MA127-2018春 期中考试答案

SUSTech Midterm II for Calculus II in Spring Semester, 2018 (Solutions)

一

判断下列命题是否正确，无需给出理由。

• F
• F
• Y
• Y
• Y
• F

二

$f(x,y)\approx 1+2x+3y$

三

$d=|\overrightarrow{QP}|\cdot \sin \theta$

$|\overrightarrow{QP}\times \overrightarrow{V}|=|\overrightarrow{QP}|\cdot |\overrightarrow{v}|\cdot \sin \theta$

$\therefore d=\frac{|\overrightarrow{QP}\times \overrightarrow{v}|}{|\overrightarrow{V}|}$

$$\begin{array}{rl}& P=(1,1,5),Q=(1,3,0)(t=0)\\ & \overrightarrow{QP}=⟨1,1,5⟩-⟨1,3,0⟩\\ & =⟨0,-2,5⟩\\ & \overrightarrow{V}=⟨1,-1,2⟩\\ & \overrightarrow{QP}\times \overrightarrow{v}=|\begin{array}{ccc}i& j& k\\ 0& -2& 5\\ 1& -1& 2\end{array}|=i+5j+2k\\ & d=\sqrt{1+5^2+2^2}/\sqrt{1+1+2^2}=\sqrt{5}\end{array}$$

四

Sol:

$\because r(0)=⟨0,0,1⟩,r(1)=⟨\sqrt{2},\sqrt{2},0⟩,\therefore 0\le t\le 1$

$r^{\prime }(t)=\sqrt{2}i+\sqrt{2}j-(2t)k$

$|r^{\prime }(t)|=\sqrt{2+2+4t^2}=2\sqrt{1+t^2}$

$L={\int }_0^1|r^{\prime }(t)|dt$

$={\int }_0^{1}2\sqrt{1+t^2}dt$

$=[t\sqrt{1+t^2}+\ln (t+\sqrt{1+t^2}){]}_0^1$

$=\sqrt{2}+\ln (1+\sqrt{2})$

五

Sol:

$r^{\prime }(t)=(-a\sin t)j+(a\cos t)j+bk=v(t)$

$$|r^{\prime }(t)|=\sqrt{a^2+b^2}=|v(t)|$$

Unit tangent vector: $T=\frac{r^{\prime }(t)}{|r^{\prime }(t)|}=\frac{1}{\sqrt{a^2+b^2}}⟨-a\sin t$, $a\cos t$, $b⟩$

$\frac{dT}{dt}=\frac{1}{\sqrt{a^2+b^2}}⟨-a\cos t,-a\sin t,0⟩$

$k=|\frac{dT}{dt}|\frac{1}{|v|}=\frac{a}{a^2+b^2}$

$\underset{―}{N}=\frac{\frac{dT}{dt}}{|\frac{dT}{dt}|}=⟨-\cos t,-\sin t,0⟩$,

六

Sol:

Along pathes $y=k{x}^2,k\ne 0$

$f(x,y)=\frac{x^2(k{x}^2)}{x^4+k^2{x}^4}=\frac{k}{1+k^2}$

$$\therefore \underset{(x,y)\to (0,0)}{lim}f(x,y)=\underset{x\to 0}{lim}\frac{k}{1+k^2}=\frac{k}{1+k^2}$$

$\therefore$ By the two -path Test for nonexistence of a limit, we see the limit DNES

七

$\frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial v}$

$=y(\frac{2v}{u})+x$

$=(u+v)(\frac{2v}{u})+\frac{v^2}{u}$

when $u=-1,v=2$

$\frac{\partial w}{\partial v}=-4+(-4)=-8$

八

Sol:

• $f_x=4x^3+4y$
• $f_y=4y^3+4x$

Critical pts:

$$\{\begin{array}{l}4x^3+4y=0\\ 4y^3+4x=0\end{array},f_{xy}=4$$$$\{\begin{array}{l}x=0\\ y=0\end{array}\{\begin{array}{l}x=1\\ y=-1\end{array}\{\begin{array}{l}x=-1\\ y=1\end{array}$$$$\{\begin{array}{l}{f}_{xx}=12x^2\\ f_{yy}=12y^2\\ f_{xy}=4\end{array}$$$$H=\left|\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{xy}& f_{yy}\end{array}\right|=144x^2{y}^2-16$$

$(0,0):H=-16<0$ saddle

$(1,-1):H>0,f_{xx}>0$ local min

$(-1,1):H>0,f_{xx}>0$ local min

九

Sol:

Minimize $f(x,y,z)=x^2+y^2+z^2$

s.t. $z^2=xy+4$

substitute $f(x,y)=x^2+y^2+xy+4,f_x=2x+y,f_y=2y+x$

critical pt. $2x+y=2y+x=0\Rightarrow x=y=0\Rightarrow z^2=4,z=\pm 2$

$\therefore$ The pts $(0,0,\pm 2)$ are the closest on $z^2=xy+4$ to $(0,0,0)$

---

By Lagrampe Multiplier: $g(x,y,z)=xy+4-z^2=0$

$\mathrm{\nabla }f=⟨2x,2y,2z⟩,\mathrm{\nabla }g=⟨y,x,-2z⟩$

$\mathrm{\nabla }f=\lambda \mathrm{\nabla }g\Rightarrow 2x=\lambda y,2y=\lambda x,2z=-2\lambda z$

case 1. $\lambda =-1:z=\pm 2,x=y=0$

Case 2. $\lambda \ne -1,z\ne 0:x=2=-y$ or $x=-2=-y$ $f(0,0,\pm 2)=4,f(\pm 2,\mp 2,0)=8$

$(0,0,\pm 2)$ closest to the origin

十

$f_x=e^y,f_y=x{e}^y,f_{xx}=0,f_{xy}=e^y,f_{yy}=x{e}^y$

quadratic appr.

$f(x,y)\approx f(0,0)+x{f}_x(0,0)+y{f}_y(0,0)$

$+\frac{1}{2}[x^2{f}_{xx}(0,0)+2xy{f}_{xy}(0,0)+y^2{f}_{yy}(0,0)]$

$=0+x\cdot 1+y\cdot 0+\frac{1}{2}[x^2\cdot 0+2xy\cdot 1+y^2\cdot 0]$

Cubic appr.

At $(0,0):f_{xxx}=0,f_{xxy}=0,f_{xyy}=e^y{|}_{y=0}=1$ $f_{yyy}=x{e}^y{|}_{x=0}=0$

$f(x,y)\approx$ quadratic appr $+\frac{1}{6}[x^3{f}_{xxx}(0,0)+3x^{2}y{f}_{xxy}(0,0)+3x{y}^2{f}_{xy}(0,0)+y^3{f}_{yyy}(0,0)]$

$f(x,y)\approx x+xy+\frac{1}{2}x{y}^2$