접평면과 선형근사

(a) The linear approximation to $f(x,y)$ at $(a,b)$ is

$$\begin{aligned} f(x,y) &\approx f(a,b) + f_x(a,b)\,(x-a) + f_y(a,b)\,(y-b) \\ &= \ln(4a^2 + b^2) + \frac{8a}{4a^2+b^2}\, (x-a) + \frac{2b}{4a^2+b^2}\, (y-b) \end{aligned}$$

In particular, for $a=0$ and $b=1$,

$$\begin{aligned} f(x,y) &\approx 2\, (y-1) \end{aligned}$$

and, for $x=0.1$ and $y=1.2$,

$$\begin{aligned} f(0.1,1.2) &\approx 0.4 \end{aligned}$$

(b) The point $(a,b,c)$ is on the surface $z=f(x,y)$ if and only if

$$c = f(a,b) = \ln(4a^2 + b^2)$$

Note that this forces $4a^2 + b^2$ to be nonzero. The tangent plane to the surface $z = f (x, y)$ at the point $(a,b,c)$ is parallel to the plane $2x + 2y - z = 3$ if and only if $\left< 2\,,\,2\,,\,-1\right>$ is a normal vector for the tangent plane. That is, there is a nonzero number $t$ such that

$$\begin{aligned} \left< 2\,,\,2\,,\,-1\right> =t\left< f_x(a,b)\,,\,f_y(a,b)\,,\ -1\right> =t\left< \frac{8a}{4a^2+b^2} \,,\, \frac{2b}{4a^2+b^2} \,,\,-1\right> \end{aligned}$$

For the $z$–coordinates to be equal, $t$ must be $1$. Then, for the $x$– and $y$–coordinates to be equal, we need

$$\begin{aligned} \frac{8a}{4a^2+b^2} &= 2 \\ \frac{2b}{4a^2+b^2} &= 2 \end{aligned}$$

Note that these equations force both $a$ and $b$ to be nonzero. Dividing these equations gives $\frac{8a}{2b}=1$ and hence $b=4a$. Substituting $b=4a$ into either of the two equations gives

$$\begin{aligned} \frac{8a}{20a^2}=2 \implies a=\frac{1}{5} \end{aligned}$$

So $a=\frac{1}{5}$, $b=\frac{4}{5}$ and

$$\begin{aligned} c=\ln\left(\frac{4}{5^2}+\frac{4^2}{5^2}\right) =\ln\frac{4}{5} \end{aligned}$$