03. 벡터 함수Vector Functions

Solution. For every $t$, $x=\cos t$ and $y=\sin t$, so $x^2+y^2=\cos^2 t+\sin^2 t=1$. Hence the curve always lies on the cylinder $x^2+y^2=1$, and since $z=t$ increases together with $t$, the curve winds counterclockwise up the cylinder as $t$ increases. The curve is sketched in coordinate space as follows.

Solution. Let $x=t+1,\ y=t^2-1,\ z=2t$. Since $t=x-1$, we get $y=(x-1)^2-1$ and $z=2(x-1)=2x-2$. Thus the curve is the intersection of the parabolic cylinder $y=(x-1)^2-1$ with the plane $z=2x-2$: as $t$ increases, the curve moves in the $x$-direction tracing a parabola while simultaneously rising linearly in the $z$-direction. The curve is sketched in coordinate space as follows.

Solution. The component functions $\cos t,\ \sin t,\ t$ of $\mathbf{r}(t)=(\cos t)\mathbf{i}+(\sin t)\mathbf{j}+t\mathbf{k}$ are all continuous at $t=\pi/4$, so the limit of the vector function can be computed componentwise.

$$\lim_{t\to\pi/4}\mathbf{r}(t)=\left(\lim_{t\to\pi/4}\cos t\right)\mathbf{i}+\left(\lim_{t\to\pi/4}\sin t\right)\mathbf{j}+\left(\lim_{t\to\pi/4}t\right)\mathbf{k}=\frac{\sqrt2}{2}\mathbf{i}+\frac{\sqrt2}{2}\mathbf{j}+\frac{\pi}{4}\mathbf{k}.$$

Proof. The limit of a vector function exists and equals the vector of componentwise limits precisely when every component has a limit; if even one component fails to have a limit, the vector function has no limit. Since $\cos t$ and $\sin t$ are continuous for all $t$, whether $\mathbf{g}(t)$ is continuous at $t=n$ depends entirely on whether the third component $\lfloor t \rfloor$ is continuous at $t=n$.

For any integer $n$: if $n-1 \leq t \lt n$ then $\lfloor t \rfloor = n-1$, so $\lim_{t\to n^-}\lfloor t \rfloor = n-1$. If $n \leq t \lt n+1$ then $\lfloor t \rfloor = n$, so $\lim_{t\to n^+}\lfloor t \rfloor = n$. Since the left- and right-hand limits $n-1$ and $n$ differ, $\lim_{t\to n}\lfloor t \rfloor$ does not exist.

Hence $\lim_{t\to n}\mathbf{g}(t)$ does not exist either, so $\mathbf{g}(t)$ is discontinuous at $t=n$. Since $n$ was an arbitrary integer, $\mathbf{g}(t)$ is discontinuous at every integer $t=n$.

Solution.

(1) Since $\mathbf{r}(t)=(2\cos t)\mathbf{i}+(2\sin t)\mathbf{j}+(5\cos^2 t)\mathbf{k}$, the velocity is

the speed is

and the acceleration is

(2) At $t=7\pi/4$, $\mathbf{r}(7\pi/4)=\left(2\cdot\frac{\sqrt2}{2}\right)\mathbf{i}+\left(2\cdot\left(-\frac{\sqrt2}{2}\right)\right)\mathbf{j}+\left(5\cdot\frac12\right)\mathbf{k}=\sqrt2\,\mathbf{i}-\sqrt2\,\mathbf{j}+\frac52\mathbf{k}$, and substituting $t=7\pi/4$ into $\mathbf{v}(t)$ from (1) gives $\mathbf{v}(7\pi/4)=\sqrt2\,\mathbf{i}+\sqrt2\,\mathbf{j}+5\mathbf{k}$. Since $\mathbf{r}(t)$ lies on the cylinder $x^2+y^2=4$ and $z=5\cos^2 t$ oscillates with period $\pi$, the curve rises and falls twice per revolution. The curve $\mathbf{r}(t)$ together with $\mathbf{v}(7\pi/4)$ at the point $\mathbf{r}(7\pi/4)$ is sketched in coordinate space as follows.

Solution. Since $\mathbf{r}(t)=\left(\ln(t^2+1)\right)\mathbf{i}+(\tan^{-1}t)\mathbf{j}+\sqrt{t^2+1}\,\mathbf{k}$,

and at $t=0$, $\mathbf{v}(0)=\mathbf{j}$.

Differentiating again,

and at $t=0$, $\mathbf{a}(0)=2\mathbf{i}+\mathbf{k}$.

Since $\mathbf{v}(0)\cdot\mathbf{a}(0)=(0)(2)+(1)(0)+(0)(1)=0$, the vectors $\mathbf{v}(0)$ and $\mathbf{a}(0)$ are perpendicular, so the angle between them has measure $\dfrac{\pi}{2}$.

Solution.

(1) For $\mathbf{r}(t)=(\sin t)\mathbf{i}+(t^2-\cos t)\mathbf{j}+e^t\mathbf{k}$, $\mathbf{r}(0)=(0,-1,1)$, and $\mathbf{r}'(t)=(\cos t)\mathbf{i}+(2t+\sin t)\mathbf{j}+e^t\mathbf{k}$, so $\mathbf{r}'(0)=(1,0,1)$. Hence the tangent line at $t=0$ passes through $(0,-1,1)$ with direction vector $(1,0,1)$, so for a parameter $s$,

(2) For $\mathbf{r}(t)=(t^2,1+t,2t-3)$, the point $(1,2,-1)$ corresponds to $t=1$ (since $1^2=1$, $1+1=2$, and $2\cdot1-3=-1$ all hold). Since $\mathbf{r}'(t)=(2t,1,2)$, $\mathbf{r}'(1)=(2,1,2)$. Hence the tangent line passes through $(1,2,-1)$ with direction vector $(2,1,2)$, so for a parameter $s$,

Solution.

(1) Integrating each component separately,

where $\mathbf{C}$ is a constant vector.

(2) Using $\mathbf{R}(t)=(\sin t)\mathbf{i}+t\mathbf{j}-t^2\mathbf{k}$ (the antiderivative from (1) with $\mathbf{C}=\mathbf{0}$),

Solution. Integrating $\mathbf{a}(t)=-(3\cos t)\mathbf{i}-(3\sin t)\mathbf{j}+2\mathbf{k}$ gives

and since $\mathbf{v}(0)=(0,3,0)$, we have $(0,3,0)+\mathbf{C}_1=(0,3,0)$, so $\mathbf{C}_1=\mathbf{0}$. Hence

Integrating again,

and since $\mathbf{r}(0)=(4,0,0)$, we have $(3,0,0)+\mathbf{C}_2=(4,0,0)$, so $\mathbf{C}_2=(1,0,0)$. Therefore the position vector of A is

Solution. Integrate each component.

First component: $\displaystyle\int\tan t\,dt=-\ln|\cos t|+C_1$, and since $r_1(0)=3$, $-\ln1+C_1=3$, so $C_1=3$. Hence $r_1(t)=3-\ln|\cos t|$.

Second component: $\displaystyle\int\cos\!\left(\frac t2\right)dt=2\sin\!\left(\frac t2\right)+C_2$, and since $r_2(0)=-2$, $2\sin0+C_2=-2$, so $C_2=-2$. Hence $r_2(t)=2\sin\!\left(\dfrac t2\right)-2$.

Third component: $\displaystyle\int(-\sec 2t)\,dt=-\frac12\ln|\sec 2t+\tan 2t|+C_3$, and since $r_3(0)=1$, $-\frac12\ln1+C_3=1$, so $C_3=1$. Hence $r_3(t)=1-\dfrac12\ln|\sec 2t+\tan 2t|$.

Therefore

Solution. Integrating $\dfrac{d^2\mathbf{r}}{dt^2}=e^t\mathbf{i}-e^{-t}\mathbf{j}+4e^{2t}\mathbf{k}$ gives

and since $\left.\dfrac{d\mathbf{r}}{dt}\right|_{t=0}=(-1,4,0)$, we have $(1,1,2)+\mathbf{C}_1=(-1,4,0)$, so $\mathbf{C}_1=(-2,3,-2)$. Hence

Integrating again,

and since $\mathbf{r}(0)=(3,1,2)$, we have $(1,-1,1)+\mathbf{C}_2=(3,1,2)$, so $\mathbf{C}_2=(2,2,1)$. Hence

Solution. Since A moves along the straight-line path from $(1,2,3)$ to $(4,1,4)$, the direction vector of this line is

Since A moves only along the line, both its velocity and acceleration vectors must point along $\overrightarrow{u}$ (or its opposite); indeed the acceleration $(3,-1,1)=\sqrt{11}\,\overrightarrow{u}$ matches this direction. As the speed at $(1,2,3)$ is $2$ and the motion is in the direction of $\overrightarrow{u}$, $\mathbf{v}(0)=2\overrightarrow{u}$.

Since the acceleration is constant,

and substituting $\overrightarrow{u}=\frac{1}{\sqrt{11}}(3,-1,1)$ gives

that is,