다변수 미분 전체 문제

(a) The specified graph is $z=f(x,y)=x^2-y^2$ or $F(x,y,z)=x^2-y^2-z=0$. Observe that $f(-2,1)=3$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(-2,1,3) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< 2x,-2y,-1\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< -4,-2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(-2,1,3)$. So the tangent plane is

$$-4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3$$

and the normal line is

$$\left< x,y,z\right>=\left< -2,1,3\right>+t\left< 4,2,1\right>$$

(b) The specified graph is $z=f(x,y)=e^{xy}$ or $F(x,y,z)=e^{xy}-z=0$. Observe that $f(2,0)=1$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(2,0,1) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< ye^{xy},xe^{xy},-1\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< 0,2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(2,0,1)$. So the tangent plane is

$$0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1$$

and the normal line is

$$\left< x,y,z\right>=\left< 2,0,1\right>+t\left< 0,2,-1\right>$$

(a) The shoreline is $f(x,y)=0$ or $x^2+4x+4y^2=32$ or $(x+2)^2+4y^2=36$, which is an ellipse centred on $(-2,0)$ with semiaxes $6$ in the $x$-direction and $3$ in the $y$-direction.

(b,c) The gradient of $f$ at $(-1,1)$ is

$${ \boldsymbol{\nabla} } f(-1,1)=\big[(-2x-4)\,\hat{\boldsymbol{\imath}}-8y\,\hat{\boldsymbol{\jmath}}\big]_{(-1,1)}=-2\,\hat{\boldsymbol{\imath}}-8\,\hat{\boldsymbol{\jmath}}$$

To remain at constant depth, he should swim perpendicular to the depth gradient. So he should swim in direction $\pm\frac{1}{\sqrt{17}}\left< 4,-1\right>$. To increase his depth as rapidly as possible, he should swim in the direction of the depth gradient, which is $-\frac{1}{\sqrt{17}}\left< 1,4\right>$.

Both of the functions $f(x, y) = \sqrt{x^2 + y^2}$ (i.e. (ii)) and $f(x, y) = x^2 + y^2$ (i.e. (iv)) are invariant under rotations around the $(0,0)$. So their level curves are circles centred on the origin. In polar coordinates $\sqrt{x^2 + y^2}$ is $r$. So the sketched level curves of the function in (ii) are $r = 0, 0.1, 0.2, \ldots , 1.9, 2$. They are equally spaced. So at this point, we know that the third picture goes with (iv) and the fourth picture goes with (ii).

Notice that the lines $x=y$, $x=-y$ and $y=0$ are all level curves of the function $f(x, y) = y(x + y)(x - y) + 1$ (i.e. of (iii)) with $f=1$. So the first picture goes with (iii). And the second picture goes with (i).

Here are the pictures with critical points marked on them. There are saddle points where level curves cross and there are local max's or min's at “bull's eyes”.

(i)

(ii)

(iii)

(iv)