Unit C4: Directional Derivatives, Tangent Planes, Linearization第 C4 单元：方向导数（directional derivative）、切平面（tangent plane）与线性化（linearization）

Let $f(x,y) = x^2 y + 3y$. Find $D_{\mathbf{u}} f$ at $(1,2)$ in the direction of $\mathbf{v} = \langle 3, 4 \rangle$.设 $f(x,y) = x^2 y + 3y$。求 $f$ 在 $(1,2)$ 处沿 $\mathbf{v} = \langle 3, 4 \rangle$ 方向的 $D_{\mathbf{u}} f$。

First the gradient: $f_x = 2xy$, $f_y = x^2 + 3$, so $\nabla f(1,2) = \langle 4, 4 \rangle$.先求梯度：$f_x = 2xy$，$f_y = x^2 + 3$，故 $\nabla f(1,2) = \langle 4, 4 \rangle$。

Normalize: $|\mathbf{v}| = 5$, so $\mathbf{u} = \langle 3/5, 4/5 \rangle$. Then归一化：$|\mathbf{v}| = 5$，故 $\mathbf{u} = \langle 3/5, 4/5 \rangle$。于是

$$ D_{\mathbf{u}} f(1,2) = \langle 4,4 \rangle \cdot \langle 3/5, 4/5 \rangle = \tfrac{12}{5} + \tfrac{16}{5} = \tfrac{28}{5}. $$

Let $f(x,y,z) = xy^2 z^3$. Find $D_{\mathbf{u}} f$ at the point $(2,-1,1)$ in the direction toward $(0,0,0)$, that is along $\mathbf{v} = \langle 0,0,0\rangle - \langle 2,-1,1\rangle = \langle -2, 1, -1\rangle$.设 $f(x,y,z) = xy^2 z^3$。求 $f$ 在点 $(2,-1,1)$ 处沿指向 $(0,0,0)$ 的方向的 $D_{\mathbf{u}} f$，即沿 $\mathbf{v} = \langle 0,0,0\rangle - \langle 2,-1,1\rangle = \langle -2, 1, -1\rangle$。

The partials are $f_x = y^2 z^3$, $f_y = 2xy z^3$, $f_z = 3xy^2 z^2$. At $(2,-1,1)$:各偏导数为 $f_x = y^2 z^3$，$f_y = 2xy z^3$，$f_z = 3xy^2 z^2$。在 $(2,-1,1)$ 处：

$$ \nabla f(2,-1,1) = \langle (1)(1),\ 2(2)(-1)(1),\ 3(2)(1)(1)\rangle = \langle 1,\ -4,\ 6\rangle. $$

Normalize the direction: $|\mathbf{v}| = \sqrt{4 + 1 + 1} = \sqrt6$, so $\mathbf{u} = \tfrac{1}{\sqrt6}\langle -2,1,-1\rangle$. Then把方向归一化：$|\mathbf{v}| = \sqrt{4 + 1 + 1} = \sqrt6$，故 $\mathbf{u} = \tfrac{1}{\sqrt6}\langle -2,1,-1\rangle$。于是

$$ D_{\mathbf{u}} f = \langle 1,-4,6\rangle \cdot \tfrac{1}{\sqrt6}\langle -2,1,-1\rangle = \frac{-2 - 4 - 6}{\sqrt6} = \frac{-12}{\sqrt6} = -2\sqrt6. $$

The negative sign tells us $f$ is decreasing as we head from $(2,-1,1)$ toward the origin, at a rate of $2\sqrt6 \approx 4.90$ units of $f$ per unit of distance.负号表明：从 $(2,-1,1)$ 朝原点方向走时 $f$ 在减小，速率为每单位距离减少 $2\sqrt6 \approx 4.90$ 个单位的 $f$。

Directional derivatives generalize partials, so the partials must reappear as special cases. Take $f(x,y) = e^{x}\sin y$ and the standard basis direction $\mathbf{u} = \mathbf{i} = \langle 1, 0\rangle$ at a general point $(x,y)$.方向导数是偏导数的推广，因此偏导数必定作为特例重新出现。取 $f(x,y) = e^{x}\sin y$，在一般点 $(x,y)$ 处沿标准基方向 $\mathbf{u} = \mathbf{i} = \langle 1, 0\rangle$。

$\nabla f = \langle e^x \sin y,\ e^x \cos y\rangle$, and since $\mathbf{i}$ is already a unit vector,$\nabla f = \langle e^x \sin y,\ e^x \cos y\rangle$，由于 $\mathbf{i}$ 已是单位向量，

$$ D_{\mathbf{i}} f = \nabla f \cdot \langle 1,0\rangle = e^x \sin y = f_x. $$

Likewise $D_{\mathbf{j}} f = f_y = e^x \cos y$. This is the sanity check that anchors the whole subject: the directional derivative in the direction of a coordinate axis is just the partial derivative for that variable. If a formula ever fails this test, it is wrong.同理 $D_{\mathbf{j}} f = f_y = e^x \cos y$。这是贯穿整个主题的检验基准：沿坐标轴方向的方向导数就是该变量的偏导数。任何公式若通不过这一检验，就是错的。

Fix a point where $\nabla f \neq \mathbf{0}$. For any unit vector $\mathbf{u}$, the Cauchy-Schwarz inequality gives固定一个满足 $\nabla f \neq \mathbf{0}$ 的点。对任意单位向量 $\mathbf{u}$，由柯西-施瓦茨不等式（Cauchy-Schwarz inequality）得

$$ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = |\nabla f|\,|\mathbf{u}|\cos\theta = |\nabla f|\cos\theta. $$

Since $-1 \le \cos\theta \le 1$, the value ranges over $[-|\nabla f|,\, |\nabla f|]$. The maximum $|\nabla f|$ is attained exactly when $\cos\theta = 1$, that is when $\mathbf{u}$ is parallel to $\nabla f$. The minimum is attained at $\mathbf{u} = -\nabla f / |\nabla f|$, and $D_{\mathbf{u}} f = 0$ precisely when $\cos\theta = 0$, i.e. $\mathbf{u}\perp\nabla f$. If $\nabla f = \mathbf{0}$ every directional derivative is zero and the point is critical.由于 $-1 \le \cos\theta \le 1$，该值取遍 $[-|\nabla f|,\, |\nabla f|]$。最大值 $|\nabla f|$ 恰在 $\cos\theta = 1$ 时取得，即 $\mathbf{u}$ 与 $\nabla f$ 平行时。最小值在 $\mathbf{u} = -\nabla f / |\nabla f|$ 处取得；而 $D_{\mathbf{u}} f = 0$ 当且仅当 $\cos\theta = 0$，即 $\mathbf{u}\perp\nabla f$。若 $\nabla f = \mathbf{0}$，则每个方向导数都为零，该点为临界点（critical point）。

Temperature is $T(x,y) = 100 - x^2 - 2y^2$. At $(2,1)$, in which direction does $T$ increase fastest, and how fast?温度为 $T(x,y) = 100 - x^2 - 2y^2$。在 $(2,1)$ 处，$T$ 沿哪个方向增长最快，增长有多快？

$\nabla T = \langle -2x, -4y \rangle$, so $\nabla T(2,1) = \langle -4, -4 \rangle$. The steepest-ascent direction is $\langle -4,-4\rangle$, or as a unit vector $\langle -1/\sqrt2, -1/\sqrt2\rangle$. The maximum rate is $|\nabla T| = \sqrt{16+16} = 4\sqrt2$ degrees per unit distance.$\nabla T = \langle -2x, -4y \rangle$，故 $\nabla T(2,1) = \langle -4, -4 \rangle$。最速上升方向是 $\langle -4,-4\rangle$，写成单位向量为 $\langle -1/\sqrt2, -1/\sqrt2\rangle$。最大速率为 $|\nabla T| = \sqrt{16+16} = 4\sqrt2$ 度每单位距离。

Let $f(x,y) = x e^{y}$. At the point $P=(2,0)$, find a unit vector $\mathbf{u}$ for which $D_{\mathbf{u}} f(P) = 1$, and explain when no such direction exists.设 $f(x,y) = x e^{y}$。在点 $P=(2,0)$ 处，求使 $D_{\mathbf{u}} f(P) = 1$ 的单位向量 $\mathbf{u}$，并说明何时不存在这样的方向。

First, $\nabla f = \langle e^y,\ x e^y\rangle$, so $\nabla f(2,0) = \langle 1, 2\rangle$ and $|\nabla f| = \sqrt5$. Writing $D_{\mathbf{u}} f = |\nabla f|\cos\theta = \sqrt5\cos\theta$, we need $\sqrt5 \cos\theta = 1$, hence $\cos\theta = 1/\sqrt5 \approx 0.447$, giving $\theta \approx 63.4^\circ$ measured from the gradient. A prescribed rate $r$ is achievable as a directional derivative precisely when $|r| \le |\nabla f|$, because $D_{\mathbf{u}} f$ ranges over $[-|\nabla f|, |\nabla f|]$. Here $1 \le \sqrt5$, so two directions work (one on each side of $\nabla f$). For a concrete one, rotate the unit gradient $\langle 1/\sqrt5, 2/\sqrt5\rangle$ by $\theta = 63.4^\circ$. Asking for a rate above $\sqrt5$ would be impossible: no direction can beat the steepest one.首先，$\nabla f = \langle e^y,\ x e^y\rangle$，故 $\nabla f(2,0) = \langle 1, 2\rangle$，$|\nabla f| = \sqrt5$。写成 $D_{\mathbf{u}} f = |\nabla f|\cos\theta = \sqrt5\cos\theta$，需要 $\sqrt5 \cos\theta = 1$，从而 $\cos\theta = 1/\sqrt5 \approx 0.447$，即从梯度方向起量得 $\theta \approx 63.4^\circ$。指定速率 $r$ 能作为方向导数实现，当且仅当 $|r| \le |\nabla f|$，因为 $D_{\mathbf{u}} f$ 取遍 $[-|\nabla f|, |\nabla f|]$。此处 $1 \le \sqrt5$，故有两个方向可行（梯度两侧各一个）。具体取一个：把单位梯度 $\langle 1/\sqrt5, 2/\sqrt5\rangle$ 旋转 $\theta = 63.4^\circ$ 即可。要求速率超过 $\sqrt5$ 则不可能：没有方向能胜过最陡的方向。

The height of a hill is $h(x,y) = 200 - \tfrac{1}{100}(3x^2 + 2y^2)$ meters, with $x,y$ in meters. A hiker stands above the ground point $(60, 40)$. Find the bearing of steepest ascent and the slope encountered in that direction.某山丘的高度为 $h(x,y) = 200 - \tfrac{1}{100}(3x^2 + 2y^2)$ 米，其中 $x,y$ 以米为单位。一名登山者站在地面点 $(60, 40)$ 正上方。求最速上升的方位以及该方向上的坡度。

$\nabla h = \langle -\tfrac{6x}{100}, -\tfrac{4y}{100}\rangle = \langle -0.06x, -0.04y\rangle$. At $(60,40)$: $\nabla h = \langle -3.6,\ -1.6\rangle$. Steepest ascent points along $\langle -3.6, -1.6\rangle$, i.e. back toward smaller $x$ and $y$, which makes sense because the summit is at the origin where $h$ is largest. The slope in that direction is$\nabla h = \langle -\tfrac{6x}{100}, -\tfrac{4y}{100}\rangle = \langle -0.06x, -0.04y\rangle$。在 $(60,40)$ 处：$\nabla h = \langle -3.6,\ -1.6\rangle$。最速上升沿 $\langle -3.6, -1.6\rangle$，即朝更小的 $x$ 和 $y$ 回退，这很合理，因为山顶在原点，那里 $h$ 最大。该方向上的坡度为

$$ |\nabla h| = \sqrt{(-3.6)^2 + (-1.6)^2} = \sqrt{12.96 + 2.56} = \sqrt{15.52} \approx 3.94. $$

So the hiker rises about $3.94$ meters of height per meter of horizontal travel in the steepest direction. Walking perpendicular to $\nabla h$, along $\langle 1.6, -3.6\rangle$, keeps elevation momentarily constant: that is a contour line of the hill.因此登山者沿最陡方向每水平前进 $1$ 米，高度约上升 $3.94$ 米。沿垂直于 $\nabla h$ 的方向 $\langle 1.6, -3.6\rangle$ 行走则瞬时保持海拔不变：那就是山丘的一条等高线。

Find the tangent plane to $z = x^2 + y^2$ at $(1, 2, 5)$.求 $z = x^2 + y^2$ 在 $(1, 2, 5)$ 处的切平面。

$f_x = 2x$, $f_y = 2y$, so $f_x(1,2) = 2$ and $f_y(1,2) = 4$. With $f(1,2) = 5$:$f_x = 2x$，$f_y = 2y$，故 $f_x(1,2) = 2$，$f_y(1,2) = 4$。又 $f(1,2) = 5$：

$$ z = 5 + 2(x-1) + 4(y-2) = 2x + 4y - 5. $$

Check: at $(1,2)$ this gives $z = 2 + 8 - 5 = 5$, matching the point on the surface.检验：在 $(1,2)$ 处得 $z = 2 + 8 - 5 = 5$，与曲面上的点一致。

Find the tangent plane to $z = \ln(x^2 + y^2)$ at the point above $(1, 0)$.求 $z = \ln(x^2 + y^2)$ 在 $(1, 0)$ 正上方那一点处的切平面。

First the base value: $f(1,0) = \ln(1) = 0$, so the surface point is $(1,0,0)$. The partials are先求基准值：$f(1,0) = \ln(1) = 0$，故曲面点为 $(1,0,0)$。各偏导数为

$$ f_x = \frac{2x}{x^2+y^2}, \qquad f_y = \frac{2y}{x^2+y^2}. $$

At $(1,0)$: $f_x = 2/1 = 2$ and $f_y = 0/1 = 0$. The tangent plane is在 $(1,0)$ 处：$f_x = 2/1 = 2$，$f_y = 0/1 = 0$。切平面为

$$ z = 0 + 2(x-1) + 0(y-0) = 2(x-1) = 2x - 2. $$

Notice the plane does not involve $y$ at all, because the surface is locally flat in the $y$ direction at this point: $f_y(1,0) = 0$. That is geometrically sensible since $(1,0)$ sits on the $x$-axis, a symmetry line of $\ln(x^2+y^2)$.注意该平面完全不含 $y$，因为在这一点曲面沿 $y$ 方向局部是平的：$f_y(1,0) = 0$。这在几何上很合理，因为 $(1,0)$ 位于 $x$ 轴上，而 $x$ 轴是 $\ln(x^2+y^2)$ 的一条对称线。

The tangent plane comes paired with a normal line. For $z = x^2 + y^2$ at $(1,2,5)$, using the graph normal $\langle f_x, f_y, -1\rangle = \langle 2, 4, -1\rangle$, the normal line through $(1,2,5)$ is切平面总与一条法线（normal line）配对出现。对 $z = x^2 + y^2$ 在 $(1,2,5)$ 处，用图像法向量 $\langle f_x, f_y, -1\rangle = \langle 2, 4, -1\rangle$，过 $(1,2,5)$ 的法线为

$$ (x,y,z) = (1,2,5) + t\langle 2, 4, -1\rangle, \quad \text{i.e.}\quad x = 1 + 2t,\ y = 2 + 4t,\ z = 5 - t. $$

This line is perpendicular to the tangent plane $2x + 4y - z = 5$ found by rearranging Worked Example 3.1. A quick check: the plane's coefficient vector $\langle 2, 4, -1\rangle$ matches the direction of the normal line, confirming they are perpendicular and parallel respectively, as they must be.该直线垂直于把例题 3.1 整理后得到的切平面 $2x + 4y - z = 5$。快速检验：平面的系数向量 $\langle 2, 4, -1\rangle$ 与法线的方向一致，证实它们分别相互垂直、相互平行，正如理应如此。

Use linearization to estimate $\sqrt{(3.02)^2 + (3.97)^2}$.用线性化估计 $\sqrt{(3.02)^2 + (3.97)^2}$。

Let $f(x,y) = \sqrt{x^2 + y^2}$ at $(a,b) = (3,4)$, where $f(3,4) = 5$. The partials are $f_x = x/\sqrt{x^2+y^2}$ and $f_y = y/\sqrt{x^2+y^2}$, so $f_x(3,4) = 3/5$, $f_y(3,4) = 4/5$.取 $f(x,y) = \sqrt{x^2 + y^2}$，基点 $(a,b) = (3,4)$，其中 $f(3,4) = 5$。偏导数为 $f_x = x/\sqrt{x^2+y^2}$，$f_y = y/\sqrt{x^2+y^2}$，故 $f_x(3,4) = 3/5$，$f_y(3,4) = 4/5$。

$$ L(3.02, 3.97) = 5 + \tfrac{3}{5}(0.02) + \tfrac{4}{5}(-0.03) = 5 + 0.012 - 0.024 = 4.988. $$

Estimate $(2.01)^3 (0.98)^4$ using a linear approximation.用线性近似估计 $(2.01)^3 (0.98)^4$。

Let $f(x,y) = x^3 y^4$ at $(a,b) = (2,1)$, where $f(2,1) = 8$. The partials are $f_x = 3x^2 y^4$ and $f_y = 4x^3 y^3$, so $f_x(2,1) = 3(4)(1) = 12$ and $f_y(2,1) = 4(8)(1) = 32$. The increments are $dx = 0.01$, $dy = -0.02$. Then取 $f(x,y) = x^3 y^4$，基点 $(a,b) = (2,1)$，其中 $f(2,1) = 8$。偏导数为 $f_x = 3x^2 y^4$，$f_y = 4x^3 y^3$，故 $f_x(2,1) = 3(4)(1) = 12$，$f_y(2,1) = 4(8)(1) = 32$。增量为 $dx = 0.01$，$dy = -0.02$。于是

$$ L(2.01, 0.98) = 8 + 12(0.01) + 32(-0.02) = 8 + 0.12 - 0.64 = 7.48. $$

For comparison the exact value is $(2.01)^3(0.98)^4 \approx 8.1206 \times 0.92237 \approx 7.491$, so the linear estimate $7.48$ is within about $0.15\%$. The size of the error scales with the second derivatives times the square of the increments, which is why small increments make the estimate sharp.作为对照，精确值为 $(2.01)^3(0.98)^4 \approx 8.1206 \times 0.92237 \approx 7.491$，故线性估计 $7.48$ 的误差约在 $0.15\%$ 以内。误差大小与二阶导数乘以增量的平方成比例，这正是增量越小估计越精确的原因。

The period of a pendulum is $T = 2\pi\sqrt{L/g}$. If $L$ is measured with a relative error up to $0.5\%$ and $g$ with a relative error up to $0.1\%$, bound the relative error in $T$.单摆的周期为 $T = 2\pi\sqrt{L/g}$。若 $L$ 的相对误差不超过 $0.5\%$，$g$ 的相对误差不超过 $0.1\%$，请给出 $T$ 的相对误差上界。

Take logarithms before differentiating, a standard trick for products and powers: $\ln T = \ln(2\pi) + \tfrac12\ln L - \tfrac12 \ln g$. The differential is微分前先取对数，这是处理乘积与幂的标准技巧：$\ln T = \ln(2\pi) + \tfrac12\ln L - \tfrac12 \ln g$。其微分为

$$ \frac{dT}{T} = \frac{1}{2}\frac{dL}{L} - \frac{1}{2}\frac{dg}{g}. $$

In the worst case the two contributions add in magnitude:在最坏情况下，两项贡献在量级上相加：

$$ \left|\frac{dT}{T}\right| \le \frac{1}{2}(0.005) + \frac{1}{2}(0.001) = 0.0025 + 0.0005 = 0.003 = 0.3\%. $$

The logarithmic differential converts each input's relative error into a weighted contribution, and the weights are exactly the exponents in the formula, here $+\tfrac12$ for $L$ and $-\tfrac12$ for $g$.对数微分把每个输入的相对误差转化为一项加权贡献，而权重正是公式中的指数，此处 $L$ 为 $+\tfrac12$，$g$ 为 $-\tfrac12$。

A rectangle is measured as $x = 30$ cm and $y = 24$ cm, each with a possible error of $\pm 0.1$ cm. Estimate the maximum error in the computed area $A = xy$.一个矩形测得 $x = 30$ cm，$y = 24$ cm，每个测量都可能有 $\pm 0.1$ cm 的误差。估计所算面积 $A = xy$ 的最大误差。

$dA = A_x\, dx + A_y\, dy = y\, dx + x\, dy$. At $(30,24)$ with $|dx|, |dy| \le 0.1$:$dA = A_x\, dx + A_y\, dy = y\, dx + x\, dy$。在 $(30,24)$ 处，且 $|dx|, |dy| \le 0.1$：

$$ |dA| \le 24(0.1) + 30(0.1) = 2.4 + 3.0 = 5.4 \ \text{cm}^2. $$

So the area $720\ \text{cm}^2$ carries an estimated uncertainty of about $5.4\ \text{cm}^2$, a relative error near $0.75\%$. The differential turns input tolerances into an output tolerance via a single linear formula.因此面积 $720\ \text{cm}^2$ 的估计不确定度约为 $5.4\ \text{cm}^2$，相对误差接近 $0.75\%$。微分通过一条线性公式把输入的容差转化为输出的容差。

Let $f(x,y) = x^2 y$ with $x = \cos t$, $y = \sin t$. Find $df/dt$ at $t = 0$.设 $f(x,y) = x^2 y$，且 $x = \cos t$，$y = \sin t$。求 $t = 0$ 处的 $df/dt$。

$f_x = 2xy$, $f_y = x^2$, and $x'(t) = -\sin t$, $y'(t) = \cos t$. So$f_x = 2xy$，$f_y = x^2$，且 $x'(t) = -\sin t$，$y'(t) = \cos t$。于是

$$ \frac{df}{dt} = (2xy)(-\sin t) + (x^2)(\cos t). $$

At $t = 0$: $x = 1$, $y = 0$, $\sin 0 = 0$, $\cos 0 = 1$, giving $df/dt = (0)(0) + (1)(1) = 1$.在 $t = 0$ 处：$x = 1$，$y = 0$，$\sin 0 = 0$，$\cos 0 = 1$，得 $df/dt = (0)(0) + (1)(1) = 1$。

Let $f(x,y) = x^2 + y^2$ with $x = r\cos\theta$ and $y = r\sin\theta$. Compute $\partial f/\partial r$ and $\partial f/\partial\theta$ by the chain rule and confirm against direct substitution.设 $f(x,y) = x^2 + y^2$，且 $x = r\cos\theta$，$y = r\sin\theta$。用链式法则求 $\partial f/\partial r$ 和 $\partial f/\partial\theta$，并与直接代入的结果相互印证。

$f_x = 2x$, $f_y = 2y$. The inner partials are $x_r = \cos\theta$, $y_r = \sin\theta$, $x_\theta = -r\sin\theta$, $y_\theta = r\cos\theta$. Then$f_x = 2x$，$f_y = 2y$。内层偏导数为 $x_r = \cos\theta$，$y_r = \sin\theta$，$x_\theta = -r\sin\theta$，$y_\theta = r\cos\theta$。于是

$$ \frac{\partial f}{\partial r} = 2x\cos\theta + 2y\sin\theta = 2r\cos^2\theta + 2r\sin^2\theta = 2r, $$ $$ \frac{\partial f}{\partial\theta} = 2x(-r\sin\theta) + 2y(r\cos\theta) = -2r^2\cos\theta\sin\theta + 2r^2\sin\theta\cos\theta = 0. $$

Direct substitution gives $f = r^2\cos^2\theta + r^2\sin^2\theta = r^2$, so indeed $f_r = 2r$ and $f_\theta = 0$. The vanishing $\theta$ derivative is the statement that $f$ is rotationally symmetric: it depends only on the radius.直接代入得 $f = r^2\cos^2\theta + r^2\sin^2\theta = r^2$，故确有 $f_r = 2r$，$f_\theta = 0$。对 $\theta$ 的导数为零，正说明 $f$ 具有旋转对称性：它只依赖于半径。

The chain rule yields the implicit-function formula. Suppose $y$ is defined implicitly by $F(x,y) = 0$. Differentiate with respect to $x$, treating $y = y(x)$:链式法则给出隐函数（implicit differentiation）公式。设 $y$ 由 $F(x,y) = 0$ 隐式确定。把 $y = y(x)$，对 $x$ 求导：

$$ F_x \cdot 1 + F_y \cdot \frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{F_x}{F_y}\ \ (F_y \neq 0). $$

Apply it to the folium-style curve $x^3 + y^3 = 6xy$ at the point $(3,3)$. Let $F = x^3 + y^3 - 6xy$. Then $F_x = 3x^2 - 6y$ and $F_y = 3y^2 - 6x$. At $(3,3)$: $F_x = 27 - 18 = 9$ and $F_y = 27 - 18 = 9$, so把它用于笛卡尔叶形线（folium）$x^3 + y^3 = 6xy$ 在点 $(3,3)$ 处。设 $F = x^3 + y^3 - 6xy$。则 $F_x = 3x^2 - 6y$，$F_y = 3y^2 - 6x$。在 $(3,3)$ 处：$F_x = 27 - 18 = 9$，$F_y = 27 - 18 = 9$，故

$$ \frac{dy}{dx} = -\frac{9}{9} = -1. $$

The tangent line to the curve at $(3,3)$ has slope $-1$. This is the multivariable chain rule doing implicit differentiation in one clean step, with no need to solve for $y$.曲线在 $(3,3)$ 处的切线斜率为 $-1$。这就是多元链式法则一步干净地完成隐函数求导，无需解出 $y$。

Let $\mathbf{r}(t)$ be any differentiable curve lying entirely in the level surface $F(x,y,z) = k$ with $\mathbf{r}(t_0) = \mathbf{a}$. Then $F(\mathbf{r}(t)) = k$ is constant, so differentiating with the chain rule:设 $\mathbf{r}(t)$ 是完全位于等值面 $F(x,y,z) = k$ 内的任一可微曲线，且 $\mathbf{r}(t_0) = \mathbf{a}$。则 $F(\mathbf{r}(t)) = k$ 为常数，用链式法则求导：

$$ \frac{d}{dt}F(\mathbf{r}(t)) = \nabla F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 0. $$

At $t = t_0$ this gives $\nabla F(\mathbf{a}) \cdot \mathbf{r}'(t_0) = 0$. Since $\mathbf{r}'(t_0)$ can be the tangent vector of any curve through $\mathbf{a}$ in the surface, $\nabla F(\mathbf{a})$ is orthogonal to every tangent vector, hence normal to the surface. The tangent plane is the plane through $\mathbf{a}$ with normal $\nabla F(\mathbf{a})$.在 $t = t_0$ 处得 $\nabla F(\mathbf{a}) \cdot \mathbf{r}'(t_0) = 0$。由于 $\mathbf{r}'(t_0)$ 可以是曲面内过 $\mathbf{a}$ 的任意曲线的切向量，故 $\nabla F(\mathbf{a})$ 与每个切向量都正交，因而是曲面的法向量。切平面就是过 $\mathbf{a}$ 且以 $\nabla F(\mathbf{a})$ 为法向量的平面。

Find the tangent plane to $x^2 + y^2 + z^2 = 9$ at $(2, 1, 2)$.求 $x^2 + y^2 + z^2 = 9$ 在 $(2, 1, 2)$ 处的切平面。

Let $F = x^2 + y^2 + z^2$. Then $\nabla F = \langle 2x, 2y, 2z \rangle$, so $\nabla F(2,1,2) = \langle 4, 2, 4 \rangle$. The plane is设 $F = x^2 + y^2 + z^2$。则 $\nabla F = \langle 2x, 2y, 2z \rangle$，故 $\nabla F(2,1,2) = \langle 4, 2, 4 \rangle$。切平面为

$$ 4(x-2) + 2(y-1) + 4(z-2) = 0, \quad \text{i.e.}\quad 2x + y + 2z = 9. $$

This is consistent with the geometry: the radius to $(2,1,2)$ is the normal to the sphere there.这与几何一致：指向 $(2,1,2)$ 的半径就是球面在该处的法向量。

Find the tangent plane and the normal line to the ellipsoid $\dfrac{x^2}{4} + y^2 + \dfrac{z^2}{9} = 3$ at $(2, 1, 3)$.求椭球面 $\dfrac{x^2}{4} + y^2 + \dfrac{z^2}{9} = 3$ 在 $(2, 1, 3)$ 处的切平面与法线。

Let $F = \tfrac{x^2}{4} + y^2 + \tfrac{z^2}{9}$. Then $\nabla F = \langle \tfrac{x}{2},\ 2y,\ \tfrac{2z}{9}\rangle$. At $(2,1,3)$: $\nabla F = \langle 1,\ 2,\ \tfrac{2}{3}\rangle$. The tangent plane is设 $F = \tfrac{x^2}{4} + y^2 + \tfrac{z^2}{9}$。则 $\nabla F = \langle \tfrac{x}{2},\ 2y,\ \tfrac{2z}{9}\rangle$。在 $(2,1,3)$ 处：$\nabla F = \langle 1,\ 2,\ \tfrac{2}{3}\rangle$。切平面为

$$ 1(x-2) + 2(y-1) + \tfrac{2}{3}(z-3) = 0 \quad\Longrightarrow\quad x + 2y + \tfrac{2}{3}z = 6, $$

or, clearing the fraction, $3x + 6y + 2z = 18$. The normal line uses the same direction $\langle 1, 2, \tfrac23\rangle$ (or its scalar multiple $\langle 3, 6, 2\rangle$):或者去分母得 $3x + 6y + 2z = 18$。法线沿同一方向 $\langle 1, 2, \tfrac23\rangle$（或其标量倍 $\langle 3, 6, 2\rangle$）：

$$ (x,y,z) = (2,1,3) + t\langle 3, 6, 2\rangle. $$

A quick verification: the point $(2,1,3)$ satisfies $3(2)+6(1)+2(3) = 6+6+6 = 18$, so it lies on the plane, as it must.快速验证：点 $(2,1,3)$ 满足 $3(2)+6(1)+2(3) = 6+6+6 = 18$，故它在该平面上，正如理应如此。

The surfaces $x^2 + y^2 + z^2 = 6$ and $x^2 + y^2 - z = 0$ both pass through $(1, 1, 2)$. Find the angle between them there, defined as the angle between their normal vectors.曲面 $x^2 + y^2 + z^2 = 6$ 与 $x^2 + y^2 - z = 0$ 都过 $(1, 1, 2)$。求它们在该处的夹角，定义为两个法向量之间的夹角。

For $F = x^2 + y^2 + z^2$: $\nabla F = \langle 2x, 2y, 2z\rangle$, so $\nabla F(1,1,2) = \langle 2, 2, 4\rangle$. For $G = x^2 + y^2 - z$: $\nabla G = \langle 2x, 2y, -1\rangle$, so $\nabla G(1,1,2) = \langle 2, 2, -1\rangle$. The angle $\phi$ between the normals satisfies对 $F = x^2 + y^2 + z^2$：$\nabla F = \langle 2x, 2y, 2z\rangle$，故 $\nabla F(1,1,2) = \langle 2, 2, 4\rangle$。对 $G = x^2 + y^2 - z$：$\nabla G = \langle 2x, 2y, -1\rangle$，故 $\nabla G(1,1,2) = \langle 2, 2, -1\rangle$。法向量之间的夹角 $\phi$ 满足

$$ \cos\phi = \frac{\nabla F \cdot \nabla G}{|\nabla F||\nabla G|} = \frac{4 + 4 - 4}{\sqrt{4+4+16}\,\sqrt{4+4+1}} = \frac{4}{\sqrt{24}\,\sqrt{9}} = \frac{4}{3\sqrt{24}} = \frac{4}{6\sqrt6} = \frac{2}{3\sqrt6}. $$

Numerically $\cos\phi = 2/(3\sqrt6) \approx 0.2722$, so $\phi \approx 74.2^\circ$. The surfaces meet at roughly $74$ degrees at that point. The gradient is doing all the geometric work: it converts each implicit surface into a concrete normal direction.数值上 $\cos\phi = 2/(3\sqrt6) \approx 0.2722$，故 $\phi \approx 74.2^\circ$。两曲面在该点约以 $74$ 度相交。梯度承担了全部几何工作：它把每个隐式曲面转化为一个具体的法向量方向。

We prove the standard sufficient condition: if $f_x$ and $f_y$ exist and are continuous on a disk around $\mathbf{a} = (a,b)$, then $f$ is differentiable at $\mathbf{a}$. The engine is the one-variable Mean Value Theorem applied one coordinate at a time.我们证明标准的充分条件：若 $f_x$ 和 $f_y$ 在 $\mathbf{a} = (a,b)$ 的某个圆盘上存在且连续，则 $f$ 在 $\mathbf{a}$ 处可微。核心引擎是逐个坐标地应用一元中值定理（Mean Value Theorem）。

Write the increment $f(a+h, b+k) - f(a,b)$ and split it through the corner point $(a+h, b)$:写出增量 $f(a+h, b+k) - f(a,b)$，并经由拐角点 $(a+h, b)$ 拆分：

$$ \Delta f = \big[f(a+h, b+k) - f(a+h, b)\big] + \big[f(a+h, b) - f(a,b)\big]. $$

Apply the Mean Value Theorem to each bracket. In the first, only the second coordinate changes, so there is some $k^\ast$ between $0$ and $k$ with $f(a+h, b+k) - f(a+h, b) = f_y(a+h, b+k^\ast)\,k$. In the second, only the first coordinate changes, so there is some $h^\ast$ between $0$ and $h$ with $f(a+h, b) - f(a,b) = f_x(a+h^\ast, b)\,h$. Hence对每个方括号应用中值定理。在第一个里只有第二个坐标变化，故存在介于 $0$ 与 $k$ 之间的某个 $k^\ast$，使 $f(a+h, b+k) - f(a+h, b) = f_y(a+h, b+k^\ast)\,k$。在第二个里只有第一个坐标变化，故存在介于 $0$ 与 $h$ 之间的某个 $h^\ast$，使 $f(a+h, b) - f(a,b) = f_x(a+h^\ast, b)\,h$。于是

$$ \Delta f = f_x(a+h^\ast, b)\,h + f_y(a+h, b+k^\ast)\,k. $$

Now subtract the proposed linear part $f_x(a,b)h + f_y(a,b)k$:现在减去拟用的线性部分 $f_x(a,b)h + f_y(a,b)k$：

$$ \Delta f - \big[f_x(a,b)h + f_y(a,b)k\big] = \underbrace{\big[f_x(a+h^\ast,b) - f_x(a,b)\big]}_{=\,\varepsilon_1}\,h + \underbrace{\big[f_y(a+h,b+k^\ast) - f_y(a,b)\big]}_{=\,\varepsilon_2}\,k. $$

Because $f_x$ and $f_y$ are continuous at $(a,b)$, and because $(a+h^\ast, b) \to (a,b)$ and $(a+h, b+k^\ast) \to (a,b)$ as $(h,k) \to (0,0)$, both $\varepsilon_1 \to 0$ and $\varepsilon_2 \to 0$. Finally, since $|h| \le |\mathbf{h}|$ and $|k| \le |\mathbf{h}|$ where $\mathbf{h} = (h,k)$,由于 $f_x$ 和 $f_y$ 在 $(a,b)$ 处连续，且当 $(h,k) \to (0,0)$ 时 $(a+h^\ast, b) \to (a,b)$、$(a+h, b+k^\ast) \to (a,b)$，故 $\varepsilon_1 \to 0$ 且 $\varepsilon_2 \to 0$。最后，由于 $|h| \le |\mathbf{h}|$ 且 $|k| \le |\mathbf{h}|$，其中 $\mathbf{h} = (h,k)$，

$$ \frac{\big|\Delta f - (f_x(a,b)h + f_y(a,b)k)\big|}{|\mathbf{h}|} \le |\varepsilon_1| + |\varepsilon_2| \longrightarrow 0. $$

That is exactly the differentiability condition with $\nabla f(\mathbf{a})\cdot\mathbf{h} = f_x h + f_y k$ as the linear term. So $C^1$ implies differentiable, which is the workhorse theorem every formula in this unit quietly relies on.这正是以 $\nabla f(\mathbf{a})\cdot\mathbf{h} = f_x h + f_y k$ 为线性项的可微性条件。于是 $C^1$ 蕴含可微，这是本单元每条公式默默依赖的主力定理。

Consider考虑

$$ f(x,y) = \begin{cases} \dfrac{x^2 y}{x^2 + y^2}, & (x,y)\neq(0,0), \\[4pt] 0, & (x,y)=(0,0). \end{cases} $$

Along any unit direction $\mathbf{u} = \langle a, b\rangle$ we compute沿任意单位方向 $\mathbf{u} = \langle a, b\rangle$ 计算

$$ D_{\mathbf{u}}f(0,0) = \lim_{t\to 0}\frac{f(ta,tb)}{t} = \lim_{t\to 0}\frac{1}{t}\cdot\frac{t^3 a^2 b}{t^2(a^2+b^2)} = \frac{a^2 b}{a^2+b^2} = a^2 b. $$

So every directional derivative exists at the origin. Yet if the gradient formula held we would need $D_{\mathbf{u}}f = \nabla f(0,0)\cdot\mathbf{u}$ to be linear in $\mathbf{u}$. Here $f_x(0,0) = 0$ and $f_y(0,0) = 0$, so the formula predicts $0$ for every direction, contradicting $a^2 b$ (take $a=b=1/\sqrt2$, giving $1/(2\sqrt2)\neq 0$). Hence $f$ is not differentiable at the origin, even though all directional derivatives exist.所以原点处每个方向导数都存在。然而若梯度公式成立，就需要 $D_{\mathbf{u}}f = \nabla f(0,0)\cdot\mathbf{u}$ 关于 $\mathbf{u}$ 是线性的。此处 $f_x(0,0) = 0$，$f_y(0,0) = 0$，故公式对每个方向都预测为 $0$，与 $a^2 b$ 矛盾（取 $a=b=1/\sqrt2$，得 $1/(2\sqrt2)\neq 0$）。因此尽管全部方向导数都存在，$f$ 在原点仍不可微。

The previous counterexample still had all directional derivatives. This one is more dramatic: both partials exist at the origin, yet the function is discontinuous there, so it cannot possibly be differentiable. Let上一个反例好歹还拥有全部方向导数。这一个更极端：两个偏导数在原点都存在，函数却在那里不连续，因而绝不可能可微。设

$$ g(x,y) = \begin{cases} \dfrac{xy}{x^2 + y^2}, & (x,y)\neq(0,0), \\[4pt] 0, & (x,y)=(0,0). \end{cases} $$

The partial $g_x(0,0)$ uses only the slice $y = 0$, where $g(x,0) = 0$ for all $x$, so $g_x(0,0) = 0$. By symmetry $g_y(0,0) = 0$. Both partials exist.偏导数 $g_x(0,0)$ 只用到切片 $y = 0$，在该切片上对所有 $x$ 都有 $g(x,0) = 0$，故 $g_x(0,0) = 0$。由对称性 $g_y(0,0) = 0$。两个偏导数都存在。

But approach the origin along $y = x$: then $g(x,x) = \dfrac{x^2}{2x^2} = \dfrac12$ for every $x \neq 0$, whereas along $y = 0$ the value is $0$. The two path limits disagree, so $\lim_{(x,y)\to(0,0)} g$ does not exist and $g$ is discontinuous at the origin. Since differentiability implies continuity, $g$ is not differentiable at $(0,0)$ despite having both partial derivatives there. This is the cleanest possible warning that "partials exist" carries almost no analytic weight on its own.但沿 $y = x$ 趋近原点：对每个 $x \neq 0$ 有 $g(x,x) = \dfrac{x^2}{2x^2} = \dfrac12$，而沿 $y = 0$ 该值为 $0$。两条路径的极限不一致，故 $\lim_{(x,y)\to(0,0)} g$ 不存在，$g$ 在原点不连续。由于可微蕴含连续，尽管 $g$ 在 $(0,0)$ 处两个偏导数都存在，它在那里仍不可微。这是最干净不过的警示：单凭"偏导数存在"几乎不带任何分析上的分量。