벡터함수의 미분과 적분

False, $\frac{d}{dt}[\text{u}(t)\,\times \,\text{u}(t)]=0$

component

displacement

It is difficult to identify the points on the graphs of $\vec r(t)$ and $\vec r\,'(t)$ that correspond to each other.

(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 $(0,0,0)$.