MA127-2018春-期中-答案

SUSTech

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

1

Determine whether the following statements are true or false? No justification is necessary

1-1

F (a) If both $f_x(x,y)$ and $f_y(x,y)$ exist at $(x_0,y_0)$, then $f(x,y)$ is continuous at $(x_0,y_0)$

1-2

F (b) Let

$$f(x,y)=\{\begin{array}{ll}\frac{xy}{x^2+y^2},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}$$

At the point $(0,0),f(x,y)$ is continuous. [Along $y=x\mathrm{\&}y=2x,f(x,y)$ as different

1-3

T (c) For the $f(x,y)$ as in $(1)$, both $f_x(0,0)$ and $f_y(0,0)$ exist. limits as $(x,y)\to (0.0)$ ]

1-4

T (d) Nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel if and only if $\mathbf{u}\times \mathbf{v}=\mathbf{0}$

1-5

T (e) The surface $y^2-x^2=z$ is a hyperbolic paraboloid

1-6

F (f) If $f(x,y)$ and its partial derivatives $f_x,f_y,f_{xy}$, and $f_{yx}$ are defined throughout an open region containing a point $(a,b)$, then $f_{xy}(a,b)=$ $f_{yx}(a,b)$

2

Suppose that the function $f(x,y)$ is differentiable, and $f(0,0)=1$, $f_x(0,0)=2,f_y(0,0)=3$. Then $f(x,y)\approx 1+2x+3y$ when both $x$ and $y$ are small (using the standard linear approximation at $(0,0)$ )

3

Find the distance from the point $(1,1,5)$ to the line

$$L:x=1+t,y=3-t,z=2t$$

$$\text{ Sud: }\begin{array}{rl}d& =|\overrightarrow{QP}|\cdot \sin \theta \\ & |\overrightarrow{QP}\times \overrightarrow{V}|=|\overrightarrow{QP}|\cdot |\overrightarrow{v}|\cdot \sin \theta \\ \therefore & d=\frac{|\overrightarrow{QP}\times \overrightarrow{v}|}{|\overrightarrow{V}|}\end{array}$$

4

Find the length of the curve

from $(0,0,1)$ to $(\sqrt{2},\sqrt{2},0)$

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

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

$$\text{begin{aligned} \& r^{prime}(t)=sqrt{2} i+sqrt{2} j-(2 t) k, quadleft|r^{prime}(t)right|=sqrt{2+2+4 t^2}=2 sqrt{1+t^2} \& L=int\_0^{1}left|r^{prime}(t)right| d t=int\_0^1 2 sqrt{1+t^2} d t=left[t sqrt{1+t^2}+ln left(t+sqrt{1+t^2}right]\_0^{1}right end{aligned}}$$$$\begin{array}{c}(2\int \sqrt{1+t^2}dt\stackrel{t=\tan \theta }{=}2\int \sqrt{1+{\tan }^2\theta }{\sec }^2\theta d\theta =\sqrt{2}+\ln (1+\sqrt{2})\\ dt={\sec }^2\theta d\theta =\frac{\pi }{2}=2\int {\sec }^3\theta d\theta =\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |+c\end{array}$$

5

Find the normal vector and the curvature for the helix

$$\mathbf{r}(t)=(a\cos t)\mathbf{i}+(a\sin t)\mathbf{j}+(bt)\mathbf{k},a,b\ge 0,a^2+b^2\ne 0$$

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 cost, $b⟩$

$\frac{dT}{dt}=\frac{1}{\sqrt{a^2+b^2}}⟨-a\cos t,-a\sin t,0⟩,k=\left|\frac{dT}{dt}\right|\frac{1}{|v|}=\frac{a}{a^{2}y}\underset{―}{a^2+b^2}\underset{―}{N}=\frac{dT}{dt}/\left|\frac{dT}{dt}\right|=⟨-\cos t$,

6

Find $\underset{(x,y)\to (0,0)}{lim}\frac{x^{2}y}{x^4+y^2}$, if it exists; otherwise give the reason why the limit does not exist

sol: Along pathes $y=k{x}^2,k\ne 0:f(x,y)=\frac{x^2\left(k{x}^2\right)}{x^4+k^2{x}^4}=\frac{k}{1+k^2}$

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

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

7

Find $\frac{\partial w}{\partial v}$ when $u=-1,v=2$, if $w=xy+\ln z,x=\frac{v^2}{u},y=u+v$,

$$\text{ sol: }\begin{array}{rl}\frac{\partial w}{\partial v}& =\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial v}\\ & =y\left(\frac{2v}{u}\right)+x\text{ when }u=-1,v=2\\ & =(u+v)\left(\frac{2v}{u}\right)+\frac{v^2}{u}\frac{\partial w}{\partial v}=-4+(-4)=-8\end{array}$$

8. (10 pts) Find the critical points of the function $f(x,y)=x^4+y^4+4xy$, and use the second derivative test to classify each point as one where a saddle, local maximum or local minimum occurs

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

9. (10 pts) Find the point on the surface $z^2=xy+4$ closest to the origin

Sol: Minimize $f(x,y,z)=x^2+y^2+z^2$ s.t. $z^2=xy+4$

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

$⩾|xy|+4⩾4$ 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 $(0,0,0)$

10. (10 pts) Use Taylor's formula for $f(x,y)=x{e}^y$ at the origin to find quadratic

and cubic approximations of $f$ near the origin

#9. By Lagrampe Multiplier: $g(x,y,z)=xy+4-z^2=0;\mathrm{\#}10.f_x=e^y,f_y=x{e}^y,f_{xx}=0,f_{xy}=e^y,f_{yy}=x{e}^y$

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

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

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

Case 2. ${\lambda }^{\ne -1},z=0=x=2=-y$ or $x=-2=-y$ $f(0,0,\pm 2)=4,f(\pm 2,\pm 2,0)=8$ closest to the origin $f(x,y)\approx x+xy+\frac{1}{2}x{y}^2$

quadratic oyer.:

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

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

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

Cubic appr. $=$ At $(0,0)=f_{xxx}=0,f_{xxy}=0,f_{xyy}={e^y|}_{y=0}=1$ $fyyy={x{e}^y|}_{x=0}=0$

$\begin{array}{rl}& \approx f(x,y)\approx \\ & +q^2\text{ advatie }+\frac{1}{6}[x^3{f}_{xxx}(0,0)+3x^{2}y{f}_{xxy}(0,0)+3x{y}^2{f}_{xy}(0,0)\end{array}$