벡터와 공간 전체 문제

(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 $(0,0,0)$.