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
Name:姓名：Date:日期：
PART I · CORE TECHNIQUES核心技术Computational fluency · 28 marks计算熟练度 · 28 分
Domain, Level Curves, and First Partial Derivatives定义域、水平曲线与一阶偏导数
Show all working. State the domain as a set with explicit inequalities. When computing partial derivatives, identify which variable you are treating as constant at each step.展示全部过程。将定义域表示为含有明确不等式的集合。计算偏导数时，在每一步指明将哪个变量视为常数。
Q1MEDIUMCOREdomain and level curves of a two-variable function二元函数的定义域与水平曲线[8 marks]
Let $f(x,y) = \ln(4 - x^2 - y^2)$.
(a)State the natural domain of $f$ as a set in the $xy$-plane. Describe the boundary and whether it is included.将 $f$ 的自然定义域表示为 $xy$ 平面上的集合。描述边界并说明是否包含在内。[3]
(b)Find the equation of the level curve $f(x,y) = k$ and describe its shape. Determine the range of values of $k$ for which a level curve exists.求水平曲线 $f(x,y) = k$ 的方程并描述其形状。确定水平曲线存在时 $k$ 的取值范围。[3]
(c)As $k$ increases from $-\infty$ toward $\ln 4$, what happens to the level curves geometrically? Relate this to the steepness of the surface $z = f(x,y)$ near the boundary.当 $k$ 从 $-\infty$ 增大趋向 $\ln 4$ 时，水平曲线的几何形态如何变化？将此与曲面 $z = f(x,y)$ 在边界附近的陡峭程度联系起来。[2]
Q2MEDIUMCOREfirst and second partial derivatives一阶与二阶偏导数[8 marks]
Let $f(x,y) = x^3 y^2 - 4xy + e^{x+2y}$.
(a)Compute $f_x$ and $f_y$.计算 $f_x$ 和 $f_y$。[3]
(b)Compute $f_{xx}$, $f_{yy}$, and both mixed partials $f_{xy}$ and $f_{yx}$.计算 $f_{xx}$、$f_{yy}$ 以及两个混合偏导数 $f_{xy}$ 和 $f_{yx}$。[4]
(c)Verify that $f_{xy} = f_{yx}$ and state the theorem that guarantees this for smooth functions.验证 $f_{xy} = f_{yx}$，并陈述对光滑函数保证此结论成立的定理。[1]
Q3MEDIUMCOREchain rule with one independent variable单自变量链式法则[6 marks]
Let $w = x^2 y - y^2 z$ with $x = t^2$, $y = \sin t$, $z = e^t$.
(a)Draw a dependency tree showing how $w$ depends on $t$ through $x$, $y$, and $z$. Write down all six elementary derivatives needed.画出依赖树，展示 $w$ 如何通过 $x$、$y$、$z$ 依赖于 $t$。写出所需的全部六个初等导数。[3]
(b)Apply the multivariable chain rule to find $dw/dt$ in terms of $t$ only. Leave the answer in simplified but unsimplified-exponential form.运用多元链式法则，求仅用 $t$ 表示的 $dw/dt$。答案保留化简后但指数项不展开的形式。[3]
Q4MEDIUMCOREgradient computation and steepest-ascent direction梯度计算与最速上升方向[6 marks]
Let $f(x,y) = x^2 - xy + 2y^2$.
(a)Compute $\nabla f$ at the point $(2, -1)$.计算点 $(2, -1)$ 处的 $\nabla f$。[2]
(b)Find the unit vector in the direction of steepest increase of $f$ at $(2,-1)$, and state the maximum rate of increase.求 $f$ 在点 $(2,-1)$ 处最速上升方向的单位向量，并给出最大增长率。[2]
(c)Write the equation of the level curve $f(x,y) = f(2,-1)$ and confirm geometrically that $\nabla f(2,-1)$ is perpendicular to it.写出水平曲线 $f(x,y) = f(2,-1)$ 的方程，并从几何角度确认 $\nabla f(2,-1)$ 垂直于该曲线。[2]
PART II · DEFINITIONS AND PROOF定义与证明Rigorous arguments · 26 marks严格论证 · 26 分
Limits in Two Variables and Clairaut's Theorem二元极限与克莱罗定理
These items are graded on the logic of the argument. A two-path proof must name both paths, substitute explicitly, and compare the values obtained. A proof of existence requires a bound that holds in all directions simultaneously, not just along selected paths.本部分按论证逻辑评分。双路径证明必须命名两条路径、显式代入并比较所得值。存在性证明需要一个在所有方向同时成立的上界，而非仅沿特定路径。
Q5HARDPROOFtwo-path test: proving a limit does not exist双路径检验：证明极限不存在[8 marks]
Investigate each limit as $(x,y) \to (0,0)$.研究以下各极限，其中 $(x,y) \to (0,0)$。
(a)Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist by testing two straight-line paths. State explicitly what each path gives and why the conclusion follows.通过检验两条直线路径，证明 $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$ 不存在。明确说明每条路径的结果以及由此得出结论的原因。[4]
(b)Show that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^4 + y^2}$ does not exist. (Hint: every straight line $y = mx$ gives $0$, so you must try a curved path. Let $y = x^2$.) Explain why checking only straight lines is insufficient to prove existence.证明 $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^4 + y^2}$ 不存在。（提示：每条直线 $y = mx$ 均给出 $0$，因此必须尝试曲线路径。令 $y = x^2$。）解释为何仅检验直线路径不足以证明极限存在。[4]
Q6HARDPROOFpolar-coordinate squeeze to prove a limit exists极坐标夹逼法证明极限存在[8 marks]
Use polar coordinates $x = r\cos\theta$, $y = r\sin\theta$ to analyse limits at the origin.使用极坐标 $x = r\cos\theta$、$y = r\sin\theta$ 分析原点处的极限。
(a)Prove that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^3}{x^2 + y^2} = 0$. Write the expression in polar form, bound it by a quantity that depends only on $r$, and apply the squeeze theorem. State clearly why the bound is independent of $\theta$.证明 $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^3}{x^2 + y^2} = 0$。将表达式化为极坐标形式，用仅依赖 $r$ 的量对其进行上界估计，并应用夹逼定理。清楚说明该上界为何与 $\theta$ 无关。[4]
(b)Now prove that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y^2}{x^2 + y^2} = 0$ by a similar polar argument.用类似的极坐标方法证明 $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^2 y^2}{x^2 + y^2} = 0$。[4]
Q7HARDPROOFClairaut's theorem: statement, verification, and counterexample insight克莱罗定理：陈述、验证与反例分析[10 marks]
This question concerns the equality of mixed partial derivatives.本题涉及混合偏导数的相等性。
(a)State Clairaut's theorem precisely, including the continuity hypothesis.精确陈述克莱罗定理，包括连续性假设。[2]
(b)For $f(x,y) = x^4 y^3 - \sin(xy^2)$, compute $f_{xy}$ and $f_{yx}$ in full and confirm they agree. Show all intermediate steps.对于 $f(x,y) = x^4 y^3 - \sin(xy^2)$，完整计算 $f_{xy}$ 和 $f_{yx}$ 并确认它们相等。展示所有中间步骤。[5]
(c)Show that $u(x,y) = e^x \cos y$ satisfies Laplace's equation $u_{xx} + u_{yy} = 0$, and as a by-product verify $u_{xy} = u_{yx}$.证明 $u(x,y) = e^x \cos y$ 满足拉普拉斯方程 $u_{xx} + u_{yy} = 0$，并同时验证 $u_{xy} = u_{yx}$。[3]
PART III · APPLICATIONS AND SYNTHESIS应用与综合Extended problems · 28 marks综合题 · 28 分
Chain Rule, Implicit Differentiation, and Gradient Applications链式法则、隐函数微分与梯度应用
Set up each problem cleanly. Carry exact symbolic expressions through intermediate steps. When using the chain rule with several intermediate variables, draw the dependency tree and label every edge before writing the formula.清晰建立每道题的框架。在中间步骤中保留精确符号表达式。使用含多个中间变量的链式法则时，先画出依赖树并标注每条边，再写出公式。
Q8HARDAPPLIEDchain rule with two independent variables and a Laplacian identity双自变量链式法则与拉普拉斯恒等式[10 marks]
Let $z = f(x,y)$ where $x = s + t$ and $y = s - t$. Assume $f$ has continuous second-order partial derivatives.设 $z = f(x,y)$，其中 $x = s + t$，$y = s - t$。假设 $f$ 具有连续的二阶偏导数。
(a)Use the chain rule to express $\partial z/\partial s$ and $\partial z/\partial t$ in terms of $f_x$ and $f_y$.使用链式法则，将 $\partial z/\partial s$ 和 $\partial z/\partial t$ 用 $f_x$ 和 $f_y$ 表示。[3]
(b)Differentiate again to find $\partial^2 z / \partial s^2$ and $\partial^2 z / \partial t^2$, each in terms of $f_{xx}$, $f_{xy}$, $f_{yx}$, $f_{yy}$. You may use $f_{xy} = f_{yx}$ by Clairaut's theorem.再次微分，求用 $f_{xx}$、$f_{xy}$、$f_{yx}$、$f_{yy}$ 表示的 $\partial^2 z / \partial s^2$ 和 $\partial^2 z / \partial t^2$。可由克莱罗定理使用 $f_{xy} = f_{yx}$。[4]
(c)Show that $\dfrac{\partial^2 z}{\partial s^2} - \dfrac{\partial^2 z}{\partial t^2} = 4\,f_{xy}$. Interpret this identity: if $f$ satisfies Laplace's equation $f_{xx} + f_{yy} = 0$, what additional relationship holds between $\partial^2 z/\partial s^2$ and $\partial^2 z/\partial t^2$?证明 $\dfrac{\partial^2 z}{\partial s^2} - \dfrac{\partial^2 z}{\partial t^2} = 4\,f_{xy}$。解释此恒等式：若 $f$ 满足拉普拉斯方程 $f_{xx} + f_{yy} = 0$，则 $\partial^2 z/\partial s^2$ 与 $\partial^2 z/\partial t^2$ 之间还有什么关系？[3]
Q9HARDAPPLIEDimplicit partial differentiation from a surface constraint曲面约束下的隐函数偏微分[9 marks]
The equation $F(x,y,z) = x^2 z + y^2 z^3 - 3xyz = 5$ implicitly defines $z$ as a function of $x$ and $y$ near a point where $F_z \ne 0$.方程 $F(x,y,z) = x^2 z + y^2 z^3 - 3xyz = 5$ 在满足 $F_z \ne 0$ 的点附近将 $z$ 隐式定义为 $x$ 和 $y$ 的函数。
(a)Compute the three partial derivatives $F_x$, $F_y$, $F_z$ of the function $F(x,y,z) = x^2 z + y^2 z^3 - 3xyz$.计算函数 $F(x,y,z) = x^2 z + y^2 z^3 - 3xyz$ 的三个偏导数 $F_x$、$F_y$、$F_z$。[3]
(b)Using the implicit differentiation formula $\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}$ and $\dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}$, find $\partial z/\partial x$ and $\partial z/\partial y$ in terms of $x$, $y$, and $z$.使用隐函数微分公式 $\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}$ 和 $\dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}$，求以 $x$、$y$、$z$ 表示的 $\partial z/\partial x$ 和 $\partial z/\partial y$。[4]
(c)Evaluate $\partial z/\partial x$ and $\partial z/\partial y$ at the point $(x,y,z) = (1,1,1)$ by substituting into the formulas from (b). Note: this point need not lie on the surface $F = 5$; the formulas in (b) give the partial derivatives in terms of $x$, $y$, $z$ at any point where $F_z \ne 0$.将 $(x,y,z) = (1,1,1)$ 代入 (b) 的公式，求此点处的 $\partial z/\partial x$ 和 $\partial z/\partial y$。注意：该点不必在曲面 $F = 5$ 上；(b) 中的公式在任意满足 $F_z \ne 0$ 的点处均给出以 $x$、$y$、$z$ 表示的偏导数。[2]
Q10HARDAPPLIEDgradient, level surfaces, and normal vector to a surface梯度、水平曲面与曲面法向量[9 marks]
Consider the surface $S$ defined by $F(x,y,z) = x^2 + 2y^2 - z = 4$ and the function $g(x,y) = x^2 + 2y^2$.考虑由 $F(x,y,z) = x^2 + 2y^2 - z = 4$ 定义的曲面 $S$ 以及函数 $g(x,y) = x^2 + 2y^2$。
(a)Compute $\nabla F(x,y,z)$ and evaluate it at the point $P = (1, 1, -1)$. Verify that $P$ lies on $S$.计算 $\nabla F(x,y,z)$ 并在点 $P = (1, 1, -1)$ 处求值。验证 $P$ 在 $S$ 上。[3]
(b)Write the equation of the tangent plane to $S$ at $P$ using the gradient as the normal vector.以梯度为法向量，写出曲面 $S$ 在点 $P$ 处的切平面方程。[2]
(c)Compute $\nabla g(1,1)$ for the two-variable function $g(x,y) = x^2 + 2y^2$. Find the maximum rate of increase of $g$ at $(1,1)$, the unit direction achieving it, and a unit vector tangent to the level curve $g = 3$ at $(1,1)$.计算二元函数 $g(x,y) = x^2 + 2y^2$ 在点 $(1,1)$ 处的 $\nabla g(1,1)$。求 $g$ 在 $(1,1)$ 处的最大增长率、实现最大增长率的单位方向向量，以及水平曲线 $g = 3$ 在点 $(1,1)$ 处的单位切向量。[4]