Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 6: directional derivatives, the gradient, steepest ascent, tangent planes, linearization, differentials, gradients and level surfaces第 1 至 6 节：方向导数、梯度、最陡上升方向、切平面、线性化、微分、梯度与等值面CALC III
Let $f(x,y)=x^{2}y+y^{3}$. (a) Compute $\nabla f(x,y)$ and evaluate at $P=(1,2)$. (b) Find $D_{\mathbf{u}}f(P)$ in the direction of $\mathbf{v}=\langle 3,4\rangle$, first normalising to a unit vector. (c) Direction and rate of maximum increase at $P$.设 $f(x,y)=x^{2}y+y^{3}$。(a) 计算 $\nabla f(x,y)$ 并在 $P=(1,2)$ 处求值。(b) 先将 $\mathbf{v}=\langle 3,4\rangle$ 化为单位向量，再求 $f$ 在 $P$ 处的方向导数 $D_{\mathbf{u}}f(P)$。(c) $f$ 在 $P$ 处的最大增长方向及最大增长率。
Differentiate: $f_{x}=2xy$ and $f_{y}=x^{2}+3y^{2}$. (M1) At $(1,2)$: $f_{x}=2(1)(2)=4$ and $f_{y}=1^{2}+3(2)^{2}=1+12=13$. (A1) Hence $\nabla f(1,2)=\langle 4,13\rangle$. (A1)求导：$f_{x}=2xy$，$f_{y}=x^{2}+3y^{2}$。(M1) 在 $(1,2)$ 处：$f_{x}=2(1)(2)=4$，$f_{y}=1^{2}+3(2)^{2}=1+12=13$。(A1) 因此 $\nabla f(1,2)=\langle 4,13\rangle$。(A1)
The direction vector $\mathbf{v}=\langle 3,4\rangle$ has magnitude $|\mathbf{v}|=\sqrt{9+16}=5$. (M1) The unit vector is $\mathbf{u}=\tfrac{1}{5}\langle 3,4\rangle=\langle\tfrac{3}{5},\tfrac{4}{5}\rangle$. (A1)方向向量 $\mathbf{v}=\langle 3,4\rangle$ 的模为 $|\mathbf{v}|=\sqrt{9+16}=5$。(M1) 单位向量为 $\mathbf{u}=\tfrac{1}{5}\langle 3,4\rangle=\langle\tfrac{3}{5},\tfrac{4}{5}\rangle$。(A1)
$$ D_{\mathbf{u}}f(1,2)=\nabla f(1,2)\cdot\mathbf{u}=4\cdot\frac{3}{5}+13\cdot\frac{4}{5}=\frac{12}{5}+\frac{52}{5}=\frac{64}{5}. $$(A1)
The maximum directional derivative equals $|\nabla f(1,2)|=\sqrt{4^{2}+13^{2}}=\sqrt{16+169}=\sqrt{185}$, attained in the direction $\mathbf{u}^{*}=\langle 4,13\rangle/\sqrt{185}$. (A1·A1)最大方向导数等于 $|\nabla f(1,2)|=\sqrt{4^{2}+13^{2}}=\sqrt{16+169}=\sqrt{185}$，在方向 $\mathbf{u}^{*}=\langle 4,13\rangle/\sqrt{185}$ 上取得。(A1·A1)
Let $f(x,y)=3x^{2}-xy+y^{2}$. (a) Find $f_{x}$, $f_{y}$ and evaluate at $Q=(1,-1)$. (b) Equation of the tangent plane at $(1,-1,f(1,-1))$. (c) Estimate $f(1.02,-0.97)$ from the tangent plane.设 $f(x,y)=3x^{2}-xy+y^{2}$。(a) 求 $f_{x}$、$f_{y}$ 并在 $Q=(1,-1)$ 处求值。(b) 切平面在 $(1,-1,f(1,-1))$ 处的方程。(c) 利用切平面估计 $f(1.02,-0.97)$。
$f_{x}=6x-y$ and $f_{y}=-x+2y$. (M1) At $(1,-1)$: $f_{x}=6(1)-(-1)=7$ and $f_{y}=-1+2(-1)=-3$. (A1·A1)$f_{x}=6x-y$，$f_{y}=-x+2y$。(M1) 在 $(1,-1)$ 处：$f_{x}=6(1)-(-1)=7$，$f_{y}=-1+2(-1)=-3$。(A1·A1)
First, $f(1,-1)=3(1)^{2}-(1)(-1)+(-1)^{2}=3+1+1=5$, so the base point is $(1,-1,5)$. (M1) The tangent plane formula $z=f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ gives (A1):首先，$f(1,-1)=3(1)^{2}-(1)(-1)+(-1)^{2}=3+1+1=5$，故基点为 $(1,-1,5)$。(M1) 切平面公式 $z=f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ 给出 (A1)：
$$ z=5+7(x-1)+(-3)(y-(-1))=5+7(x-1)-3(y+1). $$Equivalently, $z=7x-3y-5$. (A1)等价地，$z=7x-3y-5$。(A1)
Set $\Delta x=1.02-1=0.02$ and $\Delta y=-0.97-(-1)=0.03$. (M1)令 $\Delta x=1.02-1=0.02$，$\Delta y=-0.97-(-1)=0.03$。(M1)
$$ f(1.02,-0.97)\approx 5+7(0.02)+(-3)(0.03)=5+0.14-0.09=5.05. $$(A1)
Let $f(x,y)=\sqrt{x^{2}+y^{2}}$, $P_{0}=(3,4)$. (a) Write the linearization $L(x,y)$ at $P_{0}$. (b) Write $df$ at $P_{0}$ and estimate $\Delta f$ from $(3,4)$ to $(3.06,3.92)$.设 $f(x,y)=\sqrt{x^{2}+y^{2}}$，$P_{0}=(3,4)$。(a) 写出 $f$ 在 $P_{0}$ 处的线性化 $L(x,y)$。(b) 写出 $P_{0}$ 处的全微分 $df$，并估计从 $(3,4)$ 到 $(3.06,3.92)$ 时 $\Delta f$ 的值。
At $P_{0}=(3,4)$: $f(3,4)=\sqrt{9+16}=5$. (M1) Partial derivatives: $f_{x}=\dfrac{x}{\sqrt{x^{2}+y^{2}}}=\dfrac{3}{5}$ and $f_{y}=\dfrac{y}{\sqrt{x^{2}+y^{2}}}=\dfrac{4}{5}$. (A1) Linearization: (A1)在 $P_{0}=(3,4)$ 处：$f(3,4)=\sqrt{9+16}=5$。(M1) 偏导数：$f_{x}=\dfrac{x}{\sqrt{x^{2}+y^{2}}}=\dfrac{3}{5}$，$f_{y}=\dfrac{y}{\sqrt{x^{2}+y^{2}}}=\dfrac{4}{5}$。(A1) 线性化：(A1)
$$ L(x,y)=5+\frac{3}{5}(x-3)+\frac{4}{5}(y-4). $$The total differential at $P_{0}$ is (M1):$P_{0}$ 处的全微分为 (M1)：
$$ df=\frac{3}{5}\,dx+\frac{4}{5}\,dy. $$Here $dx=3.06-3=0.06$ and $dy=3.92-4=-0.08$. (A1) Hence (A1):这里 $dx=3.06-3=0.06$，$dy=3.92-4=-0.08$。(A1) 因此 (A1)：
$$ \Delta f\approx df=\frac{3}{5}(0.06)+\frac{4}{5}(-0.08)=0.036-0.064=-0.028. $$Let $g(x,y)=x^{2}+4y^{2}$, $P=(2,1)$. (a) Compute $\nabla g(2,1)$ and the max rate of change. (b) Find a unit vector $\mathbf{u}$ with $D_{\mathbf{u}}g(2,1)=0$ and explain geometrically.设 $g(x,y)=x^{2}+4y^{2}$，$P=(2,1)$。(a) 计算 $\nabla g(2,1)$ 及最大变化率。(b) 找一个单位向量 $\mathbf{u}$ 使得 $D_{\mathbf{u}}g(2,1)=0$，并作几何解释。
$g_{x}=2x=4$ and $g_{y}=8y=8$ at $P=(2,1)$. (M1) So $\nabla g(2,1)=\langle 4,8\rangle$ and the maximum rate of change is $|\nabla g|=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}$. (A1)在 $P=(2,1)$ 处：$g_{x}=2x=4$，$g_{y}=8y=8$。(M1) 因此 $\nabla g(2,1)=\langle 4,8\rangle$，最大变化率为 $|\nabla g|=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}$。(A1)
$D_{\mathbf{u}}g=\nabla g\cdot\mathbf{u}=0$ requires $\mathbf{u}$ perpendicular to $\langle 4,8\rangle$. (M1) A vector perpendicular to $\langle 4,8\rangle$ is $\langle 8,-4\rangle$ (swap and negate one component). Normalising: $|\langle 8,-4\rangle|=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$, so $\mathbf{u}=\langle 8,-4\rangle/(4\sqrt{5})=\langle 2,-1\rangle/\sqrt{5}$. (A1) (Either sign works.)$D_{\mathbf{u}}g=\nabla g\cdot\mathbf{u}=0$ 要求 $\mathbf{u}$ 与 $\langle 4,8\rangle$ 垂直。(M1) 与 $\langle 4,8\rangle$ 垂直的向量为 $\langle 8,-4\rangle$（交换分量并取反一个）。化为单位向量：$|\langle 8,-4\rangle|=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$，故 $\mathbf{u}=\langle 8,-4\rangle/(4\sqrt{5})=\langle 2,-1\rangle/\sqrt{5}$。(A1)（两个方向均可。）
Geometric meaning: $g(x,y)=8$ is an ellipse passing through $P=(2,1)$. (M1) The gradient $\nabla g$ is perpendicular to that ellipse, so the tangent to the ellipse at $P$ is exactly the direction in which $g$ does not change. Moving along the level curve keeps $g$ constant, so the directional derivative is zero. (A1)几何含义：$g(x,y)=8$ 是过 $P=(2,1)$ 的椭圆。(M1) 梯度 $\nabla g$ 垂直于该椭圆，因此椭圆在 $P$ 处的切线方向正是 $g$ 不变的方向。沿等值曲线移动保持 $g$ 为常数，故方向导数为零。(A1)
Let $f:\mathbb{R}^{2}\to\mathbb{R}$ be differentiable at $\mathbf{a}=(a_{1},a_{2})$, $\mathbf{u}=(u_{1},u_{2})$ a unit vector. (a) Write the limit definition of $D_{\mathbf{u}}f(\mathbf{a})$ and expand $f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})$ using differentiability. (b) Divide by $h$, take $h\to 0$, conclude $D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}$. (c) Apply Cauchy-Schwarz to show $D_{\mathbf{u}}f\le|\nabla f|$ with equality at $\mathbf{u}=\nabla f/|\nabla f|$.设 $f:\mathbb{R}^{2}\to\mathbb{R}$ 在 $\mathbf{a}=(a_{1},a_{2})$ 处可微，$\mathbf{u}=(u_{1},u_{2})$ 为单位向量。(a) 写出 $D_{\mathbf{u}}f(\mathbf{a})$ 的极限定义，并利用可微性展开 $f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})$。(b) 两边除以 $h$，令 $h\to 0$，得出 $D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}$。(c) 应用柯西-施瓦茨不等式证明 $D_{\mathbf{u}}f\le|\nabla f|$，且等号在 $\mathbf{u}=\nabla f/|\nabla f|$ 时成立。
By definition: (M1)由定义：(M1)
$$ D_{\mathbf{u}}f(\mathbf{a})=\lim_{h\to 0}\frac{f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})}{h}=\lim_{h\to 0}\frac{f(a_{1}+hu_{1},\,a_{2}+hu_{2})-f(a_{1},a_{2})}{h}. $$Since $f$ is differentiable at $\mathbf{a}$, the first-order Taylor condition gives (M1):由于 $f$ 在 $\mathbf{a}$ 处可微，一阶泰勒条件给出 (M1)：
$$ f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})=f_{x}(\mathbf{a})\cdot(hu_{1})+f_{y}(\mathbf{a})\cdot(hu_{2})+\varepsilon(h), $$where $\varepsilon(h)$ is the error term satisfying $\varepsilon(h)/h\to 0$ as $h\to 0$ (this is the defining property of differentiability). (A1) Collecting: (A1)其中 $\varepsilon(h)$ 为满足 $\varepsilon(h)/h\to 0$（$h\to 0$）的误差项（这是可微性的定义性质）。(A1) 整理得：(A1)
$$ f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})=h\bigl(f_{x}(\mathbf{a})u_{1}+f_{y}(\mathbf{a})u_{2}\bigr)+\varepsilon(h). $$Dividing the expansion in (a) by $h$: (M1)将 (a) 中的展开式两边除以 $h$：(M1)
$$ \frac{f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})}{h}=f_{x}(\mathbf{a})u_{1}+f_{y}(\mathbf{a})u_{2}+\frac{\varepsilon(h)}{h}. $$As $h\to 0$, the error ratio $\varepsilon(h)/h\to 0$ by the differentiability hypothesis. (R1) Therefore (A1):当 $h\to 0$ 时，由可微性假设知误差比 $\varepsilon(h)/h\to 0$。(R1) 因此 (A1)：
$$ D_{\mathbf{u}}f(\mathbf{a})=f_{x}(\mathbf{a})u_{1}+f_{y}(\mathbf{a})u_{2}=\nabla f(\mathbf{a})\cdot\mathbf{u}. $$The Cauchy-Schwarz inequality states $|\mathbf{a}\cdot\mathbf{b}|\le|\mathbf{a}||\mathbf{b}|$ for any two vectors. Apply it with $\mathbf{a}=\nabla f(\mathbf{p})$ and $\mathbf{b}=\mathbf{u}$: (M1)柯西-施瓦茨不等式指出对任意两个向量有 $|\mathbf{a}\cdot\mathbf{b}|\le|\mathbf{a}||\mathbf{b}|$。令 $\mathbf{a}=\nabla f(\mathbf{p})$，$\mathbf{b}=\mathbf{u}$ 代入：(M1)
$$ D_{\mathbf{u}}f(\mathbf{a})=\nabla f(\mathbf{a})\cdot\mathbf{u}\le|\nabla f(\mathbf{a})\cdot\mathbf{u}|\le|\nabla f(\mathbf{a})||\mathbf{u}|=|\nabla f(\mathbf{a})|, $$since $|\mathbf{u}|=1$. (A1) Equality in Cauchy-Schwarz holds when the two vectors are parallel: $\mathbf{u}=\nabla f(\mathbf{a})/|\nabla f(\mathbf{a})|$. This unit vector is the gradient direction, confirming that steepest ascent is in the gradient direction with rate $|\nabla f(\mathbf{a})|$. (R1)因为 $|\mathbf{u}|=1$。(A1) 柯西-施瓦茨不等式在两向量平行时取等：$\mathbf{u}=\nabla f(\mathbf{a})/|\nabla f(\mathbf{a})|$。该单位向量即为梯度方向，证实了最陡上升方向为梯度方向，增长率为 $|\nabla f(\mathbf{a})|$。(R1)
Let $F:\mathbb{R}^{3}\to\mathbb{R}$ be differentiable, $S: F(x,y,z)=k$, and $\mathbf{r}(t)$ a smooth curve on $S$ with $\mathbf{r}(0)=\mathbf{p}_{0}$, $\mathbf{r}'(0)=\mathbf{T}$. (a) Differentiate $F(\mathbf{r}(t))=k$ and evaluate at $t=0$. (b) Explain why $\nabla F(\mathbf{p}_{0})$ is normal to $S$. (c) Write the tangent plane equation at $\mathbf{p}_{0}$.设 $F:\mathbb{R}^{3}\to\mathbb{R}$ 可微，$S: F(x,y,z)=k$，$\mathbf{r}(t)$ 为 $S$ 上的光滑曲线，满足 $\mathbf{r}(0)=\mathbf{p}_{0}$，$\mathbf{r}'(0)=\mathbf{T}$。(a) 对 $F(\mathbf{r}(t))=k$ 求导并在 $t=0$ 处求值。(b) 解释为何 $\nabla F(\mathbf{p}_{0})$ 是 $S$ 的法向量。(c) 写出 $\mathbf{p}_{0}$ 处的切平面方程。
Since $\mathbf{r}(t)$ lies on $S$, $F(\mathbf{r}(t))=k$ for all $t$. Differentiating both sides with respect to $t$ using the (multivariable) chain rule: (M1)由于 $\mathbf{r}(t)$ 在 $S$ 上，对所有 $t$ 有 $F(\mathbf{r}(t))=k$。用（多元）链式法则对两边关于 $t$ 求导：(M1)
$$ \frac{d}{dt}F(\mathbf{r}(t))=\nabla F(\mathbf{r}(t))\cdot\mathbf{r}'(t)=0. $$(M1) Evaluating at $t=0$, where $\mathbf{r}(0)=\mathbf{p}_{0}$ and $\mathbf{r}'(0)=\mathbf{T}$: (A1)(M1) 在 $t=0$ 处求值，其中 $\mathbf{r}(0)=\mathbf{p}_{0}$，$\mathbf{r}'(0)=\mathbf{T}$：(A1)
$$ \nabla F(\mathbf{p}_{0})\cdot\mathbf{T}=0. $$This says $\nabla F(\mathbf{p}_{0})$ is orthogonal to the tangent vector $\mathbf{T}$ of every curve $\mathbf{r}$ lying on $S$. (A1)这说明 $\nabla F(\mathbf{p}_{0})$ 与 $S$ 上过该点每条曲线 $\mathbf{r}$ 的切向量 $\mathbf{T}$ 正交。(A1)
The argument in (a) applies to any smooth curve on $S$ through $\mathbf{p}_{0}$, so $\nabla F(\mathbf{p}_{0})$ is orthogonal to every tangent vector of $S$ at $\mathbf{p}_{0}$. (R1) The set of all tangent vectors at $\mathbf{p}_{0}$ spans the tangent plane of $S$, so a vector orthogonal to all of them is by definition a normal to $S$. (A1)(a) 中的论证适用于 $S$ 上过 $\mathbf{p}_{0}$ 的任意光滑曲线，故 $\nabla F(\mathbf{p}_{0})$ 与 $S$ 在 $\mathbf{p}_{0}$ 处的每一切向量正交。(R1) $\mathbf{p}_{0}$ 处所有切向量的集合张成 $S$ 的切平面，因此与所有切向量都正交的向量按定义即为 $S$ 的法向量。(A1)
The tangent plane at $\mathbf{p}_{0}=(x_{0},y_{0},z_{0})$ consists of all points $(x,y,z)$ whose displacement from $\mathbf{p}_{0}$ is perpendicular to the normal $\nabla F(\mathbf{p}_{0})$. (M1) Hence the equation is: (A1)$\mathbf{p}_{0}=(x_{0},y_{0},z_{0})$ 处的切平面由所有满足位移向量与法向量 $\nabla F(\mathbf{p}_{0})$ 垂直的点 $(x,y,z)$ 构成。(M1) 因此方程为：(A1)
$$ F_{x}(x_{0},y_{0},z_{0})(x-x_{0})+F_{y}(x_{0},y_{0},z_{0})(y-y_{0})+F_{z}(x_{0},y_{0},z_{0})(z-z_{0})=0. $$Ellipsoid $F(x,y,z)=x^{2}+2y^{2}+3z^{2}=6$, point $P_{0}=(1,1,1)$. (a) Verify $P_{0}$ lies on the ellipsoid and compute $\nabla F(1,1,1)$. (b) Tangent plane at $P_{0}$. (c) Normal line at $P_{0}$ and its intersection with $z=0$.椭球面 $F(x,y,z)=x^{2}+2y^{2}+3z^{2}=6$，点 $P_{0}=(1,1,1)$。(a) 验证 $P_{0}$ 在椭球面上并计算 $\nabla F(1,1,1)$。(b) $P_{0}$ 处的切平面。(c) $P_{0}$ 处的法线及其与 $z=0$ 的交点。
$F(1,1,1)=1^{2}+2(1)^{2}+3(1)^{2}=1+2+3=6$ ✓. (A1) The partial derivatives are $F_{x}=2x$, $F_{y}=4y$, $F_{z}=6z$, giving $\nabla F(1,1,1)=\langle 2,4,6\rangle$. (A1)$F(1,1,1)=1^{2}+2(1)^{2}+3(1)^{2}=1+2+3=6$ ✓。(A1) 偏导数为 $F_{x}=2x$，$F_{y}=4y$，$F_{z}=6z$，因此 $\nabla F(1,1,1)=\langle 2,4,6\rangle$。(A1)
Using the formula from Q6(c) with normal $\langle 2,4,6\rangle$: (M1)用第 Q6(c) 题的公式，法向量为 $\langle 2,4,6\rangle$：(M1)
$$ 2(x-1)+4(y-1)+6(z-1)=0. $$Expanding: $2x+4y+6z-2-4-6=0$, so $2x+4y+6z=12$. (A1) Dividing by $2$: (A1)展开：$2x+4y+6z-2-4-6=0$，故 $2x+4y+6z=12$。(A1) 两边除以 $2$：(A1)
$$ x+2y+3z=6. $$The normal line through $P_{0}=(1,1,1)$ in the direction $\nabla F=\langle 2,4,6\rangle$ has parametric equations: (M1)过 $P_{0}=(1,1,1)$ 沿 $\nabla F=\langle 2,4,6\rangle$ 方向的法线的参数方程为：(M1)
$$ x=1+2t,\quad y=1+4t,\quad z=1+6t. $$(A1) To find the intersection with $z=0$: set $1+6t=0$, giving $t=-\tfrac{1}{6}$. Then (A1):(A1) 求与 $z=0$ 的交点：令 $1+6t=0$，解得 $t=-\tfrac{1}{6}$。则 (A1)：
$$ x=1+2\!\left(-\tfrac{1}{6}\right)=1-\tfrac{1}{3}=\tfrac{2}{3},\quad y=1+4\!\left(-\tfrac{1}{6}\right)=1-\tfrac{2}{3}=\tfrac{1}{3}. $$The normal line hits the $xy$-plane at $\left(\tfrac{2}{3},\tfrac{1}{3},0\right)$.法线与 $xy$ 平面的交点为 $\left(\tfrac{2}{3},\tfrac{1}{3},0\right)$。
$T(x,y)=100-2x^{2}-3y^{2}$, $A=(3,2)$. (a) $\nabla T(3,2)$, steepest-ascent unit vector, and max rate. (b) Directional derivative toward the origin; physical interpretation. (c) Estimate $\Delta T$ for a step of length $0.1$ in the steepest-ascent direction.$T(x,y)=100-2x^{2}-3y^{2}$，$A=(3,2)$。(a) 求 $\nabla T(3,2)$、最陡上升方向单位向量及最大增长率。(b) 指向原点方向的方向导数及物理解释。(c) 沿最陡上升方向移动步长 $0.1$ 时估计 $\Delta T$。
$T_{x}=-4x$ and $T_{y}=-6y$. (M1) At $(3,2)$: $T_{x}=-12$ and $T_{y}=-12$, so $\nabla T(3,2)=\langle -12,-12\rangle$. (A1)$T_{x}=-4x$，$T_{y}=-6y$。(M1) 在 $(3,2)$ 处：$T_{x}=-12$，$T_{y}=-12$，故 $\nabla T(3,2)=\langle -12,-12\rangle$。(A1)
Maximum rate of increase: $|\nabla T|=\sqrt{(-12)^{2}+(-12)^{2}}=\sqrt{288}=12\sqrt{2}$. (M1) Unit steepest-ascent vector: $\mathbf{u}^{*}=\langle -12,-12\rangle/(12\sqrt{2})=\langle -1,-1\rangle/\sqrt{2}$. (A1)最大增长率：$|\nabla T|=\sqrt{(-12)^{2}+(-12)^{2}}=\sqrt{288}=12\sqrt{2}$。(M1) 最陡上升方向单位向量：$\mathbf{u}^{*}=\langle -12,-12\rangle/(12\sqrt{2})=\langle -1,-1\rangle/\sqrt{2}$。(A1)
The direction from $A=(3,2)$ toward the origin is $\langle 0,0\rangle-\langle 3,2\rangle=\langle -3,-2\rangle$. Normalise: $|\langle -3,-2\rangle|=\sqrt{13}$, so $\mathbf{u}=\langle -3,-2\rangle/\sqrt{13}$. (M1)从 $A=(3,2)$ 指向原点的方向为 $\langle 0,0\rangle-\langle 3,2\rangle=\langle -3,-2\rangle$。化为单位向量：$|\langle -3,-2\rangle|=\sqrt{13}$，故 $\mathbf{u}=\langle -3,-2\rangle/\sqrt{13}$。(M1)
$$ D_{\mathbf{u}}T(3,2)=\nabla T\cdot\mathbf{u}=\langle -12,-12\rangle\cdot\frac{\langle -3,-2\rangle}{\sqrt{13}}=\frac{36+24}{\sqrt{13}}=\frac{60}{\sqrt{13}}\approx 16.64. $$(A1) The positive sign means temperature is increasing as you move toward the origin: the hot spot is near the centre, not the rim. (R1)(A1) 正号表明向原点移动时温度升高：热点靠近中心而非边缘。(R1)
Temperature at $A$: $T(3,2)=100-2(9)-3(4)=100-18-12=70\,^{\circ}$C. (M1) Moving in the steepest-ascent direction by $\Delta s=0.1$, the linearization gives (A1):$A$ 处温度：$T(3,2)=100-2(9)-3(4)=100-18-12=70\,^{\circ}$C。(M1) 沿最陡上升方向移动 $\Delta s=0.1$，线性化给出 (A1)：
$$ \Delta T\approx |\nabla T|\cdot\Delta s=12\sqrt{2}\cdot 0.1=1.2\sqrt{2}\approx 1.70\,^{\circ}\text{C}. $$Estimated temperature after the step: $70+1.2\sqrt{2}\approx 71.70\,^{\circ}$C. (A1)移步后估计温度：$70+1.2\sqrt{2}\approx 71.70\,^{\circ}$C。(A1)
$c=\sqrt{a^{2}+b^{2}}$, $(a,b)=(6,8)$, $|da|\le 0.05$, $|db|\le 0.05$. (a) Exact $c$ and linearization $L(a,b)$. (b) Total differential $dc$ and max absolute error. (c) Max relative error as a percentage; which leg contributes more?$c=\sqrt{a^{2}+b^{2}}$，$(a,b)=(6,8)$，$|da|\le 0.05$，$|db|\le 0.05$。(a) $c$ 的精确值与线性化 $L(a,b)$。(b) 全微分 $dc$ 及最大绝对误差。(c) 最大相对误差（百分比）及哪条直角边贡献更大？
$c(6,8)=\sqrt{36+64}=\sqrt{100}=10$ cm. (M1) Partial derivatives: $c_{a}=a/\sqrt{a^{2}+b^{2}}=6/10=\tfrac{3}{5}$ and $c_{b}=b/\sqrt{a^{2}+b^{2}}=8/10=\tfrac{4}{5}$. (A1) Linearization: (A1)$c(6,8)=\sqrt{36+64}=\sqrt{100}=10$ cm。(M1) 偏导数：$c_{a}=a/\sqrt{a^{2}+b^{2}}=6/10=\tfrac{3}{5}$，$c_{b}=b/\sqrt{a^{2}+b^{2}}=8/10=\tfrac{4}{5}$。(A1) 线性化：(A1)
$$ L(a,b)=10+\frac{3}{5}(a-6)+\frac{4}{5}(b-8). $$The total differential is (M1):全微分为 (M1)：
$$ dc=\frac{3}{5}\,da+\frac{4}{5}\,db. $$(A1) The maximum absolute error occurs when each term is as large as possible in magnitude, i.e., when $|da|=|db|=0.05$ with signs chosen to make both terms positive: (M1)(A1) 当每一项在绝对值意义下尽可能大时，即当 $|da|=|db|=0.05$ 且符号使两项均为正时，绝对误差取最大：(M1)
$$ |dc|_{\max}=\frac{3}{5}\cdot 0.05+\frac{4}{5}\cdot 0.05=0.05\left(\frac{3}{5}+\frac{4}{5}\right)=0.05\cdot\frac{7}{5}=0.07\text{ cm}. $$(A1)
Maximum relative error: (M1)最大相对误差：(M1)
$$ \frac{|dc|_{\max}}{c}=\frac{0.07}{10}=0.007=0.7\%. $$(A1) Leg $b$ contributes more to the absolute error: its coefficient $c_{b}=\tfrac{4}{5}=0.8$ exceeds $c_{a}=\tfrac{3}{5}=0.6$. (R1) This is because $b=8$ is larger than $a=6$, and for a right triangle the longer leg has a steeper effect on the hypotenuse: $c_{b}=b/c$ and $c_{a}=a/c$, so the longer leg always has the larger partial derivative.(A1) 直角边 $b$ 对绝对误差的贡献更大：其系数 $c_{b}=\tfrac{4}{5}=0.8$ 大于 $c_{a}=\tfrac{3}{5}=0.6$。(R1) 这是因为 $b=8$ 大于 $a=6$，对于直角三角形，较长的直角边对斜边的影响更陡：$c_{b}=b/c$，$c_{a}=a/c$，因此较长边始终具有较大的偏导数。
$z=f(x,y)=xe^{y-1}-y^{2}+1$, $P=(1,1,1)$. (a) Compute $\nabla f(1,1)$ and the tangent plane at $P$. (b) Directional derivative of $f$ in the steepest-descent direction. (c) Rewrite as $F(x,y,z)=0$ and confirm $\nabla F(1,1,1)$ is consistent with the tangent plane in (a).$z=f(x,y)=xe^{y-1}-y^{2}+1$，$P=(1,1,1)$。(a) 计算 $\nabla f(1,1)$ 及 $P$ 处的切平面。(b) $f$ 在最陡下降方向上的方向导数。(c) 改写为 $F(x,y,z)=0$，验证 $\nabla F(1,1,1)$ 与 (a) 中切平面相符。
First verify $P$ lies on the surface: $f(1,1)=1\cdot e^{1-1}-1^{2}+1=1-1+1=1$ ✓. Partial derivatives: $f_{x}=e^{y-1}$ and $f_{y}=xe^{y-1}-2y$. (M1) At $(1,1)$: $f_{x}=e^{0}=1$ and $f_{y}=1\cdot e^{0}-2(1)=1-2=-1$. So $\nabla f(1,1)=\langle 1,-1\rangle$. (A1)首先验证 $P$ 在曲面上：$f(1,1)=1\cdot e^{1-1}-1^{2}+1=1-1+1=1$ ✓。偏导数：$f_{x}=e^{y-1}$，$f_{y}=xe^{y-1}-2y$。(M1) 在 $(1,1)$ 处：$f_{x}=e^{0}=1$，$f_{y}=1\cdot e^{0}-2(1)=1-2=-1$。故 $\nabla f(1,1)=\langle 1,-1\rangle$。(A1)
Tangent plane: (M1)切平面：(M1)
$$ z=f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)=1+1(x-1)+(-1)(y-1)=1+x-1-y+1. $$ $$ z=x-y+1. \quad\text{(A1)} $$Steepest descent is in the direction $-\nabla f(1,1)/|\nabla f(1,1)|=-\langle 1,-1\rangle/\sqrt{2}=\langle -1,1\rangle/\sqrt{2}$. (A1) The directional derivative in this direction equals $-|\nabla f(1,1)|=-\sqrt{1^{2}+(-1)^{2}}=-\sqrt{2}$. (A1)最陡下降方向为 $-\nabla f(1,1)/|\nabla f(1,1)|=-\langle 1,-1\rangle/\sqrt{2}=\langle -1,1\rangle/\sqrt{2}$。(A1) 该方向上的方向导数等于 $-|\nabla f(1,1)|=-\sqrt{1^{2}+(-1)^{2}}=-\sqrt{2}$。(A1)
Write $F(x,y,z)=xe^{y-1}-y^{2}+1-z$. Then $F_{x}=e^{y-1}$, $F_{y}=xe^{y-1}-2y$, $F_{z}=-1$. (M1) At $(1,1,1)$: $F_{x}=1$, $F_{y}=-1$, $F_{z}=-1$, so $\nabla F(1,1,1)=\langle 1,-1,-1\rangle$. (A1)令 $F(x,y,z)=xe^{y-1}-y^{2}+1-z$。则 $F_{x}=e^{y-1}$，$F_{y}=xe^{y-1}-2y$，$F_{z}=-1$。(M1) 在 $(1,1,1)$ 处：$F_{x}=1$，$F_{y}=-1$，$F_{z}=-1$，故 $\nabla F(1,1,1)=\langle 1,-1,-1\rangle$。(A1)
The tangent plane via the level-surface formula is $1(x-1)+(-1)(y-1)+(-1)(z-1)=0$, i.e. $x-y-z+1=0$, which rearranges to $z=x-y+1$: the same plane found in (a). (R1)用等值面公式得切平面为 $1(x-1)+(-1)(y-1)+(-1)(z-1)=0$，即 $x-y-z+1=0$，整理为 $z=x-y+1$：与 (a) 中所得平面相同。(R1)