的水平曲线为 (M1)
$$ x^2 - xy + 2y^2 = 8. $$Geometrically, $\nabla f$ is perpendicular to this curve because the gradient is always orthogonal to the level sets of $f$: any tangent vector $\mathbf{r}'$ to the level curve satisfies $\nabla f \cdot \mathbf{r}' = 0$ by differentiating the constant relation $f(\mathbf{r}(t)) = 8$ with the chain rule. (A1)从几何上看，$\nabla f$ 与该水平曲线垂直，因为梯度总是正交于 $f$ 的水平集：对常值关系 $f(\mathbf{r}(t)) = 8$ 用链式法则求导，可得水平曲线的任意切向量 $\mathbf{r}'$ 均满足 $\nabla f \cdot \mathbf{r}' = 0$。(A1)
(a) Show $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}$ does not exist. (b) Show $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2 y}{x^4+y^2}$ does not exist; explain why straight-line paths are insufficient.(a) 证明 $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}$ 不存在。(b) 证明 $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2 y}{x^4+y^2}$ 不存在；解释为何仅凭直线路径不足以说明问题。
Path 1:路径 1： $y = 0$ (approach along the $x$-axis). Substitute $y = 0$: (M1)（沿 $x$ 轴趋近）。代入 $y = 0$：(M1)
$$ \frac{x^2 - 0}{x^2 + 0} = \frac{x^2}{x^2} = 1. $$As $x \to 0$, the value is constantly $1$, so the limit along this path is $1$. (A1)当 $x \to 0$ 时，值恒为 $1$，故沿此路径的极限为 $1$。(A1)
Path 2:路径 2： $x = 0$ (approach along the $y$-axis). Substitute $x = 0$: (M1)（沿 $y$ 轴趋近）。代入 $x = 0$：(M1)
$$ \frac{0 - y^2}{0 + y^2} = \frac{-y^2}{y^2} = -1. $$The limit along this path is $-1$. (A1) Since the two paths give different values ($1 \ne -1$), the two-variable limit does not exist.沿此路径的极限为 $-1$。(A1) 由于两条路径给出不同的值（$1 \ne -1$），二元极限不存在。
Every line $y = mx$:每条直线 $y = mx$： Substitute $y = mx$ ($m$ any real constant): (M1)代入 $y = mx$（$m$ 为任意实常数）：(M1)
$$ \frac{x^2 \cdot mx}{x^4 + m^2 x^2} = \frac{mx^3}{x^2(x^2 + m^2)} = \frac{mx}{x^2 + m^2}. $$As $x \to 0$, this tends to $\dfrac{m \cdot 0}{0 + m^2} = 0$ (for $m \ne 0$), and for $m=0$ the numerator is identically $0$. So every straight-line path gives limit $0$. (A1)当 $x \to 0$ 时，趋向 $\dfrac{m \cdot 0}{0 + m^2} = 0$（$m \ne 0$ 时），$m = 0$ 时分子恒为 $0$。故所有直线路径均给出极限 $0$。(A1)
Parabolic path $y = x^2$:抛物线路径 $y = x^2$： Substitute $y = x^2$: (M1)代入 $y = x^2$：(M1)
$$ \frac{x^2 \cdot x^2}{x^4 + (x^2)^2} = \frac{x^4}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2}. $$The value is constantly $\tfrac12$ along this path, so the limit along the parabola is $\tfrac12 \ne 0$. (A1) Since two paths disagree, the limit does not exist. The example also demonstrates that checking only straight lines cannot prove a limit exists: path-testing can only disprove a limit, never establish it.沿此路径的值恒为 $\tfrac12$，故沿抛物线的极限为 $\tfrac12 \ne 0$。(A1) 由于两条路径结果不一致，极限不存在。此例同时说明，仅检验直线路径无法证明极限存在：路径检验只能否定极限的存在，无法证明它存在。
Prove using polar coordinates: (a) $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^3}{x^2+y^2} = 0$; (b) $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2} = 0$.使用极坐标证明：(a) $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^3}{x^2+y^2} = 0$；(b) $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2} = 0$。
Set $x = r\cos\theta$, $y = r\sin\theta$ so $x^2 + y^2 = r^2$. (M1)令 $x = r\cos\theta$，$y = r\sin\theta$，则 $x^2 + y^2 = r^2$。(M1)
$$ \left|\frac{x^3}{x^2+y^2}\right| = \left|\frac{r^3\cos^3\theta}{r^2}\right| = r\,|\cos^3\theta|. $$(A1) Since $|\cos\theta| \le 1$ for all $\theta$, we have $r\,|\cos^3\theta| \le r$. (M1) As $r \to 0$ (i.e. $(x,y)\to(0,0)$), this upper bound tends to $0$, independent of $\theta$. By the squeeze theorem (with lower bound $0$),(A1) 由于对所有 $\theta$ 均有 $|\cos\theta| \le 1$，故 $r\,|\cos^3\theta| \le r$。(M1) 当 $r \to 0$（即 $(x,y)\to(0,0)$）时，该上界趋向 $0$，与 $\theta$ 无关。由夹逼定理（下界为 $0$），
$$ 0 \le \left|\frac{x^3}{x^2+y^2}\right| \le r \to 0, $$so the limit is $0$. (A1)故极限为 $0$。(A1)
In polar form: (M1)极坐标形式：(M1)
$$ \left|\frac{x^2y^2}{x^2+y^2}\right| = \left|\frac{r^4\cos^2\theta\sin^2\theta}{r^2}\right| = r^2|\cos^2\theta\sin^2\theta|. $$(A1) Since $|\cos^2\theta\sin^2\theta| \le 1$ for all $\theta$ (product of two factors each at most $1$), we have (M1)(A1) 由于对所有 $\theta$ 均有 $|\cos^2\theta\sin^2\theta| \le 1$（两个均不超过 $1$ 的因子之积），故 (M1)
$$ \left|\frac{x^2y^2}{x^2+y^2}\right| \le r^2. $$As $r \to 0$, $r^2 \to 0$ regardless of $\theta$. By the squeeze theorem, the limit is $0$. (A1)当 $r \to 0$ 时，$r^2 \to 0$，与 $\theta$ 无关。由夹逼定理，极限为 $0$。(A1)
(a) State Clairaut's theorem. (b) For $f(x,y) = x^4y^3 - \sin(xy^2)$, compute $f_{xy}$ and $f_{yx}$ in full. (c) Show $u = e^x\cos y$ is harmonic and verify $u_{xy} = u_{yx}$.(a) 陈述克莱罗定理。(b) 对 $f(x,y) = x^4y^3 - \sin(xy^2)$，完整计算 $f_{xy}$ 和 $f_{yx}$。(c) 证明 $u = e^x\cos y$ 是调和函数并验证 $u_{xy} = u_{yx}$。
Clairaut's theorem.克莱罗定理。 If $f$ is defined on an open disk $D$ containing $(a,b)$ and if the mixed partial derivatives $f_{xy}$ and $f_{yx}$ are both continuous on $D$, then $f_{xy}(a,b) = f_{yx}(a,b)$. (A1, R1) The continuity hypothesis is necessary: counterexamples exist for functions where the mixed partials fail to be continuous.若 $f$ 在包含点 $(a,b)$ 的开圆盘 $D$ 上有定义，且混合偏导数 $f_{xy}$ 和 $f_{yx}$ 在 $D$ 上均连续，则 $f_{xy}(a,b) = f_{yx}(a,b)$。(A1, R1) 连续性假设是必要的：对混合偏导数不连续的函数，存在反例。
Compute $f_x$ first (treat $y$ as constant): (M1)先计算 $f_x$（将 $y$ 视为常数）：(M1)
$$ f_x = 4x^3 y^3 - y^2\cos(xy^2). $$Explanation: $\frac{\partial}{\partial x}\sin(xy^2) = \cos(xy^2)\cdot y^2$ by the chain rule. (A1)说明：由链式法则，$\frac{\partial}{\partial x}\sin(xy^2) = \cos(xy^2)\cdot y^2$。(A1)
Differentiate $f_x$ with respect to $y$ to get $f_{xy}$: (M1)对 $f_x$ 关于 $y$ 求导得 $f_{xy}$：(M1)
$$ f_{xy} = \frac{\partial}{\partial y}\bigl(4x^3 y^3 - y^2\cos(xy^2)\bigr). $$ $$ = 12x^3 y^2 - \bigl[2y\cos(xy^2) + y^2 \cdot (-\sin(xy^2))\cdot 2xy\bigr]. $$ $$ = 12x^3 y^2 - 2y\cos(xy^2) + 2xy^3\sin(xy^2). $$(A1) Now compute $f_y$ (treat $x$ as constant):(A1) 再计算 $f_y$（将 $x$ 视为常数）：
$$ f_y = 3x^4 y^2 - 2xy\cos(xy^2). $$Explanation: $\frac{\partial}{\partial y}\sin(xy^2) = \cos(xy^2)\cdot 2xy$. Differentiate $f_y$ with respect to $x$ to get $f_{yx}$:说明：$\frac{\partial}{\partial y}\sin(xy^2) = \cos(xy^2)\cdot 2xy$。对 $f_y$ 关于 $x$ 求导得 $f_{yx}$：
$$ f_{yx} = \frac{\partial}{\partial x}\bigl(3x^4 y^2 - 2xy\cos(xy^2)\bigr). $$ $$ = 12x^3 y^2 - \bigl[2y\cos(xy^2) + 2xy\cdot(-\sin(xy^2))\cdot y^2\bigr]. $$ $$ = 12x^3 y^2 - 2y\cos(xy^2) + 2xy^3\sin(xy^2). $$(A1) Both mixed partials agree: $f_{xy} = f_{yx} = 12x^3y^2 - 2y\cos(xy^2) + 2xy^3\sin(xy^2)$.(A1) 两个混合偏导数一致：$f_{xy} = f_{yx} = 12x^3y^2 - 2y\cos(xy^2) + 2xy^3\sin(xy^2)$。
Differentiate twice in each variable. In $x$: (M1)对每个变量各求二阶导数。关于 $x$：(M1)
$$ u_x = e^x\cos y, \qquad u_{xx} = e^x\cos y. $$In $y$:关于 $y$：
$$ u_y = -e^x\sin y, \qquad u_{yy} = -e^x\cos y. $$Sum: $u_{xx} + u_{yy} = e^x\cos y - e^x\cos y = 0$, so $u$ is harmonic. (A1)求和：$u_{xx} + u_{yy} = e^x\cos y - e^x\cos y = 0$，故 $u$ 是调和函数。(A1)
Mixed partials: $u_{xy} = \frac{\partial}{\partial y}(e^x\cos y) = -e^x\sin y$ and $u_{yx} = \frac{\partial}{\partial x}(-e^x\sin y) = -e^x\sin y$. They agree, as Clairaut guarantees. (A1)混合偏导数：$u_{xy} = \frac{\partial}{\partial y}(e^x\cos y) = -e^x\sin y$，$u_{yx} = \frac{\partial}{\partial x}(-e^x\sin y) = -e^x\sin y$。二者一致，与克莱罗定理保证一致。(A1)
Let $z = f(x,y)$, $x = s+t$, $y = s-t$. (a) Find $\partial z/\partial s$ and $\partial z/\partial t$. (b) Find $\partial^2 z/\partial s^2$ and $\partial^2 z/\partial t^2$. (c) Show $\partial^2 z/\partial s^2 - \partial^2 z/\partial t^2 = 4f_{xy}$; interpret for harmonic $f$.设 $z = f(x,y)$，$x = s+t$，$y = s-t$。(a) 求 $\partial z/\partial s$ 和 $\partial z/\partial t$。(b) 求 $\partial^2 z/\partial s^2$ 和 $\partial^2 z/\partial t^2$。(c) 证明 $\partial^2 z/\partial s^2 - \partial^2 z/\partial t^2 = 4f_{xy}$，并对调和函数 $f$ 给出解释。
The parameter derivatives are $\partial x/\partial s = 1$, $\partial y/\partial s = 1$, $\partial x/\partial t = 1$, $\partial y/\partial t = -1$. (M1)参数导数为 $\partial x/\partial s = 1$，$\partial y/\partial s = 1$，$\partial x/\partial t = 1$，$\partial y/\partial t = -1$。(M1)
$$ \frac{\partial z}{\partial s} = f_x \frac{\partial x}{\partial s} + f_y \frac{\partial y}{\partial s} = f_x(1) + f_y(1) = f_x + f_y. \quad \text{(A1)} $$ $$ \frac{\partial z}{\partial t} = f_x \frac{\partial x}{\partial t} + f_y \frac{\partial y}{\partial t} = f_x(1) + f_y(-1) = f_x - f_y. \quad \text{(A1)} $$Differentiate $\partial z/\partial s = f_x + f_y$ with respect to $s$, using the chain rule again on $f_x$ and $f_y$: (M1)对 $\partial z/\partial s = f_x + f_y$ 关于 $s$ 求导，对 $f_x$ 和 $f_y$ 再次应用链式法则：(M1)
$$ \frac{\partial^2 z}{\partial s^2} = \frac{\partial}{\partial s}(f_x + f_y) = \left(f_{xx}\frac{\partial x}{\partial s} + f_{xy}\frac{\partial y}{\partial s}\right) + \left(f_{yx}\frac{\partial x}{\partial s} + f_{yy}\frac{\partial y}{\partial s}\right). $$ $$ = (f_{xx}\cdot 1 + f_{xy}\cdot 1) + (f_{yx}\cdot 1 + f_{yy}\cdot 1) = f_{xx} + f_{xy} + f_{yx} + f_{yy}. $$By Clairaut's theorem $f_{xy} = f_{yx}$, so $\partial^2 z/\partial s^2 = f_{xx} + 2f_{xy} + f_{yy}$. (A1)由克莱罗定理 $f_{xy} = f_{yx}$，故 $\partial^2 z/\partial s^2 = f_{xx} + 2f_{xy} + f_{yy}$。(A1)
Differentiate $\partial z/\partial t = f_x - f_y$ with respect to $t$: (M1)对 $\partial z/\partial t = f_x - f_y$ 关于 $t$ 求导：(M1)
$$ \frac{\partial^2 z}{\partial t^2} = \frac{\partial}{\partial t}(f_x - f_y) = \left(f_{xx}\frac{\partial x}{\partial t} + f_{xy}\frac{\partial y}{\partial t}\right) - \left(f_{yx}\frac{\partial x}{\partial t} + f_{yy}\frac{\partial y}{\partial t}\right). $$ $$ = (f_{xx}\cdot 1 + f_{xy}\cdot(-1)) - (f_{yx}\cdot 1 + f_{yy}\cdot(-1)) = f_{xx} - f_{xy} - f_{yx} + f_{yy}. $$Using $f_{xy} = f_{yx}$: $\partial^2 z/\partial t^2 = f_{xx} - 2f_{xy} + f_{yy}$. (A1)使用 $f_{xy} = f_{yx}$：$\partial^2 z/\partial t^2 = f_{xx} - 2f_{xy} + f_{yy}$。(A1)
Subtract the two results: (M1)两式相减：(M1)
$$ \frac{\partial^2 z}{\partial s^2} - \frac{\partial^2 z}{\partial t^2} = (f_{xx}+2f_{xy}+f_{yy}) - (f_{xx}-2f_{xy}+f_{yy}) = 4f_{xy}. \quad \text{(A1)} $$If $f$ satisfies Laplace's equation $f_{xx} + f_{yy} = 0$, note that both $\partial^2 z/\partial s^2 = f_{xx}+2f_{xy}+f_{yy}$ and $\partial^2 z/\partial t^2 = f_{xx}-2f_{xy}+f_{yy}$ equal $2f_{xy}$ and $-2f_{xy}$ after using $f_{xx} = -f_{yy}$, so their sum $\partial^2 z/\partial s^2 + \partial^2 z/\partial t^2 = 2(f_{xx}+f_{yy}) = 0$. In particular if additionally $f_{xy} = 0$, then both second derivatives in $(s,t)$ vanish. More generally, for harmonic $f$ the second-order information splits cleanly between the $s$ and $t$ directions. (R1)若 $f$ 满足拉普拉斯方程 $f_{xx} + f_{yy} = 0$，利用 $f_{xx} = -f_{yy}$，$\partial^2 z/\partial s^2 = f_{xx}+2f_{xy}+f_{yy}$ 和 $\partial^2 z/\partial t^2 = f_{xx}-2f_{xy}+f_{yy}$ 分别等于 $2f_{xy}$ 和 $-2f_{xy}$，故其和 $\partial^2 z/\partial s^2 + \partial^2 z/\partial t^2 = 2(f_{xx}+f_{yy}) = 0$。特别地，若还有 $f_{xy} = 0$，则 $(s,t)$ 方向的两个二阶导数均为零。更一般地，对调和函数 $f$，二阶信息在 $s$ 和 $t$ 方向之间清晰地分离。(R1)
Surface $F(x,y,z) = x^2z + y^2z^3 - 3xyz = 5$ defines $z = z(x,y)$ implicitly. (a) Find $F_x$, $F_y$, $F_z$. (b) Find $\partial z/\partial x$ and $\partial z/\partial y$. (c) Evaluate both at $(1,1,1)$.曲面 $F(x,y,z) = x^2z + y^2z^3 - 3xyz = 5$ 将 $z = z(x,y)$ 隐式确定。(a) 求 $F_x$、$F_y$、$F_z$。(b) 求 $\partial z/\partial x$ 和 $\partial z/\partial y$。(c) 在点 $(1,1,1)$ 处求值。
Treat $F(x,y,z) = x^2z + y^2z^3 - 3xyz$ as a function of three independent variables and differentiate each one in turn, holding the other two fixed. (M1)将 $F(x,y,z) = x^2z + y^2z^3 - 3xyz$ 视为三个独立变量的函数，依次对每个变量求导，保持其余两个固定。(M1)
$$ F_x = 2xz - 3yz. \quad \text{(A1)} $$ $$ F_y = 2yz^3 - 3xz. $$ $$ F_z = x^2 + 3y^2z^2 - 3xy. \quad \text{(A1)} $$The implicit function theorem (for $F_z \ne 0$) gives (M1)隐函数定理（在 $F_z \ne 0$ 处）给出 (M1)
$$ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2xz - 3yz}{x^2 + 3y^2z^2 - 3xy}. \quad \text{(A1)} $$ $$ \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{2yz^3 - 3xz}{x^2 + 3y^2z^2 - 3xy}. \quad \text{(A1)} $$(A1 for correct sign convention and both formulas)（符号约定正确且两个公式均正确得 A1）
The formulas in (b) express $\partial z/\partial x$ and $\partial z/\partial y$ as ratios of partial derivatives of $F$, valid at any point $(x,y,z)$ where $F_z \ne 0$. Substitute $(x,y,z) = (1,1,1)$: (M1)(b) 中的公式将 $\partial z/\partial x$ 和 $\partial z/\partial y$ 表示为 $F$ 的偏导数之比，在任意满足 $F_z \ne 0$ 的点 $(x,y,z)$ 处均成立。代入 $(x,y,z) = (1,1,1)$：(M1)
$$ F_x(1,1,1) = 2(1)(1) - 3(1)(1) = 2 - 3 = -1. $$ $$ F_y(1,1,1) = 2(1)(1)^3 - 3(1)(1) = 2 - 3 = -1. $$ $$ F_z(1,1,1) = (1)^2 + 3(1)^2(1)^2 - 3(1)(1) = 1 + 3 - 3 = 1 \ne 0. $$ $$ \frac{\partial z}{\partial x}\bigg|_{(1,1,1)} = -\frac{F_x}{F_z} = -\frac{-1}{1} = 1. \qquad \frac{\partial z}{\partial y}\bigg|_{(1,1,1)} = -\frac{F_y}{F_z} = -\frac{-1}{1} = 1. \quad \text{(A1)} $$Surface $F(x,y,z) = x^2+2y^2-z = 4$; function $g(x,y) = x^2+2y^2$. (a) $\nabla F$ at $P=(1,1,-1)$; verify $P$ on $S$. (b) Tangent plane at $P$. (c) $\nabla g(1,1)$; max rate; steepest direction; tangent to level curve.曲面 $F(x,y,z) = x^2+2y^2-z = 4$；函数 $g(x,y) = x^2+2y^2$。(a) 求点 $P=(1,1,-1)$ 处的 $\nabla F$；验证 $P$ 在 $S$ 上。(b) 求 $P$ 处的切平面。(c) 求 $\nabla g(1,1)$、最大增长率、最速上升方向及水平曲线处的切向量。
The partial derivatives of $F(x,y,z) = x^2 + 2y^2 - z$ are: (M1)$F(x,y,z) = x^2 + 2y^2 - z$ 的偏导数为：(M1)
$$ F_x = 2x, \quad F_y = 4y, \quad F_z = -1. $$So $\nabla F = \langle 2x, 4y, -1\rangle$. At $P = (1,1,-1)$:故 $\nabla F = \langle 2x, 4y, -1\rangle$。在点 $P = (1,1,-1)$ 处：
$$ \nabla F(1,1,-1) = \langle 2, 4, -1\rangle. \quad \text{(A1)} $$Verify $P \in S$: $F(1,1,-1) = 1 + 2 - (-1) = 1 + 2 + 1 = 4$. Yes, $P$ lies on $S$. (A1)验证 $P \in S$：$F(1,1,-1) = 1 + 2 - (-1) = 1 + 2 + 1 = 4$。是的，$P$ 在 $S$ 上。(A1)
The gradient $\nabla F(P)$ is the normal vector to $S$ at $P$, so the tangent plane is (M1)梯度 $\nabla F(P)$ 是曲面 $S$ 在点 $P$ 处的法向量，故切平面为 (M1)
$$ F_x(P)(x - 1) + F_y(P)(y - 1) + F_z(P)(z - (-1)) = 0, $$ $$ 2(x - 1) + 4(y - 1) - 1(z + 1) = 0, $$ $$ 2x + 4y - z = 7. \quad \text{(A1)} $$The partial derivatives of $g(x,y) = x^2 + 2y^2$ are $g_x = 2x$ and $g_y = 4y$. At $(1,1)$: (M1)$g(x,y) = x^2 + 2y^2$ 的偏导数为 $g_x = 2x$，$g_y = 4y$。在点 $(1,1)$ 处：(M1)
$$ \nabla g(1,1) = \langle 2, 4\rangle. $$The maximum rate of increase is the magnitude:最大增长率为梯度的模长：
$$ |\nabla g(1,1)| = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. \quad \text{(A1)} $$The unit direction of steepest increase is:最速上升方向的单位向量为：
$$ \mathbf{u} = \frac{\nabla g}{|\nabla g|} = \frac{1}{2\sqrt{5}}\langle 2, 4\rangle = \frac{1}{\sqrt{5}}\langle 1, 2\rangle. \quad \text{(A1)} $$A unit vector tangent to the level curve $g = 3$ at $(1,1)$ is perpendicular to $\nabla g$. Rotating $\langle 2,4\rangle$ by $90°$ gives $\langle -4, 2\rangle$. Normalising:水平曲线 $g = 3$ 在点 $(1,1)$ 处的单位切向量与 $\nabla g$ 垂直。将 $\langle 2,4\rangle$ 旋转 $90°$ 得 $\langle -4, 2\rangle$。单位化：
$$ \mathbf{t} = \frac{1}{2\sqrt{5}}\langle -4, 2\rangle = \frac{1}{\sqrt{5}}\langle -2, 1\rangle. \quad \text{(A1)} $$