Unit C2: Vector-Valued Functions and Curves第 C2 单元：向量值函数与曲线

Consider $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$. The first two components satisfy $x^2+y^2=\cos^2 t+\sin^2 t=1$, so the projection of the curve onto the $xy$-plane is the unit circle, traced counterclockwise. Meanwhile $z=t$ increases steadily, lifting the point at a constant rate. The result is a helix that winds around the cylinder $x^2+y^2=1$.考虑 $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$。前两个分量满足 $x^2+y^2=\cos^2 t+\sin^2 t=1$，因此曲线在 $xy$ 平面上的投影是单位圆，沿逆时针方向描出。与此同时 $z=t$ 稳定增大，以恒定速率把点抬高。结果是一条绕在圆柱 $x^2+y^2=1$ 上的螺旋线。

One full turn corresponds to $t$ advancing by $2\pi$, during which $z$ rises by exactly $2\pi$ (the pitch of the helix).转一整圈对应 $t$ 增加 $2\pi$，在此期间 $z$ 恰好上升 $2\pi$（即螺旋线的螺距）。

$$\mathbf{r}(0)=\langle 1,0,0\rangle,\qquad \mathbf{r}(\tfrac{\pi}{2})=\langle 0,1,\tfrac{\pi}{2}\rangle.$$

Parametrize the curve of intersection of the cylinder $x^2+y^2=4$ with the plane $z=x+2$. The cylinder suggests $x=2\cos t$, $y=2\sin t$. Substituting into the plane gives $z=2\cos t+2$.把圆柱 $x^2+y^2=4$ 与平面 $z=x+2$ 的交线参数化。圆柱提示取 $x=2\cos t$、$y=2\sin t$。代入平面方程得 $z=2\cos t+2$。

$$\mathbf{r}(t)=\langle 2\cos t,\ 2\sin t,\ 2\cos t+2\rangle,\qquad 0\le t\le 2\pi.$$

This is an ellipse sitting in the tilted plane, projecting down to the circle of radius $2$.这是一个位于倾斜平面内的椭圆，向下投影为半径为 $2$ 的圆。

Parametrize the intersection of the paraboloid $z=x^2+y^2$ with the plane $y=x$. On the plane we have $y=x$, so let $x=t$. Then $y=t$ and $z=x^2+y^2=t^2+t^2=2t^2$.把抛物面 $z=x^2+y^2$ 与平面 $y=x$ 的交线参数化。在该平面上有 $y=x$，故令 $x=t$。于是 $y=t$，且 $z=x^2+y^2=t^2+t^2=2t^2$。

$$\mathbf{r}(t)=\langle t,\ t,\ 2t^2\rangle,\qquad t\in\mathbb{R}.$$

Check a point: at $t=1$, $\mathbf{r}(1)=\langle 1,1,2\rangle$, and indeed $1^2+1^2=2=z$ and $y=x$. The curve is a parabola opening upward, lying in the vertical plane $y=x$. Notice that nothing forced a trigonometric parametrization here: the constraint $y=x$ is linear, so a polynomial parameter is the natural choice. Matching the parametrization to the algebraic shape of the constraints, circles and ellipses get sines and cosines, lines get a linear parameter, is the single most useful habit in this section.验证一个点：当 $t=1$ 时 $\mathbf{r}(1)=\langle 1,1,2\rangle$，确有 $1^2+1^2=2=z$ 且 $y=x$。该曲线是一条开口向上的抛物线，位于竖直平面 $y=x$ 内。注意这里并没有什么强迫我们用三角参数化（parametrization）：约束 $y=x$ 是线性的，因此多项式参数才是自然选择。让参数化匹配约束的代数形状，即圆和椭圆配正弦余弦、直线配线性参数，是本节最有用的一个习惯。

Evaluate $\displaystyle\lim_{t\to 0}\Big\langle \frac{\sin t}{t},\ \frac{e^{t}-1}{t},\ t^2+3\Big\rangle$. Take the limit in each slot:计算 $\displaystyle\lim_{t\to 0}\Big\langle \frac{\sin t}{t},\ \frac{e^{t}-1}{t},\ t^2+3\Big\rangle$。逐个分量取极限：

$$\lim_{t\to 0}\frac{\sin t}{t}=1,\qquad \lim_{t\to 0}\frac{e^{t}-1}{t}=1,\qquad \lim_{t\to 0}(t^2+3)=3,$$ $$\Rightarrow\quad \lim_{t\to 0}\mathbf{r}(t)=\langle 1,\ 1,\ 3\rangle.$$

The first two components are not defined at $t=0$ as written, so $\mathbf{r}$ has a removable discontinuity there. Defining $\mathbf{r}(0)=\langle 1,1,3\rangle$ patches it into a continuous function, exactly as one repairs a removable discontinuity of a scalar function. This is the componentwise limit theorem doing real work.按原式书写，前两个分量在 $t=0$ 处未定义，因此 $\mathbf{r}$ 在该处有一个可去间断点。定义 $\mathbf{r}(0)=\langle 1,1,3\rangle$ 即可把它补成一个连续函数，正如修补一个标量函数的可去间断点一样。这正是按分量取极限的定理在发挥实际作用。

Find a smooth parametrization of the line segment from $P=(1,2,3)$ to $Q=(4,0,-1)$, traversed once from $P$ to $Q$. The standard device is to write $\mathbf{r}(t)=(1-t)\,\overrightarrow{OP}+t\,\overrightarrow{OQ}$ for $0\le t\le 1$, which is $P$ at $t=0$ and $Q$ at $t=1$:求从 $P=(1,2,3)$ 到 $Q=(4,0,-1)$ 的线段的光滑参数化，从 $P$ 到 $Q$ 走一次。标准手法是对 $0\le t\le 1$ 写 $\mathbf{r}(t)=(1-t)\,\overrightarrow{OP}+t\,\overrightarrow{OQ}$，它在 $t=0$ 时为 $P$，在 $t=1$ 时为 $Q$：

$$\mathbf{r}(t)=\langle 1+3t,\ 2-2t,\ 3-4t\rangle,\qquad 0\le t\le 1.$$

Here $\mathbf{r}'(t)=\langle 3,-2,-4\rangle$ is constant and nonzero, so the parametrization is smooth with the correct orientation from $P$ to $Q$. To reverse direction, use $\mathbf{r}(1-t)$ instead; to traverse the full line rather than the segment, drop the restriction on $t$.这里 $\mathbf{r}'(t)=\langle 3,-2,-4\rangle$ 是常向量且非零，故参数化光滑，且具有从 $P$ 到 $Q$ 的正确定向。若要反向，改用 $\mathbf{r}(1-t)$；若要遍历整条直线而非线段，去掉对 $t$ 的限制即可。

For $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$, differentiate componentwise:对 $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$，按分量求导：

$$\mathbf{r}'(t)=\langle -\sin t,\ \cos t,\ 1\rangle.$$

At $t=\tfrac{\pi}{2}$ the point is $\mathbf{r}(\tfrac{\pi}{2})=\langle 0,1,\tfrac{\pi}{2}\rangle$ and the tangent direction is $\mathbf{r}'(\tfrac{\pi}{2})=\langle -1,0,1\rangle$. A parametrization of the tangent line is在 $t=\tfrac{\pi}{2}$ 处，点为 $\mathbf{r}(\tfrac{\pi}{2})=\langle 0,1,\tfrac{\pi}{2}\rangle$，切线方向为 $\mathbf{r}'(\tfrac{\pi}{2})=\langle -1,0,1\rangle$。切线的一个参数化为

$$\ell(u)=\langle 0,1,\tfrac{\pi}{2}\rangle+u\langle -1,0,1\rangle=\langle -u,\ 1,\ \tfrac{\pi}{2}+u\rangle.$$

Let $\mathbf{u}(t)=\langle t,\ t^2,\ 1\rangle$ and $\mathbf{v}(t)=\langle \cos t,\ 0,\ t\rangle$. We verify the product rule $\tfrac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}'$ by computing both sides.设 $\mathbf{u}(t)=\langle t,\ t^2,\ 1\rangle$、$\mathbf{v}(t)=\langle \cos t,\ 0,\ t\rangle$。我们通过计算等式两边来验证乘积法则 $\tfrac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}'$。

Direct route: $\mathbf{u}\cdot\mathbf{v}=t\cos t+0+t=t\cos t+t$, so $\tfrac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\cos t-t\sin t+1$.直接法：$\mathbf{u}\cdot\mathbf{v}=t\cos t+0+t=t\cos t+t$，故 $\tfrac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\cos t-t\sin t+1$。

Product-rule route: $\mathbf{u}'=\langle 1,\ 2t,\ 0\rangle$, $\mathbf{v}'=\langle -\sin t,\ 0,\ 1\rangle$, so乘积法则法：$\mathbf{u}'=\langle 1,\ 2t,\ 0\rangle$、$\mathbf{v}'=\langle -\sin t,\ 0,\ 1\rangle$，于是

$$\mathbf{u}'\cdot\mathbf{v}=\cos t+0+0=\cos t,\qquad \mathbf{u}\cdot\mathbf{v}'=-t\sin t+0+1.$$

Adding gives $\cos t-t\sin t+1$, matching the direct computation. The product rule lets you differentiate $|\mathbf{r}(t)|^2=\mathbf{r}\cdot\mathbf{r}$ without first extracting a square root, which is the trick behind several later derivations.相加得 $\cos t-t\sin t+1$，与直接计算一致。乘积法则让你无需先开平方根就能对 $|\mathbf{r}(t)|^2=\mathbf{r}\cdot\mathbf{r}$ 求导，这正是后面若干推导背后的技巧。

Find $\mathbf{r}(t)$ given $\mathbf{r}'(t)=\langle 2t,\ \cos t,\ e^{t}\rangle$ and $\mathbf{r}(0)=\langle 1,\ 2,\ 0\rangle$. Integrate componentwise:已知 $\mathbf{r}'(t)=\langle 2t,\ \cos t,\ e^{t}\rangle$ 和 $\mathbf{r}(0)=\langle 1,\ 2,\ 0\rangle$，求 $\mathbf{r}(t)$。按分量积分：

$$\mathbf{r}(t)=\Big\langle t^2,\ \sin t,\ e^{t}\Big\rangle+\mathbf{C}.$$

Apply the initial condition. At $t=0$ the antiderivative without the constant is $\langle 0,\ 0,\ 1\rangle$, so $\langle 0,0,1\rangle+\mathbf{C}=\langle 1,2,0\rangle$ forces $\mathbf{C}=\langle 1,\ 2,\ -1\rangle$. Therefore代入初始条件。在 $t=0$ 处，去掉常数的原函数为 $\langle 0,\ 0,\ 1\rangle$，故 $\langle 0,0,1\rangle+\mathbf{C}=\langle 1,2,0\rangle$ 迫使 $\mathbf{C}=\langle 1,\ 2,\ -1\rangle$。因此

$$\mathbf{r}(t)=\langle t^2+1,\ \sin t+2,\ e^{t}-1\rangle.$$

The constant of integration is a single vector with three independent components, so one vector initial condition pins down all three constants at once.积分常数是一个带三个独立分量的单一向量，因此一个向量初始条件就能一次性确定全部三个常数。

Compute $\displaystyle\int_0^{1}\langle 4t^3,\ \tfrac{1}{1+t^2},\ 6t\rangle\,dt$. Integrate each slot over $[0,1]$:计算 $\displaystyle\int_0^{1}\langle 4t^3,\ \tfrac{1}{1+t^2},\ 6t\rangle\,dt$。在 $[0,1]$ 上逐分量积分：

$$\int_0^1 4t^3\,dt=\big[t^4\big]_0^1=1,\quad \int_0^1\frac{dt}{1+t^2}=\big[\arctan t\big]_0^1=\frac{\pi}{4},\quad \int_0^1 6t\,dt=\big[3t^2\big]_0^1=3.$$ $$\int_0^{1}\mathbf{r}(t)\,dt=\Big\langle 1,\ \frac{\pi}{4},\ 3\Big\rangle.$$

A definite vector integral returns a single constant vector, just as a definite scalar integral returns a number. This is the displacement vector when $\mathbf{r}(t)$ is read as a velocity.向量定积分返回一个常向量，正如标量定积分返回一个数。当把 $\mathbf{r}(t)$ 视为速度时，这就是位移向量。

Compare two parametrizations of the same parabola $y=x^2$. First, $\mathbf{r}_1(t)=\langle t,\ t^2,\ 0\rangle$ gives $\mathbf{r}_1'(t)=\langle 1,\ 2t,\ 0\rangle$, which is never zero, so $\mathbf{r}_1$ is smooth and has a well-defined tangent direction everywhere. Second, $\mathbf{r}_2(t)=\langle t^3,\ t^6,\ 0\rangle$ traces the same parabola but $\mathbf{r}_2'(t)=\langle 3t^2,\ 6t^5,\ 0\rangle$, which is $\mathbf{0}$ at $t=0$.比较同一抛物线 $y=x^2$ 的两个参数化。第一个，$\mathbf{r}_1(t)=\langle t,\ t^2,\ 0\rangle$ 给出 $\mathbf{r}_1'(t)=\langle 1,\ 2t,\ 0\rangle$，它永不为零，故 $\mathbf{r}_1$ 光滑，处处有良好定义的切线方向。第二个，$\mathbf{r}_2(t)=\langle t^3,\ t^6,\ 0\rangle$ 描出同一抛物线，但 $\mathbf{r}_2'(t)=\langle 3t^2,\ 6t^5,\ 0\rangle$ 在 $t=0$ 处为 $\mathbf{0}$。

At $t=0$ the unit tangent $\mathbf{T}=\mathbf{r}_2'/|\mathbf{r}_2'|$ is undefined for $\mathbf{r}_2$, even though the underlying curve is perfectly smooth there. The defect is in the parametrization, not the geometry: $\mathbf{r}_2$ instantaneously stops at the origin. This is why "smooth curve" should be read as "admits a smooth parametrization," and why we routinely reparametrize to remove such artificial stalls before computing $\mathbf{T}$, $\mathbf{N}$, or curvature.在 $t=0$ 处，对 $\mathbf{r}_2$ 而言单位切向量 $\mathbf{T}=\mathbf{r}_2'/|\mathbf{r}_2'|$ 没有定义，尽管底层曲线在那里完全光滑。缺陷出在参数化上，而非几何上：$\mathbf{r}_2$ 在原点处瞬间停了下来。这正是为何"光滑曲线"应读作"存在某个光滑参数化"，也是为何在计算 $\mathbf{T}$、$\mathbf{N}$ 或曲率之前，我们惯常要重新参数化以消除这种人为的停滞。

Claim.命题。 If $|\mathbf{r}(t)|=c$ is constant, then $\mathbf{r}'(t)\perp\mathbf{r}(t)$ for all $t$.若 $|\mathbf{r}(t)|=c$ 为常数，则对所有 $t$ 有 $\mathbf{r}'(t)\perp\mathbf{r}(t)$。

Proof.证明。 Constant length means $\mathbf{r}(t)\cdot\mathbf{r}(t)=|\mathbf{r}(t)|^2=c^2$ is constant. Differentiate both sides using the product rule for the dot product:长度恒定意味着 $\mathbf{r}(t)\cdot\mathbf{r}(t)=|\mathbf{r}(t)|^2=c^2$ 为常数。用点积的乘积法则对两边求导：

$$\frac{d}{dt}\big[\mathbf{r}\cdot\mathbf{r}\big]=\mathbf{r}'\cdot\mathbf{r}+\mathbf{r}\cdot\mathbf{r}'=2\,\mathbf{r}\cdot\mathbf{r}'=0.$$

Hence $\mathbf{r}\cdot\mathbf{r}'=0$, so the two vectors are orthogonal. This is exactly why the unit tangent $\mathbf{T}$, which has constant length $1$, is always perpendicular to $\mathbf{T}'$, a fact used to build the normal vector in Section 5.于是 $\mathbf{r}\cdot\mathbf{r}'=0$，两向量正交。这正是为何长度恒为 $1$ 的单位切向量 $\mathbf{T}$ 总与 $\mathbf{T}'$ 垂直，这一事实将在第 5 节用来构造法向量。

Converse and geometry.逆命题与几何意义。 The argument is reversible: if $\mathbf{r}\cdot\mathbf{r}'\equiv 0$ then $\tfrac{d}{dt}|\mathbf{r}|^2=0$, so the length is constant. Geometrically, a particle whose velocity is always perpendicular to its position vector moves on a sphere centered at the origin, never approaching or receding. This single lemma is the engine behind the orthogonality of the entire TNB frame.这个论证是可逆的：若 $\mathbf{r}\cdot\mathbf{r}'\equiv 0$，则 $\tfrac{d}{dt}|\mathbf{r}|^2=0$，故长度恒定。从几何上看，速度始终垂直于位置向量的质点在以原点为中心的球面上运动，既不靠近也不远离。这一条引理正是整个 TNB 标架正交性背后的引擎。

For $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$ on $0\le t\le 2\pi$, we have $\mathbf{r}'(t)=\langle -\sin t,\ \cos t,\ 1\rangle$, so对 $0\le t\le 2\pi$ 上的 $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$，有 $\mathbf{r}'(t)=\langle -\sin t,\ \cos t,\ 1\rangle$，故

$$|\mathbf{r}'(t)|=\sqrt{\sin^2 t+\cos^2 t+1}=\sqrt{2}.$$ $$L=\int_0^{2\pi}\sqrt{2}\,dt=2\pi\sqrt{2}.$$

The speed is constant, which is what makes the helix so convenient as a first example.速率是常数，这正是螺旋线作为首个例子如此方便的原因。

Continuing with the helix, $s(t)=\int_0^t \sqrt{2}\,du=\sqrt{2}\,t$, so $t=s/\sqrt{2}$. Substituting gives the unit-speed form继续用这条螺旋线，$s(t)=\int_0^t \sqrt{2}\,du=\sqrt{2}\,t$，故 $t=s/\sqrt{2}$。代入得到单位速率形式

$$\mathbf{r}(s)=\Big\langle \cos\tfrac{s}{\sqrt{2}},\ \sin\tfrac{s}{\sqrt{2}},\ \tfrac{s}{\sqrt{2}}\Big\rangle.$$

Check: $\mathbf{r}'(s)=\tfrac{1}{\sqrt{2}}\langle -\sin\tfrac{s}{\sqrt{2}},\ \cos\tfrac{s}{\sqrt{2}},\ 1\rangle$, and $|\mathbf{r}'(s)|=\tfrac{1}{\sqrt{2}}\cdot\sqrt{2}=1$, as required.验证：$\mathbf{r}'(s)=\tfrac{1}{\sqrt{2}}\langle -\sin\tfrac{s}{\sqrt{2}},\ \cos\tfrac{s}{\sqrt{2}},\ 1\rangle$，且 $|\mathbf{r}'(s)|=\tfrac{1}{\sqrt{2}}\cdot\sqrt{2}=1$，符合要求。

Find the length of the curve $\mathbf{r}(t)=\langle \sqrt{2}\,t,\ e^{t},\ e^{-t}\rangle$ on $0\le t\le 1$. Curves like this are designed so the square-root integrand collapses to a perfect square.求曲线 $\mathbf{r}(t)=\langle \sqrt{2}\,t,\ e^{t},\ e^{-t}\rangle$ 在 $0\le t\le 1$ 上的长度。这类曲线是经过设计的，好让根号下的被积式化为一个完全平方。

Differentiate: $\mathbf{r}'(t)=\langle \sqrt{2},\ e^{t},\ -e^{-t}\rangle$. Then求导：$\mathbf{r}'(t)=\langle \sqrt{2},\ e^{t},\ -e^{-t}\rangle$。于是

$$|\mathbf{r}'(t)|^2=2+e^{2t}+e^{-2t}=e^{2t}+2+e^{-2t}=(e^{t}+e^{-t})^2.$$

The constant $2$ is exactly the cross term of the square $(e^{t}+e^{-t})^2=e^{2t}+2+e^{-2t}$, so $|\mathbf{r}'(t)|=e^{t}+e^{-t}$ and常数 $2$ 恰好是平方 $(e^{t}+e^{-t})^2=e^{2t}+2+e^{-2t}$ 中的交叉项，故 $|\mathbf{r}'(t)|=e^{t}+e^{-t}$，从而

$$L=\int_0^1 (e^{t}+e^{-t})\,dt=\big[e^{t}-e^{-t}\big]_0^1=e-e^{-1}.$$

The designed-in perfect square is the standard trick that makes a square-root integrand integrable. When a length integral looks hopeless, check whether the radicand factors as $(\text{something})^2$ before reaching for numerical methods.这种刻意设计的完全平方是让根号被积式可积的标准技巧。当一个长度积分看起来毫无希望时，在动用数值方法之前，先检查被开方式能否因式分解为 $(\text{某式})^2$。

Find the length of $\mathbf{r}(t)=\langle 2t,\ t^2,\ \tfrac13 t^3\rangle$ on $0\le t\le 2$. Differentiate: $\mathbf{r}'(t)=\langle 2,\ 2t,\ t^2\rangle$, so求 $\mathbf{r}(t)=\langle 2t,\ t^2,\ \tfrac13 t^3\rangle$ 在 $0\le t\le 2$ 上的长度。求导：$\mathbf{r}'(t)=\langle 2,\ 2t,\ t^2\rangle$，故

$$|\mathbf{r}'(t)|^2=4+4t^2+t^4=(t^2+2)^2,$$

since $(t^2+2)^2=t^4+4t^2+4$. Therefore $|\mathbf{r}'(t)|=t^2+2$ (positive for all $t$), and因为 $(t^2+2)^2=t^4+4t^2+4$。因此 $|\mathbf{r}'(t)|=t^2+2$（对所有 $t$ 都为正），从而

$$L=\int_0^2 (t^2+2)\,dt=\Big[\tfrac{t^3}{3}+2t\Big]_0^2=\tfrac{8}{3}+4=\frac{20}{3}.$$

The lesson repeats: a square root integrand is tractable exactly when the radicand is a perfect square in $t$. Spotting the factorization is half the battle in arc-length problems set by hand.这一教训再次出现：根号被积式恰好在被开方式是关于 $t$ 的完全平方时才可处理。在手算的弧长问题中，看出这个因式分解就成功了一半。

Claim.命题。 If a curve is retraced by a smooth, increasing change of parameter $t=t(\tau)$, the arc length integral gives the same value.若一条曲线被一个光滑、递增的参数变换 $t=t(\tau)$ 重新描出，则弧长积分给出相同的值。

Proof.证明。 Let the new parametrization be $\boldsymbol{\rho}(\tau)=\mathbf{r}(t(\tau))$ with $t'(\tau)>0$ and $t(\alpha)=a$, $t(\beta)=b$. By the chain rule $\boldsymbol{\rho}'(\tau)=\mathbf{r}'(t(\tau))\,t'(\tau)$, so taking magnitudes and using $t'(\tau)>0$,设新参数化为 $\boldsymbol{\rho}(\tau)=\mathbf{r}(t(\tau))$，其中 $t'(\tau)>0$ 且 $t(\alpha)=a$、$t(\beta)=b$。由链式法则 $\boldsymbol{\rho}'(\tau)=\mathbf{r}'(t(\tau))\,t'(\tau)$，取模长并利用 $t'(\tau)>0$，得

$$|\boldsymbol{\rho}'(\tau)|=|\mathbf{r}'(t(\tau))|\,t'(\tau).$$

Now compute the new length integral and substitute $u=t(\tau)$, $du=t'(\tau)\,d\tau$:现在计算新的长度积分，并代换 $u=t(\tau)$、$du=t'(\tau)\,d\tau$：

$$\int_\alpha^\beta |\boldsymbol{\rho}'(\tau)|\,d\tau=\int_\alpha^\beta |\mathbf{r}'(t(\tau))|\,t'(\tau)\,d\tau=\int_a^b |\mathbf{r}'(u)|\,du.$$

The two integrals agree, so arc length is a genuine property of the curve, not of the schedule used to traverse it. This is precisely why reparametrizing by $s$ in the previous example was legitimate: it relabels points without changing distances. If $t'(\tau)<0$ the curve is traced backward and the limits swap, but the unsigned length is unchanged.两个积分相等，因此弧长是曲线本身的固有性质，而非遍历它所用的时间安排的性质。这正是为何上例中按 $s$ 重新参数化是合法的：它只是给点重新贴标签，并不改变距离。若 $t'(\tau)<0$，曲线被反向描出且积分上下限互换，但无符号的长度不变。

Take $\mathbf{r}(t)=\langle a\cos t,\ a\sin t,\ 0\rangle$. Then取 $\mathbf{r}(t)=\langle a\cos t,\ a\sin t,\ 0\rangle$。则

$$\mathbf{r}'(t)=\langle -a\sin t,\ a\cos t,\ 0\rangle,\qquad \mathbf{r}''(t)=\langle -a\cos t,\ -a\sin t,\ 0\rangle.$$ $$\mathbf{r}'\times\mathbf{r}''=\langle 0,\ 0,\ a^2\rangle,\quad |\mathbf{r}'\times\mathbf{r}''|=a^2,\quad |\mathbf{r}'|^3=a^3.$$ $$\kappa=\frac{a^2}{a^3}=\frac{1}{a}.$$

A circle has constant curvature equal to the reciprocal of its radius, exactly as intuition demands.圆具有恒定曲率，等于其半径的倒数，与直觉完全一致。

For the plane curve $y=x^2$, use $\kappa=\dfrac{|f''(x)|}{(1+[f'(x)]^2)^{3/2}}$ with $f'(x)=2x$ and $f''(x)=2$:对平面曲线 $y=x^2$，用公式 $\kappa=\dfrac{|f''(x)|}{(1+[f'(x)]^2)^{3/2}}$，其中 $f'(x)=2x$、$f''(x)=2$：

$$\kappa(x)=\frac{2}{(1+4x^2)^{3/2}}.$$

At the vertex $x=0$ this gives $\kappa(0)=2$, the maximum curvature, so the radius of curvature there is $\rho=\tfrac12$. As $|x|\to\infty$, $\kappa\to 0$: the parabola flattens out far from the vertex. The osculating circle at the vertex has radius $\tfrac12$ and sits inside the parabola, hugging it more tightly than any other circle.在顶点 $x=0$ 处这给出 $\kappa(0)=2$，即最大曲率，故那里的曲率半径为 $\rho=\tfrac12$。当 $|x|\to\infty$ 时 $\kappa\to 0$：抛物线在远离顶点处变得平缓。顶点处的密切圆半径为 $\tfrac12$，位于抛物线内侧，比任何其他圆都更贴合它。

Find the curvature of $\mathbf{r}(t)=\langle t,\ t^2,\ t^3\rangle$ at $t=0$ and at general $t$. Compute the derivatives:求 $\mathbf{r}(t)=\langle t,\ t^2,\ t^3\rangle$ 在 $t=0$ 处以及一般 $t$ 处的曲率。计算各阶导数：

$$\mathbf{r}'(t)=\langle 1,\ 2t,\ 3t^2\rangle,\qquad \mathbf{r}''(t)=\langle 0,\ 2,\ 6t\rangle.$$ $$\mathbf{r}'\times\mathbf{r}''=\langle (2t)(6t)-(3t^2)(2),\ (3t^2)(0)-(1)(6t),\ (1)(2)-(2t)(0)\rangle=\langle 6t^2,\ -6t,\ 2\rangle.$$

Then $|\mathbf{r}'\times\mathbf{r}''|=\sqrt{36t^4+36t^2+4}=2\sqrt{9t^4+9t^2+1}$ and $|\mathbf{r}'|=\sqrt{1+4t^2+9t^4}$, so于是 $|\mathbf{r}'\times\mathbf{r}''|=\sqrt{36t^4+36t^2+4}=2\sqrt{9t^4+9t^2+1}$，$|\mathbf{r}'|=\sqrt{1+4t^2+9t^4}$，故

$$\kappa(t)=\frac{2\sqrt{9t^4+9t^2+1}}{(1+4t^2+9t^4)^{3/2}}.$$

At $t=0$ everything but the constants drops: $\kappa(0)=\dfrac{2\sqrt{1}}{1^{3/2}}=2$. The twisted cubic is the standard example of a genuinely non-planar curve, and the cross-product formula handles it without ever forming $\mathbf{T}$.在 $t=0$ 处，除常数外的项全部消失：$\kappa(0)=\dfrac{2\sqrt{1}}{1^{3/2}}=2$。扭三次曲线是真正非平面曲线的标准例子，而叉积公式处理它时根本无需构造 $\mathbf{T}$。

The osculating circle at a point shares the curve's position, tangent, and curvature there; its radius is $\rho=1/\kappa$ and its center sits a distance $\rho$ from the point along $\mathbf{N}$. From Worked Example 4.2, $\kappa(0)=2$, so $\rho=\tfrac12$. At the vertex $(0,0)$ the parabola opens upward, so $\mathbf{N}$ points in the $+y$ direction and the center is at $\big(0,\ \tfrac12\big)$. The osculating circle is某点处的密切圆与曲线在该点共享位置、切线和曲率；其半径为 $\rho=1/\kappa$，圆心沿 $\mathbf{N}$ 方向距该点 $\rho$。由例题 4.2，$\kappa(0)=2$，故 $\rho=\tfrac12$。在顶点 $(0,0)$ 处抛物线开口向上，故 $\mathbf{N}$ 指向 $+y$ 方向，圆心在 $\big(0,\ \tfrac12\big)$。密切圆为

$$x^2+\Big(y-\tfrac12\Big)^2=\tfrac14.$$

This circle matches $y=x^2$ to second order at the origin: same point, same horizontal tangent, same bending. It is the circle a car would trace if, at the vertex, the steering wheel froze at its current angle. Approximating a curve locally by its osculating circle is the geometric meaning of curvature.这个圆在原点处与 $y=x^2$ 二阶吻合：同一点、同一水平切线、同样的弯曲程度。它就是一辆车在顶点处把方向盘锁死在当前角度时会描出的那个圆。用密切圆在局部逼近曲线，正是曲率的几何含义。

Write $\mathbf{r}'=|\mathbf{r}'|\,\mathbf{T}=\dfrac{ds}{dt}\,\mathbf{T}$. Differentiating with the product rule,写 $\mathbf{r}'=|\mathbf{r}'|\,\mathbf{T}=\dfrac{ds}{dt}\,\mathbf{T}$。用乘积法则求导：

$$\mathbf{r}''=\frac{d^2s}{dt^2}\,\mathbf{T}+\frac{ds}{dt}\,\mathbf{T}'.$$

Now take the cross product $\mathbf{r}'\times\mathbf{r}''$. The $\mathbf{T}\times\mathbf{T}$ term vanishes, leaving现在作叉积 $\mathbf{r}'\times\mathbf{r}''$。$\mathbf{T}\times\mathbf{T}$ 这一项消失，留下

$$\mathbf{r}'\times\mathbf{r}''=\Big(\frac{ds}{dt}\Big)^2\,\mathbf{T}\times\mathbf{T}'.$$

Since $\mathbf{T}$ is a unit vector, $\mathbf{T}'\perp\mathbf{T}$ (Section 2), so $|\mathbf{T}\times\mathbf{T}'|=|\mathbf{T}'|$. Taking magnitudes and using $|ds/dt|=|\mathbf{r}'|$,由于 $\mathbf{T}$ 是单位向量，$\mathbf{T}'\perp\mathbf{T}$（见第 2 节），故 $|\mathbf{T}\times\mathbf{T}'|=|\mathbf{T}'|$。取模长并利用 $|ds/dt|=|\mathbf{r}'|$，得

$$|\mathbf{r}'\times\mathbf{r}''|=|\mathbf{r}'|^2\,|\mathbf{T}'|.$$

Divide by $|\mathbf{r}'|^3$ and recall $\kappa=|\mathbf{T}'|/|\mathbf{r}'|$ to recover $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$.两边除以 $|\mathbf{r}'|^3$，并回忆 $\kappa=|\mathbf{T}'|/|\mathbf{r}'|$，即得 $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3$。

For $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$ we have $\mathbf{r}'(t)=\langle -\sin t,\ \cos t,\ 1\rangle$ and $|\mathbf{r}'|=\sqrt{2}$, so对 $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$，有 $\mathbf{r}'(t)=\langle -\sin t,\ \cos t,\ 1\rangle$ 和 $|\mathbf{r}'|=\sqrt{2}$，故

$$\mathbf{T}(t)=\frac{1}{\sqrt{2}}\langle -\sin t,\ \cos t,\ 1\rangle.$$

Differentiate: $\mathbf{T}'(t)=\tfrac{1}{\sqrt{2}}\langle -\cos t,\ -\sin t,\ 0\rangle$, with $|\mathbf{T}'|=\tfrac{1}{\sqrt{2}}$. Hence求导：$\mathbf{T}'(t)=\tfrac{1}{\sqrt{2}}\langle -\cos t,\ -\sin t,\ 0\rangle$，其中 $|\mathbf{T}'|=\tfrac{1}{\sqrt{2}}$。因此

$$\mathbf{N}(t)=\langle -\cos t,\ -\sin t,\ 0\rangle.$$

The normal always points horizontally toward the central axis of the helix.法向量始终水平地指向螺旋线的中心轴。

Continue with the helix, where $\mathbf{T}(t)=\tfrac{1}{\sqrt2}\langle -\sin t,\cos t,1\rangle$ and $\mathbf{N}(t)=\langle -\cos t,-\sin t,0\rangle$. The binormal is继续用这条螺旋线，其中 $\mathbf{T}(t)=\tfrac{1}{\sqrt2}\langle -\sin t,\cos t,1\rangle$、$\mathbf{N}(t)=\langle -\cos t,-\sin t,0\rangle$。副法向量为

$$\mathbf{B}=\mathbf{T}\times\mathbf{N}=\frac{1}{\sqrt2}\big\langle (\cos t)(0)-(1)(-\sin t),\ (1)(-\cos t)-(-\sin t)(0),\ (-\sin t)(-\sin t)-(\cos t)(-\cos t)\big\rangle.$$ $$\mathbf{B}(t)=\frac{1}{\sqrt2}\langle \sin t,\ -\cos t,\ 1\rangle.$$

Check that $|\mathbf{B}|=\tfrac{1}{\sqrt2}\sqrt{\sin^2 t+\cos^2 t+1}=1$, as a binormal must be. At $t=0$ the point is $\langle 1,0,0\rangle$ and $\mathbf{B}(0)=\tfrac{1}{\sqrt2}\langle 0,-1,1\rangle$. The osculating plane there passes through $\langle 1,0,0\rangle$ with normal $\mathbf{B}(0)$, giving $0(x-1)-1(y-0)+1(z-0)=0$, that is $z=y$. The TNB frame is right-handed: $\mathbf{T}\times\mathbf{N}=\mathbf{B}$, $\mathbf{N}\times\mathbf{B}=\mathbf{T}$, $\mathbf{B}\times\mathbf{T}=\mathbf{N}$.验证 $|\mathbf{B}|=\tfrac{1}{\sqrt2}\sqrt{\sin^2 t+\cos^2 t+1}=1$，副法向量本应如此。在 $t=0$ 处，点为 $\langle 1,0,0\rangle$，$\mathbf{B}(0)=\tfrac{1}{\sqrt2}\langle 0,-1,1\rangle$。那里的密切平面过 $\langle 1,0,0\rangle$ 且以 $\mathbf{B}(0)$ 为法向量，给出 $0(x-1)-1(y-0)+1(z-0)=0$，即 $z=y$。TNB 标架是右手系：$\mathbf{T}\times\mathbf{N}=\mathbf{B}$、$\mathbf{N}\times\mathbf{B}=\mathbf{T}$、$\mathbf{B}\times\mathbf{T}=\mathbf{N}$。

Claim.命题。 The principal normal $\mathbf{N}$ points to the side the curve is turning toward, the concave (inner) side of the bend.主法向量 $\mathbf{N}$ 指向曲线正在转向的那一侧，即弯道的凹侧（内侧）。

Argument.论证。 Parametrize by arc length so $\mathbf{T}=d\mathbf{r}/ds$ is the unit tangent. By the first Frenet equation $d\mathbf{T}/ds=\kappa\mathbf{N}$ with $\kappa\ge 0$, so $\mathbf{N}$ has the same direction as $d\mathbf{T}/ds$, the rate at which the tangent direction rotates. Since $\mathbf{T}$ is a unit vector, its tip moves on the unit sphere, and the velocity of that tip points in the direction the tangent is swinging. That is exactly the direction into the turn. Concretely, on a circle traversed counterclockwise the tangent always rotates toward the center, so $\mathbf{N}$ points radially inward, matching Worked Example 5.1 where $\mathbf{N}=\langle -\cos t,-\sin t,0\rangle$ aims at the axis. Because $\kappa$ is defined to be nonnegative, $\mathbf{N}$ is unambiguous wherever $\kappa>0$; at an inflection point ($\kappa=0$) the normal is momentarily undefined, which is the curve's way of telling you it is, for that instant, going straight.按弧长参数化，使 $\mathbf{T}=d\mathbf{r}/ds$ 为单位切向量。由第一条 Frenet 方程 $d\mathbf{T}/ds=\kappa\mathbf{N}$ 且 $\kappa\ge 0$，故 $\mathbf{N}$ 与 $d\mathbf{T}/ds$ 同向，后者是切线方向旋转的速率。由于 $\mathbf{T}$ 是单位向量，其端点在单位球面上运动，该端点的速度指向切线正在摆动的方向，那恰好就是转弯朝向的方向。具体地，在逆时针绕行的圆上，切线总是朝中心旋转，故 $\mathbf{N}$ 沿径向指向内侧，这与例题 5.1 中 $\mathbf{N}=\langle -\cos t,-\sin t,0\rangle$ 指向中心轴相吻合。由于 $\kappa$ 被定义为非负，只要 $\kappa>0$，$\mathbf{N}$ 就毫无歧义；在拐点（$\kappa=0$）处法向量一时无定义，这是曲线在告诉你它此刻正沿直线前进。

Let $\mathbf{r}(t)=\langle t,\ t^2,\ 0\rangle$. Then $\mathbf{v}=\langle 1,\ 2t,\ 0\rangle$ and $\mathbf{a}=\langle 0,\ 2,\ 0\rangle$. At $t=1$:设 $\mathbf{r}(t)=\langle t,\ t^2,\ 0\rangle$。则 $\mathbf{v}=\langle 1,\ 2t,\ 0\rangle$、$\mathbf{a}=\langle 0,\ 2,\ 0\rangle$。在 $t=1$ 处：

$$|\mathbf{v}|=\sqrt{1+4}=\sqrt{5},\qquad \mathbf{v}\cdot\mathbf{a}=4,\qquad a_T=\frac{4}{\sqrt{5}}.$$ $$\mathbf{v}\times\mathbf{a}=\langle 0,\ 0,\ 2\rangle,\qquad a_N=\frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}=\frac{2}{\sqrt{5}}.$$

Check: $a_T^2+a_N^2=\tfrac{16}{5}+\tfrac{4}{5}=4=|\mathbf{a}|^2$, confirming the decomposition.验证：$a_T^2+a_N^2=\tfrac{16}{5}+\tfrac{4}{5}=4=|\mathbf{a}|^2$，证实了该分解。

A particle has $\mathbf{a}(t)=\langle 0,\ 0,\ -g\rangle$ with $\mathbf{v}(0)=\langle v_0,\ 0,\ 0\rangle$ and $\mathbf{r}(0)=\mathbf{0}$. Integrate twice:一个质点有 $\mathbf{a}(t)=\langle 0,\ 0,\ -g\rangle$，且 $\mathbf{v}(0)=\langle v_0,\ 0,\ 0\rangle$、$\mathbf{r}(0)=\mathbf{0}$。积分两次：

$$\mathbf{v}(t)=\langle v_0,\ 0,\ -gt\rangle,\qquad \mathbf{r}(t)=\Big\langle v_0 t,\ 0,\ -\tfrac{1}{2}gt^2\Big\rangle.$$

This is projectile motion: constant horizontal velocity and a parabolic drop, the vector-calculus version of the familiar kinematic equations.这就是抛体运动：水平速度恒定，竖直方向抛物线式下落，即熟悉的运动学方程的向量微积分版本。

A particle moves on a circle of radius $R$ at constant angular rate $\omega$: $\mathbf{r}(t)=\langle R\cos\omega t,\ R\sin\omega t,\ 0\rangle$. Then一个质点以恒定角速率 $\omega$ 在半径为 $R$ 的圆上运动：$\mathbf{r}(t)=\langle R\cos\omega t,\ R\sin\omega t,\ 0\rangle$。则

$$\mathbf{v}=\langle -R\omega\sin\omega t,\ R\omega\cos\omega t,\ 0\rangle,\qquad |\mathbf{v}|=R\omega\ (\text{constant}),$$ $$\mathbf{a}=\langle -R\omega^2\cos\omega t,\ -R\omega^2\sin\omega t,\ 0\rangle=-\omega^2\,\mathbf{r}.$$

The acceleration points straight back at the center and has magnitude $|\mathbf{a}|=R\omega^2$. Since the speed is constant, $a_T=0$, so all of this acceleration is normal: $a_N=\kappa|\mathbf{v}|^2=\tfrac{1}{R}(R\omega)^2=R\omega^2$, matching $|\mathbf{a}|$ exactly. This is the centripetal acceleration of introductory physics, recovered as the pure normal component.加速度笔直指回圆心，模长为 $|\mathbf{a}|=R\omega^2$。由于速率恒定，$a_T=0$，故全部加速度都是法向的：$a_N=\kappa|\mathbf{v}|^2=\tfrac{1}{R}(R\omega)^2=R\omega^2$，恰好等于 $|\mathbf{a}|$。这就是普通物理中的向心加速度，作为纯法向分量被重新得到。

For $\mathbf{r}(t)=\langle t,\ t^2,\ \tfrac{2}{3}t^3\rangle$, find $a_T$ and $a_N$ at $t=1$. Compute对 $\mathbf{r}(t)=\langle t,\ t^2,\ \tfrac{2}{3}t^3\rangle$，求 $t=1$ 处的 $a_T$ 与 $a_N$。计算

$$\mathbf{v}=\langle 1,\ 2t,\ 2t^2\rangle,\qquad \mathbf{a}=\langle 0,\ 2,\ 4t\rangle.$$

At $t=1$: $\mathbf{v}=\langle 1,2,2\rangle$, $|\mathbf{v}|=3$, $\mathbf{a}=\langle 0,2,4\rangle$. Then在 $t=1$ 处：$\mathbf{v}=\langle 1,2,2\rangle$、$|\mathbf{v}|=3$、$\mathbf{a}=\langle 0,2,4\rangle$。则

$$a_T=\frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|}=\frac{0+4+8}{3}=4,$$ $$\mathbf{v}\times\mathbf{a}=\langle (2)(4)-(2)(2),\ (2)(0)-(1)(4),\ (1)(2)-(2)(0)\rangle=\langle 4,\ -4,\ 2\rangle,$$ $$a_N=\frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}=\frac{\sqrt{16+16+4}}{3}=\frac{6}{3}=2.$$

Check: $a_T^2+a_N^2=16+4=20=|\mathbf{a}|^2=0+4+16$. The Pythagorean check ($a_T^2+a_N^2=|\mathbf{a}|^2$) is the fastest way to catch an arithmetic slip, since $\mathbf{T}\perp\mathbf{N}$.验证：$a_T^2+a_N^2=16+4=20=|\mathbf{a}|^2=0+4+16$。由于 $\mathbf{T}\perp\mathbf{N}$，这个勾股验算（$a_T^2+a_N^2=|\mathbf{a}|^2$）是发现算术失误最快的方法。

Start from $\mathbf{v}=|\mathbf{v}|\,\mathbf{T}=\dfrac{ds}{dt}\,\mathbf{T}$. Differentiate with the product rule:从 $\mathbf{v}=|\mathbf{v}|\,\mathbf{T}=\dfrac{ds}{dt}\,\mathbf{T}$ 出发。用乘积法则求导：

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d^2s}{dt^2}\,\mathbf{T}+\frac{ds}{dt}\,\frac{d\mathbf{T}}{dt}.$$

Now convert $d\mathbf{T}/dt$ to arc length using the chain rule and the first Frenet equation $d\mathbf{T}/ds=\kappa\mathbf{N}$:现在用链式法则和第一条 Frenet 方程 $d\mathbf{T}/ds=\kappa\mathbf{N}$，把 $d\mathbf{T}/dt$ 转换到弧长：

$$\frac{d\mathbf{T}}{dt}=\frac{d\mathbf{T}}{ds}\frac{ds}{dt}=\kappa\frac{ds}{dt}\,\mathbf{N}.$$

Substitute back and write $|\mathbf{v}|=ds/dt$:代回，并记 $|\mathbf{v}|=ds/dt$：

$$\mathbf{a}=\underbrace{\frac{d^2s}{dt^2}}_{a_T}\,\mathbf{T}+\underbrace{\kappa\Big(\frac{ds}{dt}\Big)^2}_{a_N}\,\mathbf{N}=\frac{d|\mathbf{v}|}{dt}\,\mathbf{T}+\kappa|\mathbf{v}|^2\,\mathbf{N}.$$

This proves three facts at once: acceleration has no binormal component (it lives in the osculating plane), the tangential part is the rate of change of speed, and the normal part is $\kappa|\mathbf{v}|^2$. The last identity also gives the practical formula $a_N=|\mathbf{v}\times\mathbf{a}|/|\mathbf{v}|$, since crossing $\mathbf{v}=|\mathbf{v}|\mathbf{T}$ with $\mathbf{a}$ kills the tangential term and leaves $|\mathbf{v}|\,a_N\,(\mathbf{T}\times\mathbf{N})$, whose magnitude is $|\mathbf{v}|\,a_N$.这一次证明了三个事实：加速度没有副法向分量（它位于密切平面内）、切向部分是速率的变化率、法向部分为 $\kappa|\mathbf{v}|^2$。最后这个恒等式还给出实用公式 $a_N=|\mathbf{v}\times\mathbf{a}|/|\mathbf{v}|$，因为把 $\mathbf{v}=|\mathbf{v}|\mathbf{T}$ 与 $\mathbf{a}$ 作叉积会消去切向项，留下 $|\mathbf{v}|\,a_N\,(\mathbf{T}\times\mathbf{N})$，其模长为 $|\mathbf{v}|\,a_N$。

For $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$: $\mathbf{r}'=\langle -\sin t,\cos t,1\rangle$, $\mathbf{r}''=\langle -\cos t,-\sin t,0\rangle$, $\mathbf{r}'''=\langle \sin t,-\cos t,0\rangle$.对 $\mathbf{r}(t)=\langle \cos t,\ \sin t,\ t\rangle$：$\mathbf{r}'=\langle -\sin t,\cos t,1\rangle$、$\mathbf{r}''=\langle -\cos t,-\sin t,0\rangle$、$\mathbf{r}'''=\langle \sin t,-\cos t,0\rangle$。

$$\mathbf{r}'\times\mathbf{r}''=\langle \sin t,\ -\cos t,\ 1\rangle,\qquad |\mathbf{r}'\times\mathbf{r}''|^2=2.$$ $$(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=\sin^2 t+\cos^2 t+0=1\ \Rightarrow\ \tau=\frac{1}{2}.$$

Both curvature and torsion equal $\tfrac12$ for this helix: it bends and twists at equal, constant rates.这条螺旋线的曲率与挠率都等于 $\tfrac12$：它以相等、恒定的速率弯曲并扭转。

Suppose acceleration is always directed along the position vector, $\mathbf{a}=c(t)\,\mathbf{r}$ for some scalar function $c$ (a central force). Consider the vector $\mathbf{r}\times\mathbf{v}$ and differentiate:假设加速度始终沿位置向量方向，即对某个标量函数 $c$ 有 $\mathbf{a}=c(t)\,\mathbf{r}$（有心力）。考虑向量 $\mathbf{r}\times\mathbf{v}$ 并求导：

$$\frac{d}{dt}(\mathbf{r}\times\mathbf{v})=\mathbf{v}\times\mathbf{v}+\mathbf{r}\times\mathbf{a}=\mathbf{0}+\mathbf{r}\times(c\,\mathbf{r})=\mathbf{0}.$$

So $\mathbf{r}\times\mathbf{v}=\mathbf{L}$ is a constant vector. The motion is therefore planar (it stays in the plane through the origin perpendicular to $\mathbf{L}$), and the magnitude $|\mathbf{r}\times\mathbf{v}|$ being constant is exactly Kepler's statement that equal areas are swept in equal times.故 $\mathbf{r}\times\mathbf{v}=\mathbf{L}$ 是一个常向量。因此运动是平面的（它始终位于过原点且垂直于 $\mathbf{L}$ 的平面内），而 $|\mathbf{r}\times\mathbf{v}|$ 为常数恰好就是开普勒所说的"相等时间内扫过相等面积"。

Show that $\mathbf{r}(t)=\langle t,\ t^2,\ 3t+2t^2\rangle$ lies in a plane. The fast way is the torsion test. Compute证明 $\mathbf{r}(t)=\langle t,\ t^2,\ 3t+2t^2\rangle$ 位于一个平面内。最快的办法是挠率判别。计算

$$\mathbf{r}'=\langle 1,\ 2t,\ 3+4t\rangle,\quad \mathbf{r}''=\langle 0,\ 2,\ 4\rangle,\quad \mathbf{r}'''=\langle 0,\ 0,\ 0\rangle.$$

Since $\mathbf{r}'''=\mathbf{0}$, the triple product $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=0$, so $\tau\equiv 0$ and the curve is planar. To name the plane directly, notice $z=3t+2t^2=3x+2y$ because $x=t$ and $y=t^2$. So every point satisfies $3x+2y-z=0$, a single plane through the origin. The torsion test agreed with the algebra, and it would have worked even if the plane were not obvious by inspection.由于 $\mathbf{r}'''=\mathbf{0}$，三重积 $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=0$，故 $\tau\equiv 0$，曲线是平面的。要直接写出该平面，注意因为 $x=t$、$y=t^2$，所以 $z=3t+2t^2=3x+2y$。于是每个点都满足 $3x+2y-z=0$，这是一个过原点的平面。挠率判别与代数结果一致，而且即使凭观察看不出这个平面，它也照样有效。

For the general helix $\mathbf{r}(t)=\langle a\cos t,\ a\sin t,\ bt\rangle$ with $a>0$, the standard results are对一般螺旋线 $\mathbf{r}(t)=\langle a\cos t,\ a\sin t,\ bt\rangle$（$a>0$），标准结果为

$$\kappa=\frac{a}{a^2+b^2},\qquad \tau=\frac{b}{a^2+b^2}.$$

Sketch of the curvature: $\mathbf{r}'=\langle -a\sin t,\ a\cos t,\ b\rangle$ with $|\mathbf{r}'|=\sqrt{a^2+b^2}$, and $\mathbf{r}''=\langle -a\cos t,-a\sin t,0\rangle$, giving $|\mathbf{r}'\times\mathbf{r}''|=a\sqrt{a^2+b^2}$, so $\kappa=a\sqrt{a^2+b^2}/(a^2+b^2)^{3/2}=a/(a^2+b^2)$. Setting $a=b=1$ recovers $\kappa=\tau=\tfrac12$ from Worked Example 7.1. Two limits are instructive: as $b\to 0$ the helix collapses to a circle of radius $a$ and $\kappa\to 1/a$, $\tau\to 0$ (planar); as $b\to\infty$ the helix stretches into a near-straight vertical line and both $\kappa,\tau\to 0$. Curvature and torsion together, both constant here, characterize the helix as the only curve with constant nonzero $\kappa$ and constant nonzero $\tau$.曲率的推导概要：$\mathbf{r}'=\langle -a\sin t,\ a\cos t,\ b\rangle$，$|\mathbf{r}'|=\sqrt{a^2+b^2}$；$\mathbf{r}''=\langle -a\cos t,-a\sin t,0\rangle$，给出 $|\mathbf{r}'\times\mathbf{r}''|=a\sqrt{a^2+b^2}$，故 $\kappa=a\sqrt{a^2+b^2}/(a^2+b^2)^{3/2}=a/(a^2+b^2)$。取 $a=b=1$ 即可重新得到例题 7.1 中的 $\kappa=\tau=\tfrac12$。两个极限很有启发：当 $b\to 0$ 时螺旋线收缩为半径 $a$ 的圆，$\kappa\to 1/a$、$\tau\to 0$（平面）；当 $b\to\infty$ 时螺旋线拉伸成近乎竖直的直线，$\kappa,\tau$ 都 $\to 0$。这里曲率与挠率都恒定，二者合在一起把螺旋线刻画为唯一一类具有恒定非零 $\kappa$ 与恒定非零 $\tau$ 的曲线。

Claim.命题。 If $\tau\equiv 0$ on a curve with $\kappa>0$, the curve lies in a fixed plane.若一条 $\kappa>0$ 的曲线上处处 $\tau\equiv 0$，则该曲线位于一个固定平面内。

Proof.证明。 Work in arc length. The third Frenet equation reads $d\mathbf{B}/ds=-\tau\mathbf{N}$. If $\tau\equiv 0$ then $d\mathbf{B}/ds=\mathbf{0}$, so the binormal $\mathbf{B}$ is a constant vector $\mathbf{B}_0$. Now examine how the position projects onto $\mathbf{B}_0$. Consider $g(s)=(\mathbf{r}(s)-\mathbf{r}(0))\cdot\mathbf{B}_0$ and differentiate:在弧长参数下进行。第三条 Frenet 方程为 $d\mathbf{B}/ds=-\tau\mathbf{N}$。若 $\tau\equiv 0$，则 $d\mathbf{B}/ds=\mathbf{0}$，故副法向量 $\mathbf{B}$ 是常向量 $\mathbf{B}_0$。现在考察位置在 $\mathbf{B}_0$ 上的投影。设 $g(s)=(\mathbf{r}(s)-\mathbf{r}(0))\cdot\mathbf{B}_0$ 并求导：

$$g'(s)=\mathbf{r}'(s)\cdot\mathbf{B}_0=\mathbf{T}(s)\cdot\mathbf{B}_0=0,$$

because $\mathbf{T}$ is always orthogonal to $\mathbf{B}$. Since $g'(s)\equiv 0$ and $g(0)=0$, we get $g(s)\equiv 0$, that is $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\mathbf{B}_0=0$ for all $s$. This is the equation of the plane through $\mathbf{r}(0)$ with normal $\mathbf{B}_0$, and the entire curve satisfies it. Hence the curve is planar, and the converse (a planar curve has $\tau\equiv 0$) follows by reversing the argument. This completes the if-and-only-if behind the section quiz.因为 $\mathbf{T}$ 始终与 $\mathbf{B}$ 正交。由于 $g'(s)\equiv 0$ 且 $g(0)=0$，得 $g(s)\equiv 0$，即对所有 $s$ 有 $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\mathbf{B}_0=0$。这正是过 $\mathbf{r}(0)$ 且以 $\mathbf{B}_0$ 为法向量的平面方程，整条曲线都满足它。故曲线是平面的；逆命题（平面曲线有 $\tau\equiv 0$）由反向论证得出。这就补全了本节测验背后的充要条件。