단위접선·법선벡터

(a) If $\kappa (s)\equiv 0$, then $\frac{\mathrm{d}\hat{\mathbf{T}}}{\mathrm{d}s}=\kappa (s)\hat{\mathbf{N}}(s)\equiv 0$ so that $\hat{\mathbf{T}}$ is a constant. As a result $\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}s}(s)=\hat{\mathbf{T}}$ and $\mathbf{r}(s) =s\hat{\mathbf{T}}+\mathbf{r}(0)$ so that the curve is the straight line with direction vector $\hat{\mathbf{T}}$ that passes through $\mathbf{r}(0)$.

(b) If $\tau(s)\equiv 0$, then $\frac{\mathrm{d}\hat{\mathbf{B}}}{\mathrm{d}s}=-\tau(s)\hat{\mathbf{N}}(s)\equiv 0$ so that $\hat{\mathbf{B}}$ is a constant. As $\hat{\mathbf{T}}(s)\perp\hat{\mathbf{B}}$,

$$\frac{\mathrm{d}}{\mathrm{d}s} (\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}} =\hat{\mathbf{T}}(s)\cdot\hat{\mathbf{B}}=0$$

and $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}}$ must be a constant. The constant must be zero (set $s=0$), so $(\mathbf{r}(s)-\mathbf{r}(0))\cdot\hat{\mathbf{B}}=0$ and $\mathbf{r}(s)$ always lies in the plane through $\mathbf{r}(0)$ with normal vector $\hat{\mathbf{B}}$.

(c) Parametrize the curve by arc length. Define the “centre of curvature” at $s$ by

$$\mathbf{r}_c(s)=\mathbf{r}(s)+\frac{1}{\kappa (s)}\hat{\mathbf{N}}(s)$$

Since $\kappa (s)=\kappa _0$ is a constant and $\tau(s)\equiv 0$,

$$\frac{\mathrm{d}}{\mathrm{d}s}\mathbf{r}_c(s) =\hat{\mathbf{T}}(s)+\frac{1}{\kappa _0}\big[\tau(s)\hat{\mathbf{B}}-\kappa (s)\hat{\mathbf{T}}\big] =\hat{\mathbf{T}}(s)+\frac{1}{\kappa _0}\big[0\hat{\mathbf{B}}-\kappa _0\hat{\mathbf{T}}\big] =0$$

Thus $\mathbf{r}_c(s)=\mathbf{r}_c$ is a constant and $\big|\mathbf{r}(s)-\mathbf{r}_c\big|=\frac{1}{\kappa _0}$ lies on the sphere of radius $\frac{1}{\kappa _0}$ centred on $\mathbf{r}_c$. Since $\tau(s)\equiv 0$, the curve also lies on a plane, so it is a circle.

(a) As

$$\begin{aligned} \mathbf{r}'(t)&=\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +2\,\hat{\mathbf{k}} & \frac{\mathrm{d}s}{\mathrm{d}t}(t) &=|\mathbf{r}'(t)|= \sqrt{4+2e^{2t}+2e^{-2t}} =\sqrt{2}\big(e^t+e^{-t}\big)\\ \mathbf{r}''(t)&=\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\imath}} + \big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\jmath}} & \mathbf{r}'(t)\times\mathbf{r}''(t)&=-2\big(e^t + e^{-t}\big)\,\hat{\boldsymbol{\imath}} + 2\big(e^t - e^{-t}\big)\,\hat{\boldsymbol{\jmath}} +4\,\hat{\mathbf{k}} \end{aligned}$$

the curvature

$$\begin{aligned} \kappa (t) &= \frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{\big(\frac{\mathrm{d}s}{\mathrm{d}t}\big)^3} =\frac{2\sqrt{4+2e^{2t}+2e^{-2t}}}{[4+2e^{2t}+2e^{-2t}]^{3/2}} =\frac{1}{2+e^{2t}+e^{-2t}} \end{aligned}$$

(b) The length of $C$ between $\mathbf{r}(0)$ and $\mathbf{r}(1)$ is

$$\begin{aligned} \int_0^1 \frac{\mathrm{d}s}{\mathrm{d}t}(t) \ \mathrm{d}t &= \sqrt{2} \int_0^1 (e^t+e^{-t}) \ \mathrm{d}t =\sqrt{2}\Big[e^t-e^{-t}\Big]_0^1 =\sqrt{2}\Big[e-\frac{1}{e}\Big] \end{aligned}$$

The point $(2,4,8)$ occurs when $t=2$.

$$\begin{aligned} \mathbf{v}(t)&=(1,2t,3t^2) & \mathbf{v}(2)&=(1,4,12)\\ \mathbf{a}(t)&=(0,2,6t) & \mathbf{a}(2)&=(0,2,12)\\ \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)&=(0,0,6) & \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(2)&=(0,0,6)\\ && \mathbf{v}(2) \times \mathbf{a}(2)&=(24,-12,2)\\ && |\mathbf{v}(2) \times \mathbf{a}(2)|&=2\sqrt{181} \end{aligned}$$

Now, we use a formula for torsion:

$$\begin{aligned}\tau(t)&=\dfrac{(\mathbf{v}(t) \times \mathbf{a}(t))\cdot\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)}{|\mathbf{v}(t) \times \mathbf{a}(t)|^2}\\ \tau(2)&=\frac{(24,-12,2)\cdot(0,0,6)}{(2\sqrt{181})^2}=\frac{3}{{181}} \end{aligned}$$

First, some preliminaries:

$$\begin{aligned} \mathbf{r}(t)&=(t^3,t,e^{ct}) & \mathbf{r}(5)&=(5^3,5,e^{5c})\\ \mathbf{v}(t)&=(3t^2,1,ce^{ct}) & \mathbf{v}(5)&=(3\cdot 5^2,1,ce^{5c})\\ \mathbf{a}(t)&=(6t,0,c^2e^{ct}) & \mathbf{a}(5)&=(6\cdot 5,0,c^2e^{5c}) \\ \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)&= (6,0,c^3e^{ct})& \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)&= (6,0,c^3e^{5c})\\ && \mathbf{v}(5) \times \mathbf{a}(5)&=(c^2e^{5c},15ce^{5c}(2-5c),-30) \end{aligned}$$

Second, we figure out what value of $c$ makes $\tau(5)=0$.

$$\begin{aligned} 0&=\tau(5)=\dfrac{(\mathbf{v}(5) \times \mathbf{a}(5))\cdot\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)}{|\mathbf{v}(5) \times \mathbf{a}(5)|^2}\\ 0&=(\mathbf{v}(5)\times\mathbf{a}(5))\cdot \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)\\ &=(c^2e^{5c},15ce^{5c}(2-5c),-30)\cdot (6,0,c^3e^{5c})\\ &=6c^2e^{5c}(1-5c)\\ c&=0 \text{ or } c=\frac15 \end{aligned}$$

If $c=0$, then $\mathbf{r}(t)=(t^3,t,1)$, and so the entire curve is contained inside the plane $z=1$. (Its torsion is zero everywhere — not just at $t=5$.)

Consider the case $c=\frac15$. When $t=5$, our curve (and its osculating circle) passes through the point $\mathbf{r}(5)=(5^3,5,e)$. The normal vector to the plane of the osculating curve is the binormal vector $\hat{\mathbf{B}}(5)=\frac{\mathbf{v}(5)\times \mathbf{a}(5)}{|\mathbf{v}(5)\times \mathbf{a}(5)|}$. Since we don't need the normal vector to the plane to be a unit vector, we can take as the normal vector to the plane simply $\mathbf{v}(5) \times \mathbf{a}(5)$, or $(e/25,3e,-30)$. Then, an equation of the plane containing the osculating circle is $(e/25)x+(3e)y-30z=-10e$. An equivalent equation for this plane is $(1/25)x+3y-(30/e)z=-10$.

(a) The curve $x^2+y^2=1$ is a circle of radius $1$. So we can parametrize it by $x(\theta)=\cos\theta$, $y(\theta)=\sin\theta$, $0\le\theta<2\pi$. The $z$-coordinate of any point on the intersection is determined by $z=1-x-y$. So we can use

$$\mathbf{r}(\theta) = \cos\theta\,\hat{\boldsymbol{\imath}} +\sin\theta\,\hat{\boldsymbol{\jmath}} + (1-\cos\theta-\sin\theta)\,\hat{\mathbf{k}} \qquad 0\le\theta<2\pi$$

(b) As

$$\begin{aligned} \mathbf{v}(\theta)=\mathbf{r}'(\theta)&=\big(-\sin\theta,\cos\theta,\sin\theta-\cos\theta\big) \\ \mathbf{a}(\theta)=\mathbf{v}'(\theta)&=\big(-\cos\theta,-\sin\theta,\cos\theta+\sin\theta\big) \end{aligned}$$

we have

$$\begin{aligned} \frac{\mathrm{d}s}{\mathrm{d}\theta} &= |\mathbf{r}'(\theta)| =\sqrt{\sin^2\theta+\cos^2\theta+(\sin\theta-\cos\theta)^2}\\ &=\sqrt{2-2\sin\theta\cos\theta} \\ &=\sqrt{2-\sin(2\theta)} \\ \mathbf{v}(\theta)\times\mathbf{a}(\theta)&=\big(1,1,1\big) \end{aligned}$$

and the curvature

$$\begin{aligned} \kappa (\theta) &= \frac{|\mathbf{v}(\theta)\times\mathbf{a}(\theta)|}{\big(\frac{\mathrm{d}s}{\mathrm{d}\theta}\big)^3} =\frac{\sqrt{3}}{[2-\sin(2\theta)]^{3/2}} \end{aligned}$$

(c) The curvature is

• a maximum (minimum) when $2-\sin(2\theta)$ is a minimum (maximum),
• which is the case when $\sin(2\theta)=1$ ($\sin(2\theta)=-1$),
• which in turn is the case when $\theta=\frac{\pi}{4},\frac{5\pi}{4}$ ($\theta=\frac{3\pi}{4},\frac{7\pi}{4}$).

So

$$\begin{aligned} \text{maximum curvature }&= \frac{\sqrt{3}}{[2-1]^{3/2}}=\sqrt{3} && \text{at }\qquad\frac{\hat{\boldsymbol{\imath}}}{\sqrt{2}}+\frac{\hat{\boldsymbol{\jmath}}}{\sqrt{2}}+(1-\sqrt{2})\,\hat{\mathbf{k}} \\ & && \text{and }\quad-\frac{\hat{\boldsymbol{\imath}}}{\sqrt{2}}-\frac{\hat{\boldsymbol{\jmath}}}{\sqrt{2}}+(1+\sqrt{2})\,\hat{\mathbf{k}} \\ \text{minimum curvature }&= \frac{\sqrt{3}}{[2-(-1)]^{3/2}}=\frac{1}{3} && \text{at }\qquad-\frac{\hat{\boldsymbol{\imath}}}{\sqrt{2}}+\frac{\hat{\boldsymbol{\jmath}}}{\sqrt{2}}+\hat{\mathbf{k}} \\ & && \text{and }\qquad\frac{\hat{\boldsymbol{\imath}}}{\sqrt{2}}-\frac{\hat{\boldsymbol{\jmath}}}{\sqrt{2}}+\hat{\mathbf{k}} \end{aligned}$$

(a) For the specified curve $\mathbf{r}(\pi) = \big(-4,0,\frac{1}{4}\big)$ and

$$\begin{aligned} \mathbf{r}(\theta) &= \Big(4 \cos\theta\,,\, 2\sin\theta\,,\, \frac{1}{4}\cos(2\theta)\Big) \\ \mathbf{v}(\theta)=\mathbf{r}'(\theta) &= \Big(-4 \sin\theta\,,\, 2\cos\theta\,,\, -\frac{1}{2}\sin(2\theta)\Big) \\ \mathbf{a}(\theta) = \mathbf{r}''(\theta) &= \Big(-4 \cos\theta\,,\, -2\sin\theta\,,\, -\cos(2\theta)\Big) \\ \mathbf{v}(\pi) &= \big(0\,,\, -2\,,\, 0\big) \\ \mathbf{a}(\pi) &= \big(4\,,\, 0\,,\, -1\big) \\ \mathbf{v}(\pi) \times \mathbf{a}(\pi) &= \big(2\,,\, 0\,,\, 8\big) \end{aligned}$$

So the curvature at $\theta=\pi$ is

$$\begin{aligned} \kappa (\pi)&=\frac{|\mathbf{v}(\pi) \times \mathbf{a}(\pi)|}{|\mathbf{v}(\pi)|^3} =\frac{|(2,0,8)|}{|(0,-2,0)|^3} =\frac{\sqrt{17}}{4} \end{aligned}$$

(b) The radius is

$$\frac{1}{\kappa (\pi)} = \frac{4}{\sqrt{17}}$$

(c) Set $\mathbf{R}(t) = \mathbf{r}(t^2)$. Then

$$\begin{aligned} \mathbf{R}'(t)&=2t\,\mathbf{r}'(t^2) \\ \mathbf{R}''(t) &= 2\,\mathbf{r}'(t^2) +4t^2\,\mathbf{r}''(t^2) \end{aligned}$$

In particular,

$$\begin{aligned} \mathbf{R}(\sqrt{\pi}) &= \Big(-4\,,\, 0\,,\, \frac{1}{4}\Big) \\ \mathbf{R}'(\sqrt{\pi}) &= 2\sqrt{\pi}\,\mathbf{v}(\pi) =\big(0\,,\, -4\sqrt{\pi}\,,\, 0\big) \\ \text{speed} = \big|\mathbf{R}'(\sqrt{\pi})\big| &=4\sqrt{\pi} \\ \text{acceleration}=\mathbf{R}''(\sqrt{\pi}) &= 2\,\mathbf{v}(\pi)+4\pi\,\mathbf{a}(\pi) = \big( 16\pi\,,\, -4\,,\, -4\pi\big) \end{aligned}$$