벡터의 연산

$27$

$\vec u = \left\langle \cos 120^\circ,\sin 120^\circ\right\rangle = \left\langle -1/2,\sqrt{3}/2\right\rangle$.

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

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