다변수 미분 전체 문제

True, by Clairaut’s theorem

${h}_{xx}(x,y,z)=\frac{6x{e}^{2y}}{z},$ ${h}_{xy}(x,y,z)=\frac{6{x}^{2}{e}^{2y}}{z},$ ${h}_{xz}(x,y,z)=-\frac{3{x}^{2}{e}^{2y}}{{z}^{2}},$ ${h}_{yx}(x,y,z)=\frac{6{x}^{2}{e}^{2y}}{z},$ ${h}_{yy}(x,y,z)=\frac{4{x}^{3}{e}^{2y}}{z},$ ${h}_{yz}(x,y,z)=-\frac{2{x}^{3}{e}^{2y}}{{z}^{2}},$ ${h}_{zx}(x,y,z)=-\frac{3{x}^{2}{e}^{2y}}{{z}^{2}},$ ${h}_{zy}(x,y,z)=-\frac{2{x}^{3}{e}^{2y}}{{z}^{2}},$ ${h}_{zz}(x,y,z)=\frac{2{x}^{3}{e}^{2y}}{{z}^{3}}$

$2.3228$ cm³

The total differential of volume is $dV = 4\pi dr + \pi dh$. The coefficient of $dr$ is greater than the coefficient of $dh$, so the volume is more sensitive to changes in the radius.

Distance of the projectile is a function of two variables (leaving $t=3$): $D(v_0,\theta) = 3v_0\cos\theta$. The total differential of $D$ is $dD = 3\cos\theta dv_0-3v_0\sin\theta d\theta$. The coefficient of $d\theta$ has a much greater magnitude than the coefficient of $dv_0$, so a small change in the angle of elevation has a much greater effect on distance traveled than a small change in initial velocity.

One critical point at $(-4,2)$; $f_{xx} = 1$ and $D = 4$, so this point corresponds to a relative minimum.

One critical point at $(7,-6)$; $D = -5$, so this point corresponds to a saddle point.

One critical point at $(6,-3)$; $D = -4$, so this point corresponds to a saddle point.

One critical point at $(0,0)$; $f_{xx} = -2$ and $D = 4$, so this point corresponds to a relative maximum.

Two critical points: at $(0,-1)$; $f_{xx} = 2$ and $D = -12$, so this point corresponds to a saddle point;

at $(0,1)$, $f_{xx} = 2$ and $D = 12$, so this corresponds to a relative minimum.

There are 4 critical points:

$(-1,-2)$, rel. max; $(1,-2)$, saddle point;

$(-1,2)$, saddle point; $(1,2)$, rel. min.,

where $f_{xx} = 2x$ and $D = 4xy$.

There are infinite critical points, whenever $x=0$ or $y=0$. With $D = -12x^2y^2$, at each critical point $D = 0$ and the test is inconclusive. (Some elementary thought shows that each is an absolute minimum.)

Six critical points: $f_x = 0$ when $x=-1,0$ and 1; $f_y = 0$ when $y = -3,3$. Together, we get the points:

$(-1,-3)$ saddle point; $(-1,3)$ rel. min

$(0,-3)$ rel. max; $(0,3)$ saddle point

$(1,-3)$ saddle point; $(1,3)$ relative min

where $f_{xx} = 12x^2-4$ and $D = 24y(3x^2-1)$.

One critical point: $f_x = 0$ when $x=3$; $f_y = 0$ when $y = 0$, so one critical point at $(3,0)$, which is a relative maximum, where $f_{xx} = \frac{y^2-16}{(16-(x-3)^2-y^2)^{3/2}}$ and $D = \frac{16}{(16-(x-3)^2-y^2)^{2}}$.

Both $f_x$ and $f_y$ are undefined along the circle $(x-3)^2+y^2=16$; at any point along this curve, $f(x,y)=0$, the absolute minimum of the function.

One critical point: $f_x = 0$ when $x=0$; $f_y = 0$ when $y = 0$, so one critical point at $(0,0)$ (although it should be noted that at $(0,0)$, both $f_x$ and $f_y$ are undefined.) The Second Derivative Test fails at $(0,0)$, with $D=0$. A graph, or simple calculation, shows that $(0,0)$ is the absolute minimum of $f$.

Let $f(x,y)=\sqrt{x^2+5y^2}$. Then

$$\begin{aligned} f_x&=\frac{x}{\sqrt{x^2+5y^2}} & f_{xx}&=\frac{1}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(x)(2x)}{(x^2+5y^2)^{3/2}} & f_{xy}&=-\frac{1}{2}\frac{(x)(10y)}{(x^2+5y^2)^{3/2}} \\ f_y&=\frac{5y}{\sqrt{x^2+5y^2}} & f_{yy}&=\frac{5}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(5y)(10y)}{(x^2+5y^2)^{3/2}}& f_{yx}&=-\frac{1}{2}\frac{(5y)(2x)}{(x^2+5y^2)^{3/2}} \end{aligned}$$

Simplifying, and in particular using that $\frac{1}{\sqrt{x^2+5y^2}} =\frac{x^2+5y^2}{(x^2+5y^2)^{3/2}}$,

$$f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}}\qquad f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}\qquad f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}}$$

(a) The first order derivatives are

$$f_r(r,\theta)=mr^{m-1}\cos m\theta\qquad f_\theta(r,\theta)=-mr^m\sin m\theta$$

The second order derivatives are

$$f_{rr}(r,\theta)=m(m-1)r^{m-2}\cos m\theta\quad f_{r\theta}(r,\theta)=-m^2r^{m-1}\sin m\theta\quad f_{\theta\theta}(r,\theta)=-m^2r^m\cos m\theta$$

so that

$$f_{rr}(1,0)=m(m-1),\ f_{r\theta}(1,0)=0,\ f_{\theta\theta}(1,0)=-m^2$$

(b) By part (a), the expression

$$f_{rr}+\frac{\lambda }{r}f_r+\frac{1}{r^2}f_{\theta\theta} =m(m-1)r^{m-2}\cos m\theta+\lambda mr^{m-2}\cos m\theta-m^2r^{m-2}\cos m\theta$$

vanishes for all $r$ and $\theta$ if and only if

$$m(m-1)+\lambda m-m^2=0\iff m(\lambda -1)=0\iff \lambda =1$$