벡터와 공간 전체 문제

For each fixed $c\ge 0$, the isobar $p(x,y)=c$ is the curve $x^2-2cx+y^2=3c^2$, or equivalently, $(x-c)^2+y^2=4c^2$. This is a circle with centre $(c,0)$ and radius $2c$. Here is a sketch of the isobars $p(x,y)=c$ with $c=0,1,2,3$.

Since the level curves are circles centred at the origin (in the $xy$-plane), when $z$ is a constant, the equation will have the form $x^2+y^2=c$ for some constant. That is, our equation looks like $$x^2+y^2=g(z),$$ where $g(z)$ is a function depending only on $z$.

Because our cross-sections are so nicely symmetric, we know the intersection of the figure with the left side of the $yz$-plane as well: $z=3(-y-1)=-3(y+1)$ (when $z\ge0$) and $z=-3(-y-1)=3(y+1)$ (when $z<0$). Below is the intersection of our surface with the $yz$ plane.

Setting $x=0$, our equation becomes $y^2=g(z)$. Looking at the right side of the $yz$ plane, this should lead to: $\left.\begin{cases} z=3(y-1) &\text{if } z\geq 0,\ y\ge 1\\ z=-3(y-1) &\text{if } z< 0,\ y\ge 1 \end{cases}\right\}$. That is:

$$\begin{aligned} |z|&=3(y-1)\\ \frac{|z|}{3}+1&=y\\ \left(\frac{|z|}{3}+1\right)^2&=y^2 &(*) \end{aligned}$$

A quick check: when we squared both sides of the equation in $(*)$, we added another solution, $\frac{|z|}{3}+1=-y$. Let's make sure we haven't diverged from our diagram.

$$\begin{aligned} &&& \left(\frac{|z|}{3}+1\right)^2=y^2\\ &\Leftrightarrow&& \underbrace{\frac{|z|}{3}+1}_{\text{positive}}=\pm y\\ &\Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y>0\\[0.05in] \frac{|z|}{3}+1 =-y & y<0 \end{cases}\\[0.1in] & \Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y\ge1\\[0.05in] \frac{|z|}{3}+1 =-y & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} |z| =3(y-1) & y\ge1\\[0.05in] |z|=-3(y+1) & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =\pm \underbrace{3(y-1)}_{\text{positive}} & y\ge1\\ z=\pm \underbrace{3(y+1)}_{\text{negative}} & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =3(y-1) & y\ge1,\ z\ge0\\ z =-3(y-1) & y\ge1,\ z\le0\\ z=-3(y+1) & y\le-1,\ z \ge 0\\ z=3(y+1) & y\le-1,\ z\le 0 \end{cases} \end{aligned}$$

This matches our diagram eactly. So, all together, the equation of the surface is $$x^2+y^2=\left( \frac{|z|}{3}+1\right)^2$$

The curve is $y=y(x)={x^3}/{3}$. Since $y'(x) = x^2$ and $y''(x) = 2x$, the curvature is

$$\begin{aligned} \kappa (x) &=\frac{\big|\frac{\mathrm{d^2}y}{\mathrm{d}{x}^2}(x)\big|} {\Big[1+\big(\frac{\mathrm{d}y}{\mathrm{d}x}(x)\big)^2\Big]^{3/2}} =\frac{\big|2x\big|} {\big[1+x^4\big]^{3/2}} \end{aligned}$$

We'd like to find the critical points of $\kappa (x)$, but differentiating it looks messy. Since $\kappa (x)$ has only nonnegative values, its maxima correspond the maxima of the function $\kappa ^2(x)$. So, we find the critical points of $\kappa ^2(x)$ instead, to save ourselves some computational toil.

$$\begin{aligned} 0&=\frac{\mathrm{d}}{\mathrm{d}x}\kappa (x)^2 =\frac{\mathrm{d}}{\mathrm{d}x}\frac{4x^2}{{(1+x^4)}^3} =\frac{8x}{{(1+x^4)}^3} - 3\frac{16x^5}{{(1+x^4)}^4} =\frac{8x(1+x^4)-3\times 16x^5}{{(1+x^4)}^4} \\ &=\frac{8x(1-5x^4)}{{(1+x^4)}^4} \end{aligned}$$

Note that $\kappa (0)=0$ and $\kappa (x)\rightarrow 0$ as $x\rightarrow\pm \infty$. So the maximum occurs when $x=\pm 1/\sqrt[4]{5}$.