벡터함수의 미분과 적분

(a) By definition,

$$\mathbf{v}'(t)=\mathbf{a}(t)=\left< \cos t,\sin t,0\right> \implies \mathbf{v}(t)=\left< \sin t+c_1, -\cos t+c_2, c_3\right>$$

for some constants $c_1$, $c_2$, $c_3$. To satisfy $\mathbf{v}(0)=\left< 0,-1,1\right>$, we need $c_1=0$, $c_2=0$ and $c_3=1$. So $\mathbf{v}(t)=\left< \sin t, -\cos t, 1\right>$. Similarly,

$$\mathbf{r}'(t)=\mathbf{v}(t)=\left< \sin t,-\cos t,1\right> \implies \mathbf{r}(t)=\left< -\cos t+d_1, -\sin t+d_2, t+d_3\right>$$

for some constants $d_1$, $d_2$, $d_3$. To satisfy $\mathbf{r}(0)=\left< -1,0,0\right>$, we need $d_1=0$, $d_2=0$ and $d_3=0$. So $\mathbf{r}(t)=\left< -\cos t, -\sin t, t\right>$.

(b) To test for orthogonality, we compute the dot product

$$\mathbf{v}(t)\cdot\mathbf{a}(t)=\left< \sin t, -\cos t, 1\right>\cdot\left< \cos t,\sin t,0\right> =\sin t\cos t-\cos t \sin t+1\times 0=0$$

so $\mathbf{v}(t)\perp\mathbf{a}(t)$ for all $t$.

(c) At $t=-\frac{\pi}{2}$ the particle is at $\mathbf{r}\big(-\frac{\pi}{2}\big)=\left< 0,1,-\frac{\pi}{2}\right>$ and has velocity $\mathbf{v}\big(-\frac{\pi}{2}\big)=\left< -1,0,1\right>$. So the tangent line must pass through $\left< 0,1,-\frac{\pi}{2}\right>$ and have direction vector $\left< -1,0,1\right>$. Here is a vector parametric equation for the tangent line.

$$\mathbf{r}(u)=\left< 0,1,-\frac{\pi}{2}\right>+u\left< -1,0,1\right>$$

(d) True. Look at the path followed by the particle from the top so that we only see $x$ and $y$ coordinates. The path we see (call this the projected path) is $x(t)=-\cos t$, $y(t)=-\sin t$, which is a circle of radius one centred on the origin. Any tangent line to any circle always remains outside the circle. So no tangent line to the projected path can pass through the $(0,0)$. So no tangent line to the path followed by the particle can pass through the $z$–axis and, in particular, through $(0,0,0)$.

(a) Since

$$x(t)^2+y(t)^2=t^2\cos^2\big(\tfrac{\pi t}{2}\big) +t^2\sin^2\big(\tfrac{\pi t}{2}\big)=t^2\qquad\text{and}\qquad z(t)^2=t^2$$

are the same, the path of the particle lies on the cone $z^2=x^2+y^2$.

(b) By definition,

$$\begin{aligned} \text{velocity}&=\mathbf{r}'(t)&&= \big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]\hat{\boldsymbol{\imath}} +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]\hat{\boldsymbol{\jmath}} +\hat{\mathbf{k}} \\ \text{speed}&=|\mathbf{r}'(t)|&&= \sqrt{\big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]^2 +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]^2 +1^2} \\ & &&= \Big[\cos^2\big(\tfrac{\pi t}{2}\big) -2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\sin^2\big(\tfrac{\pi t}{2}\big) \\ & && +\sin^2\big(\tfrac{\pi t}{2}\big) +2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\cos^2\big(\tfrac{\pi t}{2}\big) +1\Big]^{1/2}\\ & &&=\sqrt{2+\frac{\pi^2 t^2}{4}} \end{aligned}$$

(c) At $t=1$, the particle is at $\mathbf{r}(1)=(0,1,1)$ and has velocity $\mathbf{r}'(1)=\left< -\frac{\pi}{2},1,1\right>$. So for $t\ge 1$, the particle is at

$$\left< x,y,z\right> = \left< 0,1,1\right>+(t-1)\left< -\frac{\pi}{2},1,1\right>$$

This is also a vector parametric equation for the line.

(d) Assume that the particle's speed remains constant as it flies along $L$. Then the $x$-coordinate of the particle at time $t$ (for $t\ge 1$) is $-\frac{\pi}{2}(t-1)$. This takes the value $-1$ when $t-1=\frac{2}{\pi}$. So the particle hits $x=-1$, $\frac{2}{\pi}$ seconds after it flew off the cone.

(a) The tangent vectors to the two curves are

$$\mathbf{r}'_1(t)=\left< 1, 2t, 3t^2\right>\qquad \mathbf{r}'_2(t)=\left<-\sin t, \cos t, 1\right>$$

Both curves pass through $P$ at $t=0$ and then the tangent vectors are

$$\mathbf{r}'_1(0)=\left< 1, 0, 0\right>\qquad \mathbf{r}'_2(0)=\left< 0, 1, 1\right>$$

So the angle of intersection, $\theta$, is determined by

$$\begin{aligned} \mathbf{r}'_1(0)\cdot\mathbf{r}'_2(0)=|\mathbf{r}'_1(0)|\,|\mathbf{r}'_2(0)|\,\cos\theta &\implies \left< 1, 0, 0\right>\cdot\left< 0, 1, 1\right>=1\cdot\sqrt{2}\cdot\cos\theta \\ &\implies\cos\theta=0\implies \theta=90^\circ \end{aligned}$$

(b) Our strategy will be to

• find a vector $\mathbf{v}$ whose tail is on one line and whose head is on the other line and then
• find a vector $\mathbf{n}$ that is perpendicular to both lines.
• Then, if we denote by $\theta$ the angle between $\mathbf{v}$ and $\mathbf{n}$, the distance between the two lines is $|\mathbf{v}|\cos\theta = \frac{|\mathbf{v}\cdot\mathbf{n}|}{|\mathbf{n}|}$

Here we go

• So the first step is to find a $\mathbf{v}$.

• One point on the line $\mathbf{r}(t)=\left< t, -1+2t, 1+3t\right>$ is $\mathbf{r}(0)=\left< 0,-1,1\right>$.

• $(x,y,z)$ is on the other line if and only if $x+y-z=4$ and $2x-z=4$. In particular, if $z=0$ then $x+y=4$ and $2x=4$ so that $x=2$ and $y=2$.

• So the vector $\mathbf{v}=\left< 2-0\,,\,2-(-1)\,,\,0-1\right>=\left< 2,3,-1\right>$ has its head on one line and its tail on the other line.

• Next we find a vector $\mathbf{n}$ that is perpendicular to both lines.

• First we find a direction vector for the line $x+y-z=4$, $2x-z=4$. We already know that $x=y=2$, $z=0$ is on that line. We can find a second point on that line by choosing, for example, $z=2$ and then solving $x+y=6$, $2x=6$ to get $x=3$, $y=3$. So one direction vector for the line $x+y-z=4$, $2x-z=4$ is $\mathbf{d}_1=\left<3-2\,,\,3-2\,,\,2-0 \right> = \left< 1,1,2\right>$.

• A second way to get a direction vector for the line $x+y-z=4$, $2x-z=4$ is to observe that $\left< 1,1,-1\right>$ is normal to $x+y-z=4$ and so is perpendicular to the line and $\left< 2,0,-1\right>$ is normal to $2x-z=4$ and so is also perpendicular to the line. So $\left< 1,1,-1\right>\times\left< 2,0,-1\right>$ is a direction vector for the line.

• A direction vector for the line $\mathbf{r}(t)=\left< t, -1+2t, 1+3t\right>$ is $\mathbf{d}_2=\mathbf{r}'(t)=\left< 1,2,3\right>$.

• So

$$\mathbf{n}=\mathbf{d}_2\times\mathbf{d}_1 =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{matrix}\right] = \hat{\boldsymbol{\imath}} + \hat{\boldsymbol{\jmath}} -\hat{\mathbf{k}}$$

is perpendicular to both lines.

The distance between the two lines is then

$$|\mathbf{v}|\cos\theta = \frac{|\mathbf{v}\cdot\mathbf{n}|}{|\mathbf{n}|} =\frac{\left< 2,3,-1\right> \cdot\left< 1,1,-1\right>}{|\left< 1,1,-1\right>|} =\frac{6}{\sqrt{3}} =2\sqrt{3}$$

We have

$$\mathbf{v}(t) = \mathbf{r}'(t) = 5 \sqrt{2}\,\hat{\boldsymbol{\imath}} + 5e^{5t}\,\hat{\boldsymbol{\jmath}} +5 e^{-5t}\,\hat{\mathbf{k}}$$

and hence

$$|\mathbf{v}(t)| = |\mathbf{r}'(t)| = 5 \big|\sqrt{2}\,\hat{\boldsymbol{\imath}} + e^{5t}\,\hat{\boldsymbol{\jmath}} + e^{-5t}\,\hat{\mathbf{k}}\big| = 5\sqrt{2+ e^{10t}+ e^{-10t}}$$

Since $2+ e^{10t}+ e^{-10t} = \big(e^{5t}+e^{-5t}\big)^2$, that's (d).

We are told that

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

So, by definition,

$$\begin{aligned} \text{velocity}&= \mathbf{v}(t)= \mathbf{r}'(t)=-a \sin t\,\hat{\boldsymbol{\imath}}+a\cos t\,\hat{\boldsymbol{\jmath}}+c\,\hat{\mathbf{k}}\\ \text{speed}&=\frac{\mathrm{d}s}{\mathrm{d}t}(t) = |\mathbf{r}'(t)| = \sqrt{a^2+c^2} \\ \text{acceleration}&=\mathbf{a}(t)= \mathbf{r}''(t)=-a \cos t\,\hat{\boldsymbol{\imath}}-a\sin t\,\hat{\boldsymbol{\jmath}} \end{aligned}$$

As $t$ runs over an interval of length $2\pi$, $(x,y)$ traces out a circle of radius $a$ and $z$ increases by $2\pi c$. The path is a helix with radius $a$ and with each turn having height $2\pi c$.

Since

$$\begin{aligned} x'(t)&=a\big[\cos^2 t-\sin^2 t\big]=a\cos 2t\\ y'(t)&=2a\sin t\cos t=a\sin 2t\\ z'(t)&=b \end{aligned}$$

we have

$$\frac{\mathrm{d}s}{\mathrm{d}t}(t) =\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}=\sqrt{a^2+b^2}$$

As the speed $\frac{\mathrm{d}s}{\mathrm{d}t}(t)$ is constant, the length is just $\frac{\mathrm{d}s}{\mathrm{d}t}\,T=\sqrt{a^2+b^2}\,T$.

(a) We can parametrize the circle $\ x^2+y^2=9\$ as $x(\theta)=3\cos\theta$, $y(\theta)=3\sin\theta$ with $\theta$ running from $0$ to $2\pi$. As $z=2x+3y$, the ellipse can be parametrized by

$$x(\theta)=3\cos\theta,\ y(\theta)=3\sin\theta,\ z(\theta)=2x(\theta)+3 y(\theta) =6\cos\theta+9\sin\theta,\ 0\le\theta\le 2\pi$$

(b) As

$$\begin{aligned} \frac{\mathrm{d}s}{\mathrm{d}\theta} &=\sqrt{x'(\theta)^2+y'(\theta)^2+z'(\theta)^2}\\ &=\sqrt{9\sin^2\theta+9\cos^2\theta+36\sin^2\theta +81\cos^2\theta-108\sin\theta\cos\theta}\\ &=\sqrt{45+45\cos^2\theta-108\sin\theta\cos\theta} \end{aligned}$$