§14.1 다변수함수Functions of Several Variables

Solution. (a) The quantity under the square root must be nonnegative, so $x+y+1\ge0$, and the denominator must not vanish, so $x\ne1$.

Thus $D$ consists of every point on or above the line $x+y+1=0$ (equivalently $y=-x-1$) with the vertical line $x=1$ deleted, exactly the shaded region and dashed line in the first figure above. Since $3+2+1=6\ge0$ and $3\ne1$, the point $(3,2)$ lies in $D$, and

(b) The argument of the logarithm must be positive, so $y^2-x\gt0$, i.e. $x\lt y^2$.

This is the open region lying to the left of the parabola $x=y^2$, with the parabola itself excluded, as the second figure above shows. At $(3,2)$ we have $y^2-x=4-3=1\gt0$, so the point lies in $D$, and

Hence $f(3,2)=\dfrac{\sqrt6}{2}$ in (a) and $f(3,2)=0$ in (b).

Solution. The expression under the radical must be nonnegative, so $9-x^2-y^2\ge0$, that is, $x^2+y^2\le9$.

the closed disk of radius $3$ centered at the origin, shown as the shaded region above. On $D$ the quantity $x^2+y^2$ runs over $[0,9]$, so $9-x^2-y^2$ also runs over $[0,9]$ and its square root runs over $[0,3]$, attaining $3$ at the origin and $0$ on the boundary circle $x^2+y^2=9$.

Hence the range is $[0,3]$.

Solution. Reading Table 1 at the row $T=-5$ and the column $v=50$ gives $f(-5,50)=-15$.

This means that when the actual air temperature is $-5^\circ\mathrm{C}$ and the wind is blowing at $50\ \mathrm{km/h}$, the cold feels the same as it would with no wind at a temperature of $-15^\circ\mathrm{C}$.

Solution. Substituting $L=147$, $K=208$ gives

close to the recorded value $159$. Substituting $L=194$, $K=407$ gives

close to the recorded value $231$. Since labor and capital cannot be negative, the domain of this production function is

Solution. Writing $z=6-3x-2y$ with all three variables together gives $3x+2y+z=6$, the same form as the equation of a plane from §12.5.

Setting two variables to zero at a time gives the intercepts $x=2$ (when $y=z=0$), $y=3$ (when $x=z=0$), and $z=6$ (when $x=y=0$), so the plane meets the axes at $(2,0,0)$, $(0,3,0)$, $(0,0,6)$. The domain of $f$ is all of $\mathbb R^2$, so the full graph is the entire plane, but by convention only the triangular first-octant piece bounded by these three intercepts, shown above, is drawn.

Solution. The graph consists of the points satisfying $z=\sqrt{9-x^2-y^2}$. Squaring both sides gives $z^2=9-x^2-y^2$, that is,

the equation of the sphere of radius $3$ centered at the origin. Since a square root is never negative, every point of the graph must also satisfy $z\ge0$.

Hence the graph of $g$ is not the full sphere but only its upper half, the hemisphere with $z\ge0$.

Solution. Fixing $L$ and letting $K$ range over $[0,300]$ gives a vertical trace of $P=1.01L^{0.75}K^{0.25}$; since the exponent $0.25$ is positive, $P$ increases as $K$ increases. Likewise, fixing $K$ and letting $L$ vary gives a vertical trace that increases with $L$, because the exponent $0.75$ is also positive.

So over $0\le L,K\le300$ the surface increases monotonically in both directions: it starts at $P=0$ along the edges $L=0$ and $K=0$ and climbs toward larger values as $(L,K)$ moves toward the corner $(300,300)$.

Solution. Since $4x^2+y^2$ is defined for every real $x,y$, the domain is $D=\mathbb R^2$. As $4x^2\ge0$ and $y^2\ge0$, with both terms vanishing only at $x=y=0$, and the expression is unbounded above, the range is $[0,\infty)$.

A horizontal trace at a fixed level $z=k\gt0$ is $4x^2+y^2=k$, that is, $\dfrac{x^2}{k/4}+\dfrac{y^2}{k}=1$, an ellipse; a vertical trace, say at fixed $y=b$, is $z=4x^2+b^2$, a parabola.

Hence the graph $z=4x^2+y^2$ is an elliptic paraboloid opening upward with vertex at the origin.

Solution. In the map above, the point $(1,3)$ lies between the level curves $70$ and $80$, positioned somewhat closer to the $70$-curve than exactly halfway. The plain average of the two labels is

but since the point actually sits closer to the $70$-side, interpolating its position gives the estimate $f(1,3)\approx73$. The point $(4,5)$ lies between the level curves $50$ and $60$, closer to the $60$-curve, so the same kind of interpolation gives $f(4,5)\approx56$.

Solution. The level curve $f(x,y)=k$ is $6-3x-2y=k$, that is,

a family of lines all with the same slope $-\dfrac32$. Substituting $k=-6,0,6,12$ gives

four mutually parallel lines. As $k$ increases in equal steps of $6$, the constant term takes the equally spaced values $-12,-6,0,6$, so the perpendicular spacing between consecutive lines is also constant. Since $f$ is linear, equally spaced values of $k$ always produce a family of equally spaced parallel lines.

Solution. Squaring $g(x,y)=k$ (with $k\ge0$) gives $9-x^2-y^2=k^2$, that is,

a circle centered at the origin with radius $\sqrt{9-k^2}$, which exists precisely when $9-k^2\ge0$, i.e. $0\le k\le3$. For $k=0,1,2,3$ the radii are

respectively, so as shown above these curves form a family of concentric circles shrinking as $k$ increases, and at $k=3$ the circle degenerates to the single point at the origin.

Solution. The level curve $h(x,y)=k$ is $4x^2+y^2+1=k$, that is, $4x^2+y^2=k-1$, which describes an actual curve only when $k\gt1$. Dividing by $k-1$ puts it in standard form:

an ellipse with semi-axis $\dfrac12\sqrt{k-1}$ in the $x$-direction and $\sqrt{k-1}$ in the $y$-direction. As $k$ increases both semi-axes grow together, so, for instance, the curves for $k=2,5,10$ form a family of ellipses centered at the origin and growing outward.

Lifting each such ellipse to the height $z=k$ reproduces exactly the surface $z=4x^2+y^2+1$, an elliptic paraboloid — the surface of Example 8 shifted up by $1$ along the $z$-axis.

Solution. The level curve of $P(L,K)=1.01L^{0.75}K^{0.25}$ at a fixed output level $c$ is $1.01L^{0.75}K^{0.25}=c$; solving for $K$ gives

Since $L$ appears with the negative exponent $-3$, increasing $L$ while holding $c$ fixed forces $K$ to decrease.

So the contour plot consists of decreasing curves, with the output level $c$ increasing away from the origin, and each curve reflects a trade-off: to keep production constant while using more labor, less capital must be used.

Solution. The term $xy\sin z$ is defined for every $(x,y,z)$, but the term $\ln(z-y)$ requires a positive argument, so $z-y\gt0$ is needed.

which is the open half-space bounded by the plane $z=y$, specifically the side on which $z$ exceeds $y$.

Solution. The level surface $f(x,y,z)=k$ is $x^2+y^2+z^2=k$; since the left side is never negative, a surface exists only when $k\ge0$.

For $k\gt0$ this is the sphere of radius $\sqrt k$ centered at the origin, and for $k=0$ it degenerates to the single point at the origin. As $k$ increases through $0,1,2,3,\dots$, the radius $\sqrt k$ also increases (though at a slowing rate), so the level surfaces form a family of concentric spheres about the origin, growing outward as $k$ grows.

Solution. The level surface $f(x,y,z)=k$ is $x^2-y-z^2=k$; solving for $y$ gives

This has the same shape as the standard form $\dfrac{z}{c}=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}$ recorded in §12.6, with the roles of the axes permuted; since $x$ and $z$ enter the squared terms with opposite signs, the surface is a hyperbolic paraboloid.

For $k=0$ it is the saddle-shaped surface $y=x^2-z^2$ through the origin; for $k=5$ it is $y=x^2-z^2-5$, and for $k=-5$ it is $y=x^2-z^2+5$ — each a copy of the same saddle shape translated by $5$ along the $y$-axis. So the level surfaces form a family of congruent hyperbolic paraboloids stacked along the $y$-axis.