벡터와 공간 전체 문제

In general a helix can be parametrized by

$$\mathbf{r}(\theta)=a\cos\theta\,\hat{\boldsymbol{\imath}}+a\sin\theta\,\hat{\boldsymbol{\jmath}}+b\theta\,\hat{\mathbf{k}}$$

Our first task is to determine $a$ and $b$. The radius of the helix is $3$ cm, so $a=3$ cm. After 10 turns (i.e. $\theta=20\pi$) the height, $b\theta$, is 1 cm. So $b(20\pi)=1$ and $b=\frac{1}{20\pi}$cm/rad. Thus $\mathbf{r}(\theta)=3\cos\theta\,\hat{\boldsymbol{\imath}}+3\sin\theta\,\hat{\boldsymbol{\jmath}}+\frac{1}{20\pi}\theta\hat{\mathbf{k}}$.

With each full turn of the helix (i.e. each increase of $\theta$ by $2\pi$) the height of the helix increases by $2\pi b=\frac{1}{10}\textrm{cm}$. So if we can determine the length of wire in one full turn of the helix, we can easily determine how many turns the helix goes through in total, and from that we can determine the total height of the helix.

As $\mathbf{r}'(\theta)=-3\sin\theta\,\hat{\boldsymbol{\imath}}+3\cos\theta\,\hat{\boldsymbol{\jmath}}+\frac{1}{20\pi}\hat{\mathbf{k}}$ we have $\frac{\mathrm{d}s}{\mathrm{d}\theta}=\big|\mathbf{r}'(\theta)\big|=\sqrt{9+\frac{1}{400\pi^2}}$. So the length of one full turn of the helix is

$$\begin{aligned} \int_0^{2\pi}\sqrt{9+\frac{1}{400\pi^2}}\ \mathrm{d}\theta &=2\pi\sqrt{9+\frac{1}{400\pi^2}} \end{aligned}$$

and 1000cm of wire generates

$$\frac{1000}{2\pi\sqrt{9+\frac{1}{400\pi^2}}} =\frac{500}{\pi\sqrt{9+\frac{1}{400\pi^2}}}$$

turns. Each turn adds $\frac{1}{10}\textrm{cm}$ to the height, so the total height is

$$\frac{500}{\pi\sqrt{9+\frac{1}{400\pi^2}}}\cdot \frac{1}{10} =\frac{50}{\pi\sqrt{9+\frac{1}{400\pi^2}}} \approx 5.3\textrm{ cm}$$

Remark. We can check that this answer is reasonable by taking advantage of the fact that each coil adds only a very small height (relative to the radius). So we expect the length of one coil to be about the same as the circumference of a circle of the same radius, namely $6\pi$. If we were making actual circles of the wire, there would be $\frac{1000}{6\pi}$ of them. Stacking up at 10 per centimetre, this would make a pile of height $\frac{1000}{6\pi\cdot 10}$cm. Since this number is also approximately $5.3$cm, we feel our result is reasonable.

The arclength from $P$ to $P$ will be 0, so $P$ is the point where $s=0$. That is, $\mathbf{r}(0)$, or $\left( \sin(1/2),\cos(1/2),\sqrt{3}/2\right)$.

• Solution 1: We consider the situation geometrically. If we plot $\mathbf{R}$ in space (of the relevant dimension), regardless of its parametrization, the derivative at a point will give a vector tangent to $\mathbf{R}$, in the direction the curve moves when the parameter is increasing. Since $\mathbf{a}(t_0)$ and $\mathbf{b}(s_0)$ describe the same spot on the curve, $\mathbf{a}'(t_0)$ and $\mathbf{b}'(s_0)$ will be parallel (Since we specified the derivatives are nonzero, there's no messiness about vectors being parallel to a zero vector.) — they're both tangent to the same piece of curve. Furthermore, as $t$ increases, so does $s$, so the direction of increasing $t$ is the same as the direction of increasing $s$. Therefore, A. holds.

Now we consider the magnitudes of the vectors, to rule out E. Recall $|\mathbf{a}'(t)|$ is the speed at which the curve changes relative to $t$; this could be any (nonnegative) number. By the same token, $|\mathbf{b}'(s)|=1$. So, $\mathbf{b}'(s_0)$ is a unit vector, while $\mathbf{a}'(t_0)$ may or may not be. Then the two vectors are not necessarily equal (although they could be).

So, the best answer is A.

• Solution 2: The chain rule gives us a relationship between $\mathbf{b}'(s)$ and $\mathbf{a}'(t)$.

$$\begin{aligned} \frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}&=\frac{\mathrm{d}}{\mathrm{d}s}[\mathbf{a}(t(s))]=\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}\,\frac{\mathrm{d}t}{\mathrm{d}s} \end{aligned}$$

So, the vectors $\frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}$ and $\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}$ differ only by the scalar function $\frac{\mathrm{d}t}{\mathrm{d}s}$. So, at any point along the curve, these vectors are parallel.

Furthermore, we know that $t$ and $s$ are positively correlated: as $t$ increases, so does $s$, because we're covering more arclength. So, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonnegative. Furthermore, since the derivatives are nonzero, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonzero. So, $\mathbf{b}'(s_0)$ and $\mathbf{a}'(t_0)$ are positive scalar multiples of each other. That is, they are parallel, and pointing in the same direction. However, unless $\frac{\mathrm{d}t}{\mathrm{d}s}=1$ (that is, $t(s)=s+C$ for some constant $C$), the vectors do not have the same magnitude, and hence are not equal.

So, A is the best solution.

Using $\arctan t = z$, and so $t=\tan z$:

$$\begin{aligned} \mathbf{r}(t)&=\left(\frac{1}{\sqrt{1+t^2}},~ \frac{\arctan t}{\sqrt{1+t^{-2}}},~\arctan t\right)\\ &=\left(\frac{1}{\sqrt{1+\tan^2z}},~ \frac{z}{\sqrt{1+\cot^{2}z}},~z\right)\\ &=\left(\frac{1}{|\sec z|},~ \frac{z}{|\csc z|},~z\right)\\ &=\left(|\cos z|,~ z|\sin z|,~z\right) \\ \text{Since $0 \le t $, and $\arctan t < \pi/2$ we have $0 \le z < \pi/2$, so $\cos z$ and $\sin z$ are both nonnegative.} \\ &=\left(\cos z,~ z\sin z,~z\right) \end{aligned}$$

If we didn't have the restricted domain, this would make a spiral going up: $z$ is both the height of the spiral and a radian measure. The $\hat{\boldsymbol{\imath}}$-component of the spiral stays between $-1$ and $1$, while the $\hat{\boldsymbol{\jmath}}$-component increases. So, our spiral gets increasingly “wide," while staying the same “thickness."

Due to the restricted domain, our actual curve is only one-quarter of a “turn" of this spiral, indicated in red above.

The parameter $z$ is a measure of height, and it is also a radian measure as the spiral turns.

$$\begin{aligned} \mathbf{r}(t)&=(\tfrac12 t^2 , \tfrac13 t^3)\\ \mathbf{r}'(t)&=(t,t^2)\\ |\mathbf{r}'(t)|&=\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}\\ s(t)&=\int_{-1}^t|x|\sqrt{1+x^2}\mathrm{d}x=\begin{cases} \int_{-1}^t -x\sqrt{1+x^2}\mathrm{d}x & \text{when } t \le 0\\ \int_{-1}^0 -x\sqrt{1+x^2}\mathrm{d}x +\int_{0}^t x\sqrt{1+x^2}\mathrm{d}x & \text{when } t > 0\\ \end{cases} \\ \text{Let $u=1+x^2$, $\frac12\mathrm{d}u=x\mathrm{d}x$} \\ &=\begin{cases} -\int_{2}^{1+t^2} \frac{1}{2}\sqrt{u}\mathrm{d}u & \text{when } t \le 0\\ -\int_{2}^1 \frac12\sqrt{u}\mathrm{d}u+ \int_{1}^{1+t^2} \frac12\sqrt{u}\mathrm{d}u & \text{when } t > 0\\ \end{cases} \\&=\begin{cases} -\frac{1}{3}u^{3/2}|_{2}^{1+t^2} & \text{when } t \le 0\\ -\frac{1}{3}u^{3/2}|_{2}^{1}+ \frac13u^{3/2}|_{1}^{1+t^2} & \text{when } t > 0\\ \end{cases} \\&=\begin{cases} \frac{2^{3/2}}{3}-\frac{1}{3}(1+t^2)^{3/2} & \text{when } t \le 0\\ -\frac23+\frac{2^{3/2}}{3}+\frac13(1+t^2)^{3/2} & \text{when } t > 0\\ \end{cases} \\ \text{Solving for $t$ in terms of $s$:} \\ 1+t^2&=\begin{cases}(2\sqrt2-3s)^{2/3} &\text{when } t \le 0\\ (3s+2-2\sqrt{2})^{2/3} &\text{when }t>0\end{cases}\\ t^2&=\begin{cases}(2\sqrt2-3s)^{2/3}-1 &\text{when } t \le 0\\ (3s+2-2\sqrt{2})^{2/3}-1 &\text{when }t>0 \end{cases} \\ \text{Remembering that $\sqrt{t^2}=|t|$:} \\ t&=\begin{cases}-\sqrt{(2\sqrt2-3s)^{2/3}-1} &\text{when } t \le 0\\ \sqrt{(3s+2-2\sqrt{2})^{2/3}-1} &\text{when }t>0 \end{cases} \\ \text{Noting that $t=0$ when $s=\tfrac{1}{3}(2\sqrt2-1)$, we find our reparametrization of $(\tfrac12t^2,\tfrac13t^3)$.} \\ \mathbf{R}(s)&=\begin{cases} \Big(\frac12\left[ (2\sqrt2-3s)^{2/3}-1\right],-\frac13\left[(2\sqrt2-3s)^{2/3}-1\right]^{3/2}\Big)&\text{ when }s\le \frac13(2\sqrt2-1)\\ \Big(\frac12\left[(3s+2-2\sqrt2)^{2/3}-1 \right],\frac13\left[(3s+2-2\sqrt2)^{2/3}-1\right]^{3/2}\Big)&\text{ when }s>\frac13(2\sqrt2-1) \end{cases} \end{aligned}$$

Remark: after a computation with this much detail, it's nice to find a few points to check, to verify that our answer is reasonable. For instance, when $s=0$, $t$ should be $-1$, and vice-versa. Also, we found that $t=0$ corresponds to $s=\frac13(2\sqrt2-1)$. So, we should be able to verify that $\mathbf{r}(0)=\mathbf{R}\left(\frac13(2\sqrt2-1)\right)$ and $\mathbf{r}(-1)=\mathbf{R}(0)$.

• Solution 1: Note that $(\cos t+\sin t )^2+(\sin t - \cos t)^2=2$ for all $t$. So, the points $(x,y)$ of our curve lie on $x^2+y^2=2$, which is a circle of radius $\sqrt{2}$. Indeed

$$\begin{aligned} x(t)&= \cos t+\sin t =\sqrt{2}\big[\cos t \cos\tfrac{\pi}{4} +\sin t \sin\tfrac{\pi}{4}\big] \\ &=\sqrt{2}\cos\big(t-\tfrac{\pi}{4}\big) \\ y(t)&= \sin t-\cos t =\sqrt{2}\big[\sin t \cos\tfrac{\pi}{4}-\cos t\sin \tfrac{\pi}{4}\big] \\ &=\sqrt{2}\sin\big(t-\tfrac{\pi}{4}\big) \end{aligned}$$

So, $\mathbf{r}(t)$ circumnavigates a circle of radius $\sqrt{2}$ and consequently has curvature $\kappa =\frac{1}{\sqrt{2}}$.

• Solution 2: We use the formula $\kappa = \dfrac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{\left| \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^3\right|}$, remembering that $\mathbf{v}(t)=\mathbf{r}'(t)$, $\mathbf{a}(t)=\mathbf{r}''(t)$, and $\frac{\mathrm{d}s}{\mathrm{d}t}=\left| \mathbf{r}'(t)\right|$.

$$\begin{aligned} \mathbf{v}(t)&=\mathbf{r}'(t)=(-\sin t +\cos t, \cos t + \sin t)\\ \mathbf{a}(t)&=\mathbf{r}''(t)=(-\cos t -\sin t, -\sin t + \cos t)\\ \mathbf{v}(t) \times \mathbf{a}(t)&=\big[(-\sin t + \cos t)^2+(\cos t + \sin t)^2\big]\hat{\mathbf{k}}=2\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}t}&=\left|\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} \right|=\sqrt{(-\sin t + \cos t)^2+(\cos t + \sin t )^2}=\sqrt2\\ \kappa &=\left|\frac{\mathbf{v}(t) \times \mathbf{a}(t)}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^3}\right|=\left|\frac{2\hat{\mathbf{k}}}{\sqrt{2}^3}\right|=\frac{1}{\sqrt 2} \end{aligned}$$

For the given ellipse

$$\begin{aligned} \mathbf{r}(t)&= a\cos t\ \hat{\boldsymbol{\imath}} +b\sin t\ \hat{\boldsymbol{\jmath}} \\ \mathbf{v}(t)&= -a\sin t\ \hat{\boldsymbol{\imath}} +b\cos t\ \hat{\boldsymbol{\jmath}} \\ |\mathbf{v}(t)|& = \sqrt{a^2\sin^2t+b^2\cos^2 t}\\ \mathbf{a}(t)&= -a\cos t\ \hat{\boldsymbol{\imath}} -b\sin t\ \hat{\boldsymbol{\jmath}} \\ \mathbf{v}(t)\times\mathbf{a}(t) &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -a\sin t & b\cos t & 0 \\ -a\cos t & -b\sin t & 0 \end{matrix}\right] = ab\,\hat{\mathbf{k}}\\ \kappa (t)&=\frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{|\mathbf{v}(t)|^3} =\frac{ab}{{[a^2\sin^2t+b^2\cos^2 t]}^{3/2}} \end{aligned}$$

Hence the maximum (minimum) curvature is achieved when the denominator is a minimum (maximum) which is the case when $\sin t =0$ ($\cos t=0$). So $\kappa _{\rm max}=\frac{a}{b^2}$ and $\kappa _{\rm min}=\frac{b}{a^2}$.

The velocity vector is

$$\mathbf{v}(\theta)=x'(\theta)\,\hat{\boldsymbol{\imath}}+y'(\theta)\,\hat{\boldsymbol{\jmath}} = \cos\big(\tfrac{1}{2}\pi\theta^2\big)\,\hat{\boldsymbol{\imath}} + \sin\big(\tfrac{1}{2}\pi\theta^2\big)\,\hat{\boldsymbol{\jmath}}$$