벡터와 공간 전체 문제

The average velocity is found by dividing the displacement by the time traveled – it is a vector. The average speed is found by dividing the distance traveled by the time traveled – it is a scalar.

arc length

One example is traveling at a constant speed $s$ in a circle, ending at the starting position. Since the displacement is $\vec 0$, the average velocity is $\vec 0$, hence $\left\|\vec 0\right\|=0$. But traveling at constant speed $s$ means the average speed is also $s >0$.

Distance traveled is always greater than or equal to the magnitude of displacement, therefore average speed will always be at least as large as the magnitude of the average velocity.

1

0

$\vec T(t)$ and $\vec N(t)$.

the speed

time and/or distance

curvature

Answers may include lines, circles, helixes

Answers will vary; they should mention the circle is tangent to the curve and has the same curvature as the curve at that point.

$\kappa$

$a_{\text{T}}$ is not affected by curvature; the greater the curvature, the larger $a_{\text{N}}$ becomes.

(a) The curve intersects $E$ when

$$2\big(t\sin(\pi t)\big)^2+2\big(t\cos(\pi t)\big)^2 +\big(t^2\big)^2=24 \iff 2t^2+t^4=24 \iff (t^2-4)(t^2+6)=0$$

Since we need $t>0$, the desired time is $t=2$ and the corresponding point is $\mathbf{r}(2)=2\,\hat{\boldsymbol{\jmath}}+4\,\hat{\mathbf{k}}$.

(b) Since

$$\mathbf{r}'(t)=\big[\sin(\pi t)+\pi t\cos(\pi t)\big]\hat{\boldsymbol{\imath}} +\big[\cos(\pi t)-\pi t\sin(\pi t)\big]\hat{\boldsymbol{\jmath}} + 2t\hat{\mathbf{k}}$$

a tangent vector to $\Gamma$ at $P$ is any nonzero multiple of

$$\mathbf{r}'(2)=2\pi\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\,\hat{\mathbf{k}}$$

(c) A normal vector to $E$ at $P$ is

$${ \boldsymbol{\nabla} }(2x^2+2y^2+z^2)\big|_{(0,2,4)}= \left< 4x,4y,2z\right>\big|_{(0,2,4)} =\left< 0,8,8\right>$$

Since $\mathbf{r}'(2)$ and $\left< 0,8,8\right>$ are not parallel, $\Gamma$ and $E$ do not intersect at right angles.

(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 $(0,0,0)$.