접평면과 선형근사

$z=0$

(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

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

We are going to need the first order partial derivatives of $f(x,y)=ye^{-x}$ at $(x,y)=(1,e)$. Here they are.

$$\begin{aligned} f_x(x,y)&= -ye^{-x} & f_x(1,e)&=-e\,e^{-1}=-1 \\ f_y(x,y)&=e^{-x} & f_y(1,e)&= e^{-1} \end{aligned}$$

(a) The linear approximation to $f(x,y)$ at $(x,y)=(1,e)$ is

$$\begin{aligned} f(x,y) \approx f(1,e) + f_x(1,e)\,(x-1) +f_y(1,e)\,(y-e) = 1 -(x-1) +e^{-1}(y-e) \end{aligned}$$

The maximum error is then approximately

$$\begin{aligned} -1(-0.1) +e^{-1}(0.1) =\frac{1+e^{-1}}{10} \end{aligned}$$

(b) The equation of the graph is $g(x,y,z) = f(x,y) -z =0$. Any vector that is a nonzero constant times

$$\begin{aligned} { \boldsymbol{\nabla} } g(1,e,1) =\left< f_x(1,e)\,,\,f_y(1,e)\,,\,-1\right> =\left< -1\,,\,e^{-1}\,,\,-1\right> \end{aligned}$$

is perpendicular to $g=0$ at $(1,e,1)$.

Since

$$\begin{aligned} f_x(2,1) & = -2x e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = -4 \\ f_y(2,1) & = \phantom{-}8y e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = 8 \end{aligned}$$

The tangent plane to $z=f(x,y)$ at $(2,1)$ is

$$\begin{aligned} z &= f(2,1) +f_x(2,1)\,(x-2) +f_y(2,1)\,(y-1) = 1 -4 (x-2) +8(y-1) \\ &= 1-4x +8y \end{aligned}$$

and the tangent plane approximation to the value of $f(1.99, 1.01)$ is

$$\begin{aligned} f(1.99, 1.01) \approx 1 -4 (1.99-2) +8(1.01-1) = 1.12 \end{aligned}$$

(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

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