Vector-Valued Functions · Solutions向量值函数 · 解答

(a) Unit tangent(a) 单位切向量 M1·A1

$\mathbf{r}'(t)=\langle -\sin t,\cos t,1\rangle$. (M1) Speed: $|\mathbf{r}'|=\sqrt{\sin^{2}t+\cos^{2}t+1}=\sqrt{2}$. Therefore$\mathbf{r}'(t)=\langle -\sin t,\cos t,1\rangle$。(M1) 速率：$|\mathbf{r}'|=\sqrt{\sin^{2}t+\cos^{2}t+1}=\sqrt{2}$。因此

$$ \mathbf{T}(t)=\frac{1}{\sqrt{2}}\langle -\sin t,\;\cos t,\;1\rangle. \quad\text{(A1)} $$

(b) Principal unit normal(b) 主单位法向量 M1·A1·M1·A1·R1

Differentiate $\mathbf{T}$: (M1)对 $\mathbf{T}$ 求导：(M1)

$$ \mathbf{T}'(t)=\frac{1}{\sqrt{2}}\langle -\cos t,\;-\sin t,\;0\rangle. $$

Compute the magnitude: (A1)计算模长：(A1)

$$ |\mathbf{T}'(t)|=\frac{1}{\sqrt{2}}\sqrt{\cos^{2}t+\sin^{2}t+0}=\frac{1}{\sqrt{2}}\cdot 1=\frac{1}{\sqrt{2}}. $$

Normalize: (M1)规范化：(M1)

$$ \mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=\frac{\tfrac{1}{\sqrt{2}}\langle -\cos t,-\sin t,0\rangle}{\tfrac{1}{\sqrt{2}}}=\langle -\cos t,\;-\sin t,\;0\rangle. \quad\text{(A1)} $$

Verify $|\mathbf{N}|=\sqrt{\cos^{2}t+\sin^{2}t+0}=1$. (R1) Verify orthogonality:验证 $|\mathbf{N}|=\sqrt{\cos^{2}t+\sin^{2}t+0}=1$。(R1) 验证正交性：

$$ \mathbf{N}\cdot\mathbf{T}=\frac{1}{\sqrt{2}}\bigl[(-\cos t)(-\sin t)+(-\sin t)(\cos t)+0\cdot1\bigr]=\frac{1}{\sqrt{2}}[\sin t\cos t-\sin t\cos t]=0.\checkmark $$

(c) Binormal vector(c) 副法向量 M1·A1·M1·A1·R1

Compute $\mathbf{B}=\mathbf{T}\times\mathbf{N}$ using the determinant form. Write $\mathbf{T}=\tfrac{1}{\sqrt{2}}\langle T_1,T_2,T_3\rangle$ with $T_1=-\sin t$, $T_2=\cos t$, $T_3=1$, and $\mathbf{N}=\langle -\cos t,-\sin t,0\rangle$: (M1)用行列式形式计算 $\mathbf{B}=\mathbf{T}\times\mathbf{N}$。令 $\mathbf{T}=\tfrac{1}{\sqrt{2}}\langle T_1,T_2,T_3\rangle$，其中 $T_1=-\sin t$，$T_2=\cos t$，$T_3=1$，$\mathbf{N}=\langle -\cos t,-\sin t,0\rangle$：(M1)

$$ \mathbf{T}\times\mathbf{N}=\frac{1}{\sqrt{2}}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-\sin t&\cos t&1\\-\cos t&-\sin t&0\end{vmatrix}. $$

Expanding: (A1 for the expansion)展开：(A1 用于展开)

$$ \mathbf{i}\!\left[\cos t\cdot 0 - 1\cdot(-\sin t)\right]-\mathbf{j}\!\left[(-\sin t)\cdot 0 - 1\cdot(-\cos t)\right]+\mathbf{k}\!\left[(-\sin t)(-\sin t)-\cos t(-\cos t)\right] $$ $$ =\mathbf{i}(\sin t)-\mathbf{j}(\cos t)+\mathbf{k}(\sin^{2}t+\cos^{2}t)=\langle \sin t,\;-\cos t,\;1\rangle. $$

Therefore (M1 for scaling):因此（M1 用于比例缩放）：

$$ \mathbf{B}(t)=\frac{1}{\sqrt{2}}\langle \sin t,\;-\cos t,\;1\rangle. \quad\text{(A1)} $$

Verify $|\mathbf{B}|=\tfrac{1}{\sqrt{2}}\sqrt{\sin^{2}t+\cos^{2}t+1}=\tfrac{1}{\sqrt{2}}\cdot\sqrt{2}=1$. (R1) The TNB frame is a right-handed orthonormal basis at every point of the helix.验证 $|\mathbf{B}|=\tfrac{1}{\sqrt{2}}\sqrt{\sin^{2}t+\cos^{2}t+1}=\tfrac{1}{\sqrt{2}}\cdot\sqrt{2}=1$。(R1) TNB标架在螺旋线上每一点处均构成右手正交归一基。

(a) Constant magnitude implies orthogonality(a) 模长恒定推出正交性 M1·M1·A1·R1

Suppose $|\mathbf{r}(t)|=c$ for all $t$. Squaring: $|\mathbf{r}(t)|^{2}=c^{2}$ is constant. (M1) Rewrite the left side as $\mathbf{r}(t)\cdot\mathbf{r}(t)$ and differentiate both sides with respect to $t$. Apply the product rule for the dot product: (M1)设对所有 $t$ 均有 $|\mathbf{r}(t)|=c$。两边平方：$|\mathbf{r}(t)|^{2}=c^{2}$ 为常数。(M1) 将左侧改写为 $\mathbf{r}(t)\cdot\mathbf{r}(t)$，对两侧关于 $t$ 求导，应用点积的乘积法则：(M1)

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

Since $c^{2}$ is constant, its derivative is zero: $2\,\mathbf{r}\cdot\mathbf{r}'=0$. (A1) Dividing by $2$ gives $\mathbf{r}(t)\cdot\mathbf{r}'(t)=0$, so $\mathbf{r}$ and $\mathbf{r}'$ are orthogonal for all $t$. (R1)由于 $c^{2}$ 为常数，其导数为零：$2\,\mathbf{r}\cdot\mathbf{r}'=0$。(A1) 两侧除以 $2$ 得 $\mathbf{r}(t)\cdot\mathbf{r}'(t)=0$，即 $\mathbf{r}$ 与 $\mathbf{r}'$ 对所有 $t$ 正交。(R1)

(b) Converse(b) 逆命题 M1·A1

Suppose $\mathbf{r}(t)\cdot\mathbf{r}'(t)=0$ for all $t$. Then $\tfrac{d}{dt}|\mathbf{r}|^{2}=\tfrac{d}{dt}[\mathbf{r}\cdot\mathbf{r}]=2\,\mathbf{r}\cdot\mathbf{r}'=0$. (M1) A function with zero derivative is constant, so $|\mathbf{r}(t)|^{2}$ is constant and hence $|\mathbf{r}(t)|$ is constant. (A1)设对所有 $t$ 均有 $\mathbf{r}(t)\cdot\mathbf{r}'(t)=0$。则 $\tfrac{d}{dt}|\mathbf{r}|^{2}=\tfrac{d}{dt}[\mathbf{r}\cdot\mathbf{r}]=2\,\mathbf{r}\cdot\mathbf{r}'=0$。(M1) 导数为零的函数是常数，故 $|\mathbf{r}(t)|^{2}$ 为常数，从而 $|\mathbf{r}(t)|$ 为常数。(A1)

(c) Application to the unit tangent(c) 应用于单位切向量 R1·A1

The unit tangent $\mathbf{T}$ satisfies $|\mathbf{T}(t)|=1$ (constant) for all $t$. By part (a), $\mathbf{T}\cdot\mathbf{T}'=0$ everywhere, i.e. $\mathbf{T}'$ is perpendicular to $\mathbf{T}$. (R1) Therefore $\mathbf{T}'/|\mathbf{T}'|$ is a unit vector that is automatically orthogonal to $\mathbf{T}$: this is exactly the definition of the principal unit normal $\mathbf{N}=\mathbf{T}'/|\mathbf{T}'|$, and the orthogonality $\mathbf{N}\perp\mathbf{T}$ requires no separate verification. (A1)单位切向量 $\mathbf{T}$ 满足对所有 $t$ 有 $|\mathbf{T}(t)|=1$（常数）。由(a)，$\mathbf{T}\cdot\mathbf{T}'=0$ 处处成立，即 $\mathbf{T}'$ 垂直于 $\mathbf{T}$。(R1) 因此 $\mathbf{T}'/|\mathbf{T}'|$ 是自动与 $\mathbf{T}$ 正交的单位向量：这正是主单位法向量 $\mathbf{N}=\mathbf{T}'/|\mathbf{T}'|$ 的定义，正交性 $\mathbf{N}\perp\mathbf{T}$ 无需另行验证。(A1)

(a) Proof of the dot-product rule(a) 点积乘积法则的证明 M1·M1·A1·R1

Write out $\mathbf{u}\cdot\mathbf{v}=u_1 v_1+u_2 v_2+u_3 v_3$. (M1) Differentiate the scalar sum termwise using the ordinary product rule:写出 $\mathbf{u}\cdot\mathbf{v}=u_1 v_1+u_2 v_2+u_3 v_3$。(M1) 对标量和逐项用普通乘积法则求导：

$$ \frac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\frac{d}{dt}(u_1 v_1)+\frac{d}{dt}(u_2 v_2)+\frac{d}{dt}(u_3 v_3) $$ $$ =(u_1'v_1+u_1 v_1')+(u_2'v_2+u_2 v_2')+(u_3'v_3+u_3 v_3'). \quad\text{(M1)} $$

Group the terms: the first and third of each pair form $\mathbf{u}'\cdot\mathbf{v}$; the second form $\mathbf{u}\cdot\mathbf{v}'$. (A1)将各项分组：每对中第一、三项构成 $\mathbf{u}'\cdot\mathbf{v}$，第二项构成 $\mathbf{u}\cdot\mathbf{v}'$。(A1)

$$ =(u_1'v_1+u_2'v_2+u_3'v_3)+(u_1 v_1'+u_2 v_2'+u_3 v_3')=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}'. \quad\text{(R1)} $$

(b) Cross-product rule and order sensitivity(b) 叉积乘积法则与顺序敏感性 A1·A1

The product rule for the cross product is (A1)叉积的乘积法则为 (A1)

$$ \frac{d}{dt}[\mathbf{u}\times\mathbf{v}]=\mathbf{u}'\times\mathbf{v}+\mathbf{u}\times\mathbf{v}'. $$

Order cannot be reversed in the first term: $\mathbf{u}'\times\mathbf{v}=-\mathbf{v}\times\mathbf{u}'$, so swapping gives a sign error. (A1) For example, take constant $\mathbf{u}=\mathbf{i}$ and $\mathbf{v}(t)=t\mathbf{j}$. Then $\mathbf{u}'=\mathbf{0}$, $\mathbf{v}'=\mathbf{j}$, and $\tfrac{d}{dt}[\mathbf{i}\times t\mathbf{j}]=\mathbf{i}\times\mathbf{j}=\mathbf{k}$; the product rule gives $\mathbf{0}\times t\mathbf{j}+\mathbf{i}\times\mathbf{j}=\mathbf{k}$. Reversing the first term to $t\mathbf{j}\times\mathbf{0}$ still gives zero, but if the derivative were nonzero, the sign would flip.第一项中因子顺序不可对调：$\mathbf{u}'\times\mathbf{v}=-\mathbf{v}\times\mathbf{u}'$，对调会产生符号错误。(A1) 例如，取常向量 $\mathbf{u}=\mathbf{i}$，$\mathbf{v}(t)=t\mathbf{j}$。则 $\mathbf{u}'=\mathbf{0}$，$\mathbf{v}'=\mathbf{j}$，$\tfrac{d}{dt}[\mathbf{i}\times t\mathbf{j}]=\mathbf{i}\times\mathbf{j}=\mathbf{k}$；乘积法则给出 $\mathbf{0}\times t\mathbf{j}+\mathbf{i}\times\mathbf{j}=\mathbf{k}$。将第一项对调为 $t\mathbf{j}\times\mathbf{0}$ 此处结果仍为零，但若导数不为零，符号将会改变。

(c) Numerical verification at $t=0$(c) 在 $t=0$ 处的数值验证 M1·A1

Direct route: $\mathbf{u}\cdot\mathbf{v}=t\sin t+0+0=t\sin t$, so $\tfrac{d}{dt}[t\sin t]=\sin t+t\cos t$. At $t=0$: value $=0+0=0$. (M1)直接法：$\mathbf{u}\cdot\mathbf{v}=t\sin t+0+0=t\sin t$，故 $\tfrac{d}{dt}[t\sin t]=\sin t+t\cos t$。在 $t=0$ 处：值 $=0+0=0$。(M1)

Product-rule route: $\mathbf{u}'=\langle 1,2t,0\rangle$, $\mathbf{v}'=\langle \cos t,0,0\rangle$. Then $\mathbf{u}'\cdot\mathbf{v}=\sin t$ and $\mathbf{u}\cdot\mathbf{v}'=t\cos t$. Sum $=\sin t+t\cos t$. At $t=0$: value $=0+0=0$. Both sides agree. (A1)乘积法则法：$\mathbf{u}'=\langle 1,2t,0\rangle$，$\mathbf{v}'=\langle \cos t,0,0\rangle$。则 $\mathbf{u}'\cdot\mathbf{v}=\sin t$，$\mathbf{u}\cdot\mathbf{v}'=t\cos t$。两项之和 $=\sin t+t\cos t$。在 $t=0$ 处：值 $=0+0=0$。两侧一致。(A1)

(a) Deriving the decomposition(a) 推导分解式 M1·M1·M1·A1·M1·A1·B1

Since $\mathbf{T}=\mathbf{v}/v$, we have $\mathbf{v}=v\mathbf{T}$ where $v=|\mathbf{v}|=|\mathbf{r}'|$. (M1) Differentiate with respect to $t$ using the scalar-vector product rule:由于 $\mathbf{T}=\mathbf{v}/v$，有 $\mathbf{v}=v\mathbf{T}$，其中 $v=|\mathbf{v}|=|\mathbf{r}'|$。(M1) 对 $t$ 使用标量-向量乘积法则求导：

$$ \mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dv}{dt}\mathbf{T}+v\frac{d\mathbf{T}}{dt}=\dot{v}\mathbf{T}+v\mathbf{T}'. \quad\text{(M1)} $$

Now $\mathbf{T}'=\tfrac{d\mathbf{T}}{dt}$. Since $\mathbf{T}$ is a unit vector, $\mathbf{T}'$ is perpendicular to $\mathbf{T}$ (Q4). Define curvature $\kappa=|\mathbf{T}'|/v$ (the rate of turning per unit arc length), so $|\mathbf{T}'|=\kappa v$. (M1) Because $\mathbf{N}=\mathbf{T}'/|\mathbf{T}'|$, we have $\mathbf{T}'=|\mathbf{T}'|\mathbf{N}=\kappa v\mathbf{N}$. (A1) Substituting:其中 $\mathbf{T}'=\tfrac{d\mathbf{T}}{dt}$。由于 $\mathbf{T}$ 是单位向量，$\mathbf{T}'$ 垂直于 $\mathbf{T}$（第4题）。定义曲率 $\kappa=|\mathbf{T}'|/v$（单位弧长的转向速率），故 $|\mathbf{T}'|=\kappa v$。(M1) 由于 $\mathbf{N}=\mathbf{T}'/|\mathbf{T}'|$，得 $\mathbf{T}'=|\mathbf{T}'|\mathbf{N}=\kappa v\mathbf{N}$。(A1) 代入：

$$ \mathbf{a}=\dot{v}\mathbf{T}+v(\kappa v\mathbf{N})=\underbrace{\dot{v}}_{a_T}\mathbf{T}+\underbrace{\kappa v^{2}}_{a_N}\mathbf{N}. \quad\text{(M1·A1)} $$

So $a_T=dv/dt$ (tangential: rate of change of speed) and $a_N=\kappa v^{2}$ (normal: centripetal). (B1 for naming both)因此 $a_T=dv/dt$（切向：速率的变化率），$a_N=\kappa v^{2}$（法向：向心加速度）。(B1 用于命名两者)

(b) Expressing $a_T$ and $a_N$ via $\mathbf{r}'$, $\mathbf{r}''$(b) 用 $\mathbf{r}'$ 与 $\mathbf{r}''$ 表达 $a_T$ 与 $a_N$ M1·A1·M1·A1

For $a_T$: take the dot product of $\mathbf{a}=\mathbf{r}''$ with $\mathbf{T}=\mathbf{r}'/|\mathbf{r}'|$. Since $\mathbf{T}\cdot\mathbf{T}=1$ and $\mathbf{N}\cdot\mathbf{T}=0$, (M1)对于 $a_T$：将 $\mathbf{a}=\mathbf{r}''$ 与 $\mathbf{T}=\mathbf{r}'/|\mathbf{r}'|$ 作点积。由于 $\mathbf{T}\cdot\mathbf{T}=1$ 且 $\mathbf{N}\cdot\mathbf{T}=0$，(M1)

$$ \mathbf{r}''\cdot\mathbf{T}=a_T\underbrace{\mathbf{T}\cdot\mathbf{T}}_{1}+a_N\underbrace{\mathbf{N}\cdot\mathbf{T}}_{0}=a_T,\quad\text{so}\quad a_T=\mathbf{r}''\cdot\frac{\mathbf{r}'}{|\mathbf{r}'|}=\frac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|}. \quad\text{(A1)} $$

For $a_N$: take the cross product of $\mathbf{r}'=v\mathbf{T}$ with $\mathbf{r}''=a_T\mathbf{T}+a_N\mathbf{N}$. Since $\mathbf{T}\times\mathbf{T}=\mathbf{0}$ and $|\mathbf{T}\times\mathbf{N}|=|\mathbf{B}|=1$, (M1)对于 $a_N$：将 $\mathbf{r}'=v\mathbf{T}$ 与 $\mathbf{r}''=a_T\mathbf{T}+a_N\mathbf{N}$ 作叉积。由于 $\mathbf{T}\times\mathbf{T}=\mathbf{0}$ 且 $|\mathbf{T}\times\mathbf{N}|=|\mathbf{B}|=1$，(M1)

$$ \mathbf{r}'\times\mathbf{r}''=v\mathbf{T}\times(a_T\mathbf{T}+a_N\mathbf{N})=v\,a_T(\mathbf{T}\times\mathbf{T})+v\,a_N(\mathbf{T}\times\mathbf{N})=v\,a_N\mathbf{B}. $$

Taking magnitudes, $|\mathbf{r}'\times\mathbf{r}''|=v\,a_N|\mathbf{B}|=v\,a_N$ (since $a_N\ge0$). Thus $a_N=|\mathbf{r}'\times\mathbf{r}''|/v=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|$. (A1)取模长，$|\mathbf{r}'\times\mathbf{r}''|=v\,a_N|\mathbf{B}|=v\,a_N$（因 $a_N\ge0$）。因此 $a_N=|\mathbf{r}'\times\mathbf{r}''|/v=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|$。(A1)

(a) Plane-curve curvature formula(a) 平面曲线曲率公式 M1·A1·M1·A1

For $y=x^{2}$: $y'=2x$ and $y''=2$. The plane-curve formula is $\kappa=|y''|/(1+y'^{2})^{3/2}$. (M1)对于 $y=x^{2}$：$y'=2x$，$y''=2$。平面曲线公式为 $\kappa=|y''|/(1+y'^{2})^{3/2}$。(M1)

At $x=0$: $\kappa(0)=2/(1+0)^{3/2}=2$. (A1)在 $x=0$ 处：$\kappa(0)=2/(1+0)^{3/2}=2$。(A1)

At $x=1$: $y'=2$, so $\kappa(1)=2/(1+4)^{3/2}=2/5^{3/2}=2/(5\sqrt{5})$. (M1·A1)在 $x=1$ 处：$y'=2$，故 $\kappa(1)=2/(1+4)^{3/2}=2/5^{3/2}=2/(5\sqrt{5})$。(M1·A1)

(b) Curvature of the general helix by the cross-product formula(b) 用叉积公式求一般螺旋线的曲率 M1·A1·M1·A1

Compute $\mathbf{r}'$ and $\mathbf{r}''$: (M1)计算 $\mathbf{r}'$ 与 $\mathbf{r}''$：(M1)

$$ \mathbf{r}'(t)=\langle -a\sin t,\;a\cos t,\;b\rangle,\qquad \mathbf{r}''(t)=\langle -a\cos t,\;-a\sin t,\;0\rangle. $$

Compute the cross product $\mathbf{r}'\times\mathbf{r}''$:计算叉积 $\mathbf{r}'\times\mathbf{r}''$：

$$ \mathbf{r}'\times\mathbf{r}''=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-a\sin t&a\cos t&b\\-a\cos t&-a\sin t&0\end{vmatrix} $$ $$ =\mathbf{i}(a\cos t\cdot 0-b(-a\sin t))-\mathbf{j}((-a\sin t)\cdot0-b(-a\cos t))+\mathbf{k}((-a\sin t)(-a\sin t)-(a\cos t)(-a\cos t)) $$ $$ =\mathbf{i}(ab\sin t)-\mathbf{j}(ab\cos t)+\mathbf{k}(a^{2}\sin^{2}t+a^{2}\cos^{2}t)=\langle ab\sin t,\;-ab\cos t,\;a^{2}\rangle. $$

(A1) Compute the magnitudes:(A1) 计算模长：

$$ |\mathbf{r}'\times\mathbf{r}''|=\sqrt{a^{2}b^{2}\sin^{2}t+a^{2}b^{2}\cos^{2}t+a^{4}}=\sqrt{a^{2}b^{2}+a^{4}}=a\sqrt{a^{2}+b^{2}}, $$ $$ |\mathbf{r}'|=\sqrt{a^{2}\sin^{2}t+a^{2}\cos^{2}t+b^{2}}=\sqrt{a^{2}+b^{2}}. \quad\text{(M1)} $$

Apply the formula $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^{3}$:应用公式 $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^{3}$：

$$ \kappa=\frac{a\sqrt{a^{2}+b^{2}}}{\bigl(\sqrt{a^{2}+b^{2}}\bigr)^{3}}=\frac{a\sqrt{a^{2}+b^{2}}}{(a^{2}+b^{2})^{3/2}}=\frac{a}{a^{2}+b^{2}}. \quad\text{(A1)} $$

(a) Derivatives(a) 导数 A1

$$ \mathbf{r}'(t)=\langle 1,\;2t,\;2t^{2}\rangle,\qquad \mathbf{r}''(t)=\langle 0,\;2,\;4t\rangle. \quad\text{(A1)} $$

(b) Cross product $\mathbf{r}'\times\mathbf{r}''$(b) 叉积 $\mathbf{r}'\times\mathbf{r}''$ M1·A1·A1

$$ \mathbf{r}'\times\mathbf{r}''=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2t&2t^{2}\\0&2&4t\end{vmatrix}. \quad\text{(M1)} $$

Expanding: (A1 for expansion)展开：(A1 用于展开)

$$ \mathbf{i}(2t\cdot 4t-2t^{2}\cdot 2)-\mathbf{j}(1\cdot 4t-2t^{2}\cdot 0)+\mathbf{k}(1\cdot 2-2t\cdot 0) $$ $$ =\mathbf{i}(8t^{2}-4t^{2})-\mathbf{j}(4t)+\mathbf{k}(2)=\langle 4t^{2},\;-4t,\;2\rangle. \quad\text{(A1)} $$

(c) Curvature $\kappa(t)$(c) 曲率 $\kappa(t)$ M1·A1·A1

Compute the magnitudes: (M1)计算模长：(M1)

$$ |\mathbf{r}'\times\mathbf{r}''|=\sqrt{16t^{4}+16t^{2}+4}=2\sqrt{4t^{4}+4t^{2}+1}=2(2t^{2}+1), $$

since $(2t^{2}+1)^{2}=4t^{4}+4t^{2}+1\ge0$. (A1)因为 $(2t^{2}+1)^{2}=4t^{4}+4t^{2}+1\ge0$。(A1)

$$ |\mathbf{r}'|=\sqrt{1+4t^{2}+4t^{4}}=\sqrt{(1+2t^{2})^{2}}=1+2t^{2}. $$

Apply $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^{3}$:应用 $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^{3}$：

$$ \kappa(t)=\frac{2(2t^{2}+1)}{(1+2t^{2})^{3}}=\frac{2}{(1+2t^{2})^{2}}. \quad\text{(A1)} $$

(d) Value and interpretation at $t=0$(d) $t=0$ 处的值与解释 A1

$\kappa(0)=2/(1+0)^{2}=2$. At $t=0$ the curve passes through the origin with $\mathbf{r}(0)=\mathbf{0}$ and $\mathbf{r}'(0)=\mathbf{i}$, so it is moving in the $x$-direction. The curvature $\kappa=2$ means the osculating circle at the origin has radius $1/2$, the same radius of curvature as the parabola $y=x^{2}$ at its vertex (Q7a, where $\kappa=2$). (A1)$\kappa(0)=2/(1+0)^{2}=2$。在 $t=0$ 时，曲线过原点，$\mathbf{r}(0)=\mathbf{0}$，$\mathbf{r}'(0)=\mathbf{i}$，即沿 $x$ 方向运动。曲率 $\kappa=2$ 意味着原点处密切圆半径为 $1/2$，与抛物线 $y=x^{2}$ 在顶点处的曲率半径相同（第7(a)题，$\kappa=2$）。(A1)

(a) Arc-length function and its inverse(a) 弧长函数及其反函数 M1·A1

From Q2, $|\mathbf{r}'(t)|=5$ is constant. The arc-length function starting at $t=0$ is (M1)由第2题，$|\mathbf{r}'(t)|=5$ 为常数。从 $t=0$ 出发的弧长函数为 (M1)

$$ s(t)=\int_0^t|\mathbf{r}'(u)|\,du=\int_0^t 5\,du=5t. $$

Solving for $t$: $t=s/5$. (A1) Because the speed is constant, $s$ and $t$ are proportional; the reparametrization is purely a rescaling of time.解出 $t$：$t=s/5$。(A1) 由于速率恒定，$s$ 与 $t$ 成正比，重参数化仅是时间的重新缩放。

(b) Arc-length parametrization(b) 弧长参数化 M1·A1

Substitute $t=s/5$ into $\mathbf{r}(t)$: (M1)将 $t=s/5$ 代入 $\mathbf{r}(t)$：(M1)

$$ \mathbf{r}(s)=\Big\langle 3\cos\tfrac{s}{5},\;3\sin\tfrac{s}{5},\;\tfrac{4s}{5}\Big\rangle. \quad\text{(A1)} $$

(c) Verify unit speed(c) 验证单位速率 M1·A1

Differentiate $\mathbf{r}(s)$ with respect to $s$: (M1)将 $\mathbf{r}(s)$ 关于 $s$ 求导：(M1)

$$ \frac{d\mathbf{r}}{ds}=\Big\langle -\tfrac{3}{5}\sin\tfrac{s}{5},\;\tfrac{3}{5}\cos\tfrac{s}{5},\;\tfrac{4}{5}\Big\rangle. $$

Compute the magnitude:计算模长：

$$ \left|\frac{d\mathbf{r}}{ds}\right|=\sqrt{\tfrac{9}{25}\sin^{2}\tfrac{s}{5}+\tfrac{9}{25}\cos^{2}\tfrac{s}{5}+\tfrac{16}{25}}=\sqrt{\tfrac{9}{25}+\tfrac{16}{25}}=\sqrt{\tfrac{25}{25}}=1. \quad\text{(A1)} $$

(a) Velocity, acceleration, speed(a) 速度、加速度与速率 M1·A1

Differentiate componentwise: (M1)逐分量求导：(M1)

$$ \mathbf{v}(t)=\mathbf{r}'(t)=\langle 2t,\;2,\;2t\rangle,\qquad \mathbf{a}(t)=\mathbf{r}''(t)=\langle 2,\;0,\;2\rangle. $$ $$ v(t)=|\mathbf{v}|=\sqrt{4t^{2}+4+4t^{2}}=\sqrt{8t^{2}+4}=2\sqrt{2t^{2}+1}. \quad\text{(A1)} $$

(b) Tangential and normal components at $t=1$(b) $t=1$ 处的切向与法向分量 M1·A1·M1·A1

At $t=1$: $\mathbf{v}=\langle 2,2,2\rangle$, $v=2\sqrt{3}$, $\mathbf{a}=\langle 2,0,2\rangle$. (M1)在 $t=1$ 处：$\mathbf{v}=\langle 2,2,2\rangle$，$v=2\sqrt{3}$，$\mathbf{a}=\langle 2,0,2\rangle$。(M1)

$$ a_T=\frac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|}=\frac{4+0+4}{2\sqrt{3}}=\frac{8}{2\sqrt{3}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}. \quad\text{(A1)} $$

Compute $\mathbf{v}\times\mathbf{a}$: (M1)计算 $\mathbf{v}\times\mathbf{a}$：(M1)

$$ \mathbf{v}\times\mathbf{a}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&2&2\\2&0&2\end{vmatrix}=\mathbf{i}(4-0)-\mathbf{j}(4-4)+\mathbf{k}(0-4)=\langle 4,\;0,\;-4\rangle. $$ $$ |\mathbf{v}\times\mathbf{a}|=\sqrt{16+0+16}=4\sqrt{2}. $$ $$ a_N=\frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|}=\frac{4\sqrt{2}}{2\sqrt{3}}=\frac{2\sqrt{2}}{\sqrt{3}}=\frac{2\sqrt{6}}{3}. \quad\text{(A1)} $$

(c) Pythagorean identity check(c) 勾股恒等式验证 A1

$|\mathbf{a}|^{2}=4+0+4=8$. Compute $a_T^{2}+a_N^{2}$:$|\mathbf{a}|^{2}=4+0+4=8$。计算 $a_T^{2}+a_N^{2}$：

$$ a_T^{2}+a_N^{2}=\frac{48}{9}+\frac{24}{9}=\frac{72}{9}=8=|\mathbf{a}|^{2}.\checkmark \quad\text{(A1)} $$