Ii. (15pts) Determine whether the following statements are true or false? No $=$ justification is necessary

(1) If $a_n<0$ for $n>1000$, and $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges., then $\sum _{n=1}^{\mathrm{\infty }}{a}_n^2$ may diverges

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, $a_n<0$ for $n>1000\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|=\sum _{n=1}^{1000}\left|a_n\right|-\sum _{n=000}^{1000}{a}_n$ converges and $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$

$\therefore \underset{n\to \mathrm{\infty }}{lim}\left|a_n\right|=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n^2}{\left|a_n\right|}=0,\sum \left|a_n\right|$ converges, $\Rightarrow \sum a_n^2$ converges. False

(2) Suppose that the power series $\sum _{n=1}^{\mathrm{\infty }}{a}_n(x-1{)}^n$ converges at $x=0$, and diverges at $x=2$, then the interval of convergence of this series is $[0,2)$

True

$\Rightarrow i.\sum a_n(x-1{)}^n$ converges at $x=0\Rightarrow \sum (\ln (x-1{)}^n$ converges absolutely $|x-1|\mid$ so. $x-1=-1$. i.e $0<x<2$

diverges at $x=2$ ie $x-1=1\Rightarrow$ diverges for $|x-1|>1\Rightarrow x>$ or $x<0$ diverges So converges for $0⩽x<2$

*since. $\sum a_n{x}^n$ converges at $x=c\ne 0\Rightarrow$ converges absolutely for all $|x|<|c|$ diverges at $x=d\Rightarrow$ diverges for $|x|>|d|$

or $2^{\circ }$. $\sum a_n(x-1{)}^n$ converges at $x=0\Rightarrow \underset{n\to \mathrm{\infty }}{lim}{a}_n=0\Rightarrow \mathrm{\exists }\epsilon =\mid$ sit $\left|a_n\right|<1\mathrm{\forall }n>N$

$$\begin{array}{rl}& \Rightarrow |a_n(x-1{)}^n|<|x-1{|}^n\text{ for }\mathrm{\forall }|x-1|<1\\ & \text{ and }\sum (x-{1|}^n\text{ converges }\\ & {\Rightarrow \sum \left|a_n\right|x-1)}^n\mid \text{ converges for }(x-1\mid \ll \Rightarrow 0<x<2\\ & \Rightarrow \text{ converges at }0\le x<2\end{array}$$

3. If $f(x,y)$ has partial derivative's $f_x(x,y)$ and $f_y(x,y)$ ot $(x_0,y_0)$, then $\underset{x\to x_0}{lim}f(x,y_0)=\underset{y\to y_0}{lim}f(x_0,y)=f(x_0,y_0)$

$\stackrel{\text{ True }}{\to }$ Since $f_x(x_0,y_0)=\underset{x\to x_0}{lim}\frac{f(x_1,y_0)-f(x_0,y_0)}{x-x_0}=$, exist, and $\stackrel{x-x_0\to 0}{\Rightarrow }\underset{x\to x_0}{lim}f(x,y_0)=f(x_0,y_0)$

$f_y(x_0,y_0)=\underset{y\to y_0}{lim}\frac{f(x_0,y)-f(x_0,y_0)}{y-y_0}$ exist, and $\underset{y\to y_0}{lim}(y-y_0)=0\Rightarrow \underset{y\to y_0}{lim}f(x_0,y)=f(x_0,y_0)$

$\therefore \underset{x\to x_0}{lim}f(x,y_0)=\underset{y\to y_0}{lim}f(x_0,y)=f(x_0,y_0)$

(4) If a vector function $\overrightarrow{H}t$ ) is always perpendicular to its derivative $\frac{d\overrightarrow{F}}{dt}$ then $|\overrightarrow{r}(t)|$ must be constant

Thu e

$\stackrel{\text{ True. }}{\Rightarrow }$ since $\overrightarrow{F}(t)\cdot \frac{d\overrightarrow{r}}{dt}=0\Rightarrow (\overrightarrow{r(t)}\cdot \overrightarrow{H}(t){)}^{\prime }=2\overrightarrow{r}(t)\cdot \frac{d\overrightarrow{F}}{dt}=0$

$$\Rightarrow \mid \overrightarrow{Hf})\mid =c$$

(5) The curvative of a unit circle is greater than the curvature of the parabola $y=x^2$ at the origin

$\stackrel{\text{ False }}{\Rightarrow }$ For unit circe: $x^2+y^2=1\Rightarrow \{\begin{array}{l}x=\cos t\\ y=\text{ int. }\end{array}\Rightarrow \overrightarrow{r}(t)=(\cos t)\overrightarrow{t}+(\sin t)\overrightarrow{i}$

$\overrightarrow{T}=\overrightarrow{V}=(-\sin t)\overrightarrow{i}+(\cos t)\overrightarrow{j}$

$\left|\frac{d\overrightarrow{T}}{dt}\right|=1,|\overrightarrow{v}(t)|=1\Rightarrow k=1$

For $y=x^2\Rightarrow y^{\prime \prime }=2x,y^{\prime \prime }=2\Rightarrow y^{\prime }(0,0)=2$

$$k=\frac{|f^{\prime \prime }(x)|}{{\left(\sqrt{1+{(f^{\prime }(x))}^2}\right)}^3}=\frac{2}{1}=2$$

2. Multiple choice Questions: (only one correct answer for each of the following questions)

(i) Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two nonzero orthogonal vectors, which of the following must be the?

(A) $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|+|\overrightarrow{b}|$

(B) $|\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{a}|-|\overrightarrow{b}|$

(C) $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$

(D) $\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{a}-\overrightarrow{b}$

$\Rightarrow$ C. $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{|\overrightarrow{a}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2}=\sqrt{|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2},|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{|\overrightarrow{a}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2}=\sqrt{|\overrightarrow{a}{|}^2+\mid \overrightarrow{b}}$ $\therefore |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$

$F^{(2)}$ The equations of two lines are $4:x=t,y=2t,z=-t$ and $k:x=1-2t$ $y=t,z=-1+t$. Then 4 and $l_2$ are

(A) parallel ; (B) orthogonal ; (C) intersect with each offer: (D) skew ter $\stackrel{\Rightarrow }{\to }{\overrightarrow{v}}_1=(1,2,-1),\overrightarrow{v_2}=(-2,1,1)\therefore \overrightarrow{v_1}\cdot \overrightarrow{v_2}=-1,\left|\overrightarrow{v_1}\right|=\left|\overrightarrow{v_2}\right|=\sqrt{6}$ $\Rightarrow \cos \theta =-\frac{1}{6}$

but if it intersect $\{\begin{array}{l}x=t=1-2t\\ y=2t=t\end{array}\Rightarrow \{\begin{array}{l}t=\frac{1}{3}\\ t=0\end{array}x$

(3) Suppose $0⩽a_n<\frac{1}{n},(n=1,2,\dots )$, then which of the following series mast $\begin{array}{l}\text{ converges ? }\\ D(A)& \sum _{i=1}^{\mathrm{\infty }}{a}_n,& (B)\sum _{i=1}^{\mathrm{\infty }}(-1{)}^n{a}_n,& (C)\sum _{i=1}^{\mathrm{\infty }}\sqrt{a_n}\\ a_n& (D)\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n^2\text{. }\end{array}$

$\Rightarrow (A)\sum _n^{\mathrm{\infty }}<\frac{1}{n}\Rightarrow a_n^2<\frac{1}{n^2}\Rightarrow \sum a_n^2$ converges

$$\Rightarrow \mathrm{\Sigma }(-1{)}^n{a}_n^2\text{ converges. }$$

3. (9pts) Does the following series absolutely converge, conditionally converge, or diverge? Give reasons for your answer

(I) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{(1-\frac{1}{h})}^{h^2}$

$\Rightarrow ={\ln }_{n\to \mathrm{\infty }}\sqrt[n]{|an|}=\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{(-1{)}^n{(1-\frac{1}{n})}^{n^2}}=\underset{n\to \mathrm{\infty }}{lim}{(1-\frac{1}{n})}^n=e^{-1}<1$

$\Rightarrow$ converges absolutely

(2) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{e}^{-\frac{1}{n}}$

$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}{e}^{-\frac{1}{n}}=e^0=1\ne 0\Rightarrow$ diverges

(3) $\sum _{n=0019}^{\mathrm{\infty }}(-1{)}^n\frac{1}{\sqrt{n^2-2019n+1}}$

$\Rightarrow$ For $\sum _{n=019}^{\mathrm{\infty }}\frac{1}{\sqrt{n^2-2019n+1}},\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{\sqrt{n^2-2019n+1}}}{\frac{1}{n}}=1,\sum \frac{1}{n}$ diverges $\Rightarrow$ diverges

$u_n=\frac{1}{\sqrt{n^2-209n+1}}>0,u_n>u_{n+1},u_n\to 0\Rightarrow$ anverqes conditionally

4. 10 pts)

(i) Find the radius and interval of convergence of the series $\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{2}^n{x}^n}{\sqrt{n^2+n+1}}$

(2) For what values of $x$ does the series converge absolutely, or conditionally?

$\Rightarrow \left|\frac{u_{n+1}}{u_n}\right|=\frac{2^{n+1}\sqrt{n^2+n+1}}{2^n\sqrt{(n+1{)}^2+(n+1)+1}}|x|\to 2|x|<1\Rightarrow |x|<\frac{1}{2}\Rightarrow \frac{1}{1}$

(2) when $x=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{\sqrt{n^2+n+1}}$ converges conditionally since $\{\begin{array}{l}\frac{1}{\sqrt{n^2+n+1}}\ge 0\\ \frac{1}{n},\underset{n\to 0}{lim}\frac{1}{\sqrt{n^2+n+1}}0\end{array}$

$\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt{n^2+n+1}}{\frac{1}{n}}=1,\sum \frac{1}{n}$ diverges

$\Rightarrow \sum \frac{1}{\sqrt{n^2+n+1}}$ diverges

when $x=-\frac{1}{2},\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+}\cdot (-1{)}^n}{\sqrt{n^2+n+1}}=\sum _{n=0}^{\mathrm{\infty }}\frac{-1}{\sqrt{n^2+n+1}}$ diverges since $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n^2+n+1}}\frac{1}{n}=1,\sum \frac{1}{n}$ diverges

$\Rightarrow$ converges absolutely: $(-\frac{1}{2},\frac{1}{2})$, converges: $(-\frac{1}{2},\frac{1}{2}]$

5. (1 pts) converges conditionally: $x=\frac{1}{2}$

Let $x={\cos }^{3}t,y={\sin }^{3}t$, where $0⩽t⩽\frac{\pi }{2}$, be a parametrization of a curve

(i) Find the length of the curve

(2) Find the area of the surface generated by revolving the carve about th

$\Rightarrow$ (1) $\frac{dx}{dt}=-3{\cos }^{2}t\sin t,\frac{dy}{dt}=3{\sin }^{2}t\cos t\Rightarrow {\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=9{\sin }^{2}t{\cos }^{2}t$ $x$ axis

$$\begin{array}{rl}L& ={\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt={\int }_0^{\frac{\pi }{2}}\sqrt{9{\sin }^{2}t{\cos }^{2}t}dt={\int }_0^{\frac{\pi }{2}}3{\sin }^{2}t\cos tdt={\int }_0^{\frac{\pi }{2}}3\sin td\sin t\\ & {=\frac{3}{2}{\sin }^{2}t]}_0^{\frac{\pi }{2}}=\frac{3}{2}\end{array}$$

(2) $L={\int }_0^{\frac{\pi }{2}}2\pi y\sqrt{{\left(\frac{d(x)}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt={\int }_0^{\frac{\pi }{2}}2\pi {\sin }^{3}t\cdot 3\sin t\cos tdt={\int }_0^{\frac{\pi }{2}}6\pi {\sin }^{4}td\sin t$

${=\frac{6}{5}\pi {\sin }^{5}t]}_0^{\frac{\pi }{2}}=\frac{6}{5}\pi$

6.c10pts) Find the equation of the plane trough the points $(2,-1,-1)$ and $(1,0,-7$ perpendicular to the plane $2x+3y-5z+6=0$

$\Rightarrow$ let $P(2,-1,-1),Q(1,0,-1)\Rightarrow \overrightarrow{PQ}=(-1,1,0),\overrightarrow{n_1}=(2,3,-5)$

$$\begin{array}{rl}\Rightarrow \overrightarrow{k}=\overrightarrow{pa}\times \overrightarrow{n}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{i}& \overrightarrow{k}\\ -1& 1& 0\\ 2& 3& -5\end{array}\right|=-5\overrightarrow{i}-5\overrightarrow{j}-5\overrightarrow{k}& \Rightarrow \text{ the plane: }-5(x-2)-5(x+1)-5(x+\\ & \Rightarrow x+y+z=0\end{array}$$

k 7. ( 10 pts ) A particle is located at the point $(1,0,-2)$. Its initial speed $F$ is $|\overrightarrow{v}(0)|=3$ at time $t=0$, and the direction of its initial velocity is toward the point $(2,-1,3)$. The particle moves with constant acceleration $\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$. Find its position vector $\overrightarrow{r}(t)$ at time $t$

$\Rightarrow$ let $P(1,0,-2),Q(2,-1,3)\Rightarrow \overrightarrow{P}=(1,-1,5),\overrightarrow{r_0})=\overrightarrow{i}-2\overrightarrow{k}$

the direction of $\overrightarrow{v(0)}$ is $\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{1}{\sqrt{27}}(1,-1,5)$

So $\overrightarrow{v(\theta )}=|\overrightarrow{v(0)}|\cdot \frac{\overrightarrow{\beta \mathrm{\infty }}}{|\overrightarrow{PQ}|}=3\cdot \frac{1}{\sqrt[3]{3}}(1,-1,5)=\frac{1}{\sqrt{3}}(1,-1,5)$

$$\begin{array}{rl}\overrightarrow{a}(t)& =(\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}\\ \Rightarrow \overrightarrow{V(t)}& =\int (\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})dt=t\overrightarrow{i}+2t\overrightarrow{j}+t\overrightarrow{k}+\overrightarrow{c}\\ \overrightarrow{V(0)}& =\frac{1}{\sqrt{3}}\overrightarrow{i}-\frac{1}{\sqrt{3}}\overrightarrow{j}+\frac{5}{\sqrt{3}}\overrightarrow{k}\\ \Rightarrow \overrightarrow{c}& =\frac{1}{\sqrt{3}}\overrightarrow{i}-\frac{1}{\sqrt{3}}\overrightarrow{j}+\frac{5}{\sqrt{3}}\overrightarrow{k}\\ \Rightarrow \overrightarrow{V(t)}& =(t+\frac{1}{\sqrt{3}})\overrightarrow{i}+(2t-\frac{1}{\sqrt{3}})\overrightarrow{j}+(t+\frac{5}{\sqrt{3}})\overrightarrow{k}\\ \overrightarrow{r(t)}& =\int [(t+\frac{1}{\sqrt{3}})\overrightarrow{i}+(2t-\frac{1}{\sqrt{3}})\overrightarrow{j}+(t+\frac{5}{\sqrt{3}})\overrightarrow{k}]dt\\ & =(\frac{1}{2}{t}^2+\frac{1}{\sqrt{3}}t)\overrightarrow{i}+(t^2-\frac{1}{\sqrt{3}}t)\overrightarrow{j}+(\frac{1}{2}{t}^2+\frac{5}{\sqrt{3}}t)\overrightarrow{k}+\overrightarrow{c}\\ \overrightarrow{V(0)}& =\overrightarrow{i}-2\overrightarrow{k}\Rightarrow \overrightarrow{c}=\overrightarrow{r(0)}=\overrightarrow{i}-2\overrightarrow{k}\\ \overrightarrow{F(t)}& =(\frac{1}{2}{t}^2+\frac{1}{\sqrt{3}}t+1)\overrightarrow{i}+(t^2-\frac{1}{\sqrt{3}}t)\overrightarrow{j}+(\frac{1}{2}{t}^2+\frac{5}{\sqrt{3}}t-2)\overrightarrow{k}\end{array}$$

8. ( 10 pts )

Is the following function, $f(x,y)$ continuous at $(0,0)$ ? Give reasons for your answer. $f(x,y)=\{\begin{array}{ll}\frac{{\sin }^3+y^3)}{x^2+y^2}& ,(x,y)\ne (0,0)\\ 0& ,(x,y)=(0,0)\end{array}$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}f(x,y)=\underset{(x,y)\to (0,0)}{lim}\frac{\sin (x^3+y^3)}{x^2+y^2}=\underset{(x,y)\to (0,0)}{lim}\frac{\sin (x^3+y^3)}{x^3+y^3}\cdot \frac{x^3+y^3}{x^2+y^2}=\underset{(x,y)\to (0,0)}{lim}\frac{x^3+y^3}{x^2+y^2}$

$$=\underset{r\to 0}{lim}\frac{r^3{\cos }^3\theta +r^2{\sin }^3\theta }{r^2}=\underset{r\to 0}{lim}r({\cos }^3\theta +{\sin }^3\theta )=0$$

or $0⩽\left|\frac{x^3+y^3}{x^2+y^2}\right|0=|\frac{x\cdot x^2}{x^2+y^2}+\frac{y\cdot y^2}{x^2+y^2}|=|x\cdot \frac{x^2}{x^2+y^2}+y\cdot \frac{y^2}{0x^2+y^2}|⩽|x|+|y|$

$$\underset{(x,y)\to (0,0)}{lim}(|x|+|y|)=0$$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}\left|\frac{x^3+y^3}{x^2+y^2}\right|=0\Rightarrow \underset{(x,y)\to (0,0)}{lim}\frac{x^3+y^3}{x^2+y^2}=0$

$\Rightarrow \underset{f(x,y)}{lim}f(x)=0=f(0,0)$

$(x,y)\to (0,0)$

$f(x,y)$ is continuous at $(0,0)$

9.(9pts) Let $f(u)$ be differentiable, $z=f\left(e^{x}y\right)(y\ne 0)$, and $\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=1$

If $f(1)=0$, find $f(u)$

$\Rightarrow \frac{\partial z}{\partial x}=f^{\prime }\left(e^{x}y\right)\cdot e^{x}y,\frac{\partial z}{\partial y}=f^{\prime }\left(e^{x}y\right)e^x$

$\Rightarrow \frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2f^{\prime }\left(e^{x}y\right)e^{x}y=1$, let $u=e^{x}y$

$$\Rightarrow 2f^{\prime }(u)u=1\Rightarrow \frac{df}{du}=\frac{1}{2u}\Rightarrow f(u)=\frac{1}{2}\ln |u|+c$$$$\text{ or }=\frac{1}{2}\ln |2u|+$$$$\begin{array}{l}\Rightarrow f(u)=\frac{1}{2}\ln |u|+c\\ f(1)=c=0\Rightarrow f(u)=\frac{1}{2}\ln |u||f(u)=\frac{1}{2}|n|,\end{array}$$$$f(1)=\frac{1}{2}\ln 2+c_1=0$$

$10.(8\mathrm{pts})$

Find the Taylor series for $f(x)=\ln (x+\sqrt{x^2+1})$ at $x=0$

$$\begin{array}{rl}& \Rightarrow f^{\prime }(x)=\frac{1+\frac{x}{\sqrt{2^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}}={(1+x^2)}^{-\frac{1}{2}}=1+\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-\frac{1}{-2}}{k})x^{2k},-1<x<1\\ & \Rightarrow f(x)=x+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k+1}(\genfrac{}{}{0ex}{}{-\frac{1}{2}}{k})x^{2k+1},-1<x<1\end{array}$$