벡터와 공간 전체 문제

The stone, while whirling, can be modeled by $\vec r(t) = \left\langle 3\cos(8\pi t),3\sin (8\pi t)\right\rangle$.

(a) For $t$-values $t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4$, where $n$ is an integer. (b) $\left\|\vec r\,'(t)\right\| = 24\pi\approx 51.4$ft/s (c) At $t=0.006$, the stone is approximately $19.77$ft from Goliath. Using the formula for projectile motion, we want the angle of elevation that lets a projectile starting at $\left\langle 0,6\right\rangle$ with a initial velocity of $51.4$ft/s arrive at $\left\langle 19.77,9\right\rangle$. The desired angle is $0.27$ radians, or $15.69^\circ$.

The position function of the ball is $\vec r(t) = \left\langle (v_0\cos\theta)t,-16t^2+(v_0\sin\theta)t+6\right\rangle$, where $\theta$ is the angle of elevation and $v_0$ is the initial ball speed.

(a) With $v_0 = 73.33$ft/s, there are two angles of elevation possible. An angle of $\theta = 9.47^\circ$ delivers the ball in 0.83s, while an angle of $79.57^\circ$ delivers the ball in $4.5$s. (b) With $\theta=8^\circ$, the initial speed must be 53.8mph$\approx 78.9$ft/s.

Call the centre of the circumscribing circle $(\bar x,\bar y)$. This centre must be equidistant from the three vertices. So

$$\bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2$$

or, subtracting $\bar x^2+\bar y^2$ from the three equal expressions,

$$0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y$$

which implies

$$\bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c}$$

The radius is the distance from the vertex $(0,0)$ to the centre $(\bar x,\bar y)$, which is $\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}$.

Note that

• the point on $W$ with $x=0$, $y=0$ obeys $-0+3(0)+3z=6$ and so has $z=2$
• the point on $W$ with $x=0$, $y=2$ obeys $-0+3(2)+3z=6$ and so has $z=0$
• the point on $W$ with $x=3$, $y=0$ obeys $-3+3(0)+3z=6$ and so has $z=3$
• the point on $W$ with $x=3$, $y=2$ obeys $-3+3(2)+3z=6$ and so has $z=1$

So the four corners of the parallelogram are $(0,0,2)$, $(0,2,0)$, $(3,0,3)$ and $(3,2,1)$. The vectors

$$\begin{aligned} \mathbf{d}_1&=\left< 0-0 \,,\, 2-0 \,,\, 0-2 \right> = \left< 0 \,,\, 2 \,,\, -2\right> \\ \mathbf{d}_2&=\left< 3-0 \,,\, 0-0 \,,\, 3-2 \right> = \left< 3 \,,\, 0 \,,\, 1\right> \end{aligned}$$

form two sides of the paralleogram. So the area of the parallelogram is

$$\begin{aligned} \big|\mathbf{d}_1\times\mathbf{d}_2\big| =\left|\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 2 & -2 \\ 3 & 0 & 1 \end{matrix}\right]\right| =\left| 2\,\hat{\boldsymbol{\imath}} - 6\,\hat{\boldsymbol{\jmath}} -6\hat{\mathbf{k}} \right| =\sqrt{76} =2\sqrt{19} \end{aligned}$$

Denote by $\big(x(t),y(t)\big)$ the position of the skier at time $t$. As long as the skier remains on the surface of the hill

$$\begin{aligned} y(t)&=h\big(x(t)\big) \\ \implies y'(t)&=h'\big(x(t)\big)\,x'(t) \\ \implies y''(t)&=h''\big(x(t)\big)\, x'(t)^2+h'\big(x(t)\big)\,x''(t) \end{aligned}$$

So the velocity and acceleration vectors of the skier are

$$\begin{aligned} \mathbf{v}(t)&=\left< 1,h'\big(x(t)\big)\right> x'(t) \\ \mathbf{a}(t)&=\left< 1,h'\big(x(t)\big)\right> x''(t) +\left< 0,h''\big(x(t)\big)\right> x'(t)^2 \end{aligned}$$

The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill and has a magnitude such that the skier remains on the surface of the hill. From the velocity vector of the skier (which remain tangential to the hill as long as the skier remains of the surface of the hill),we see that one vector normal to the hill at $\big(x(t),y(t)\big)$ is $$\mathbf{n}(t)=\left<-h'\big(x(t)\big),1\right>$$ This vector is not a unit vector, but that's ok. By Newton's law of motion

$$m\mathbf{a}=-mg\,\hat{\boldsymbol{\jmath}}+p(t)\,\mathbf{n}(t)$$

for some function $p(t)$. Dot both sides of this equation with $\mathbf{n}(t)$.

$$m\mathbf{a}(t)\cdot\mathbf{n}(t)=-mg\hat{\boldsymbol{\jmath}}\cdot\mathbf{n}(t)+p(t)|\mathbf{n}(t)|^2$$

Substituting in

$$\begin{aligned} mh''\big(x(t)\big)\,x'(t)^2&=-mg+p(t)\left[1+h'\big(x(t)\big)^2\right] \\ \implies p(t)\left[1+h'\big(x(t)\big)^2\right] &=m\Big(g+h''\big(x(t)\big)\,x'(t)^2\Big) \end{aligned}$$

As long as $p(t)\ge 0$, the hill is pushing up in order to keep the skier on the surface. When $p(t)$ becomes negative, the hill has to pull on the skier in order to keep her on the surface. But the hill can't pull, so the skier becomes airborne instead. This happens when

$$g+h''\big(x(t)\big)x'(t)^2=0$$

That is when $x'(t)=\sqrt{-g/h''\big(x(t)\big)}$. At this time $x(t)=x_0$, $y(t)=y_0$ and the speed of the skier is

$$\sqrt{x'(t)^2+y'(t)^2} =\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)}$$

(a) One side of the triangle is $\overrightarrow{AB}=\left< 1,0,1\right> - \left< 0,1,1\right> = \left< 1,-1,0\right>$. A second side of the triangle is $\overrightarrow{AC}=\left< 1,3,0\right> - \left< 0,1,1\right> = \left< 1,2,-1\right>$. If the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $\theta$ and if we take $\overrightarrow{AB}$ as the base of the triangle, then the triangle has base length $b=|\overrightarrow{AB}|$ and height $h=|\overrightarrow{AC}|\sin\theta$ and hence

$$\text{area}=\tfrac{1}{2} bh =\tfrac{1}{2} |\overrightarrow{AB}|\,|\overrightarrow{AC}|\sin\theta =\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| =\tfrac{1}{2}|\left< 1,-1,0\right> \times \left< 1,2,-1\right>|$$

As

$$\left< 1,-1,0\right>\times\left< 1,2,-1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 2 & -1 \end{matrix}\right] =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3\hat{\mathbf{k}}$$

we have

$$\text{area}=\tfrac{1}{2}|\left< 1,1,3\right>|=\tfrac{1}{2}\sqrt{11}\approx 1.658$$

(b) A unit vector perpendicular to the plane containing the triangle is

$$\hat{\mathbf{n}}=\frac{\overrightarrow{AB}\times\overrightarrow{AC}} {|\overrightarrow{AB}\times\overrightarrow{AC}|} =\frac{\left< 1,1,3\right>}{\sqrt{11}}$$

The distance from $P$ to the plane containing the triangle is the length of the projection of $\overrightarrow{AP}=\left< 5,-10,2\right>-\left< 0,1,1\right> =\left< 5,-11,1\right>$ on $\hat{\mathbf{n}}$. If $\theta$ the angle between $\overrightarrow{AP}$ and $\hat{\mathbf{n}}$, then this is

$$\text{distance}= |\overrightarrow{AP}|\,|\cos\theta| =\big|\overrightarrow{AP}\cdot\hat{\mathbf{n}}\big| =\left|\left< 5,-11,1\right>\cdot\frac{\left< 1,1,3\right>}{\sqrt{11}}\right| =\frac{3}{\sqrt{11}}\approx 0.9045$$

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).

(a) By definition,

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

The $(x,y)= a(\cos t,\sin t)$ coordinates go around a circle of radius $a$ and centre $(0,0)$ counterclockwise. One circle is completed for each increase of $t$ by $2\pi$. At the same time, the $z$ coordinate increases at a constant rate. Each time the $(x,y)$ coordinates complete one circle, the $z$ coordinate increases by $2 \pi c$. The path is a helix with radius $a$ and with each turn having height $2\pi c$.

(b) By definition,

$$\begin{aligned} \mathbf{r}(t)&= a \cos t\sin t\,\hat{\boldsymbol{\imath}} +a\sin^2 t\,\hat{\boldsymbol{\jmath}}+a\cos t\,\hat{\mathbf{k}}\\ &= \frac{a}{2} \sin 2t\,\hat{\boldsymbol{\imath}}+a\frac{1-\cos 2t}{2}\,\hat{\boldsymbol{\jmath}}+a\cos t\,\hat{\mathbf{k}}\\ \mathbf{v}(t)=\mathbf{r}'(t)&= a \cos 2t\,\hat{\boldsymbol{\imath}} +a\sin 2t\,\hat{\boldsymbol{\jmath}}-a\sin t\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}t}(t)=|\mathbf{v}(t)|&= a\sqrt{1+\sin^2t}\\ \mathbf{a}(t)=\mathbf{r}''(t)&= -2a \sin 2t\,\hat{\boldsymbol{\imath}} +2a\cos 2t\,\hat{\boldsymbol{\jmath}}-a\cos t\,\hat{\mathbf{k}} \end{aligned}$$

Write

$$\begin{aligned} x(t)=\tfrac{a}{2} \sin 2t \qquad y(t)=a\tfrac{1-\cos 2t}{2} \qquad z(t)=a\cos t \end{aligned}$$

Then

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

Thus the $(x,y)$ coordinates go around a circle of radius $\frac{a}{2}$ and centre $\big(0,\frac{a}{2}\big)$ counterclockwise. At the same time the $z$ coordinate oscillates over the interval between $1$ and $-1$ half as fast. In fact, one can say more about the curve.