접평면과 선형근사

A normal vector to the surface $x^2 + 9y^2 + 4z^2 = 17$ at the point $(x,y,z)$ is $\left< 2x\,,\, 18y\,,\,8z\right>$. A normal vector to the plane $x - 8z = 0$ is $\left< 1\,,\,0\,,\,-8\right>$. So we want $\left< 2x\,,\, 18y\,,\,8z\right>$ to be parallel to $\left< 1\,,\,0\,,\,-8\right>$, i.e. to be a nonzero constant times $\left< 1\,,\,0\,,\,-8\right>$. This is the case whenever $y=0$ and $z=-2x$ with $x\ne 0$. In addition, we want $(x,y,z)$ to lie on the surface $x^2 + 9y^2 + 4z^2 = 17$. So we want $y=0$, $z=-2x$ and

$$\begin{aligned} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{aligned}$$

So the allowed points are $\pm(1,0,-2)$.

Let $(x_0,y_0,z_0)$ be a point on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0$. A normal vector to the plane $2x-y+z=0$ is $\left< 2,-1,1\right>$. Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and ${ \boldsymbol{\nabla} } G(x,y,z) = \left< 8x,2y,-2z\right>$, a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=\left< 8x_0,2y_0,-2z_0\right>$. So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

$$\begin{aligned} &\left< 8x_0,2y_0,-2z_0\right> =t\left< 2,-1,1\right> \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\tfrac{1}{2},-1,-1\big) \end{aligned}$$

One vector normal to the surface $F(x,y,z)=4x^2+9y^2-z^2=0$ at $(2,1,-5)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } F(2,1,-5) = \left< 8x,18y,-2z\right>\Big|_{(2,1,-5)}=\left< 16,18,10\right> \end{aligned}$$

One vector normal to the surface $G(x,y,z)=6x+3y+2z=5$ at $(2,1,-5)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } G(2,1,-5) = \left< 6,3,2\right> \end{aligned}$$

Now

• The curve lies in the surface $z^2=4x^2+9y^2$. So the tangent vector to the curve at $(2,1,-5)$ is perpendicular to the normal vector $\frac{1}{2}\left< 16,18,10\right>=\left< 8,9,5\right>$.
• The curve also lies in the surface $6x+3y+2z=5$. So the tangent vector to the curve at $(2,1,-5)$ is also perpendicular to the normal vector $\left< 6,3,2\right>$.
• So the tangent vector to the curve at $(2,1,-5)$ is parallel to

$$\begin{aligned} \left< 8,9,5\right>\times \left< 6,3,2\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 8 & 9 & 5 \\ 6 & 3 & 2 \end{matrix}\right] =\left< 3,14,-30\right> \end{aligned}$$

The desired vectors are

$$\pm\sqrt{3}\frac{\left< 3,14,-30\right>}{|\left< 3,14,-30\right>|} =\pm\sqrt{\frac{3}{1105}}\left< 3,14,-30\right>$$

Apply the linear approximation $f(0.01,1.05)\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05)$, with

$$\begin{aligned} f(x,y)&=\sin(\pi xy+\ln y) & f(0,1)&=\sin 0=0\\ f_x(x,y)&=\pi y\cos(\pi xy+\ln y) & f_x(0,1)&=\pi\cos 0=\pi\\ f_y(x,y)&=\left(\pi x+\frac{1}{y}\right)\cos(\pi xy+\ln y) & f_y(0,1)&=\cos 0=1 \end{aligned}$$

This gives

$$\begin{aligned} f(0.01,1.05)&\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05) =0+\pi(0.01)+1(0.05) \\ &=0.01\,\pi + 0.05 \approx 0.0814 \end{aligned}$$

We are going to need the first order derivatives of $f(x,y)$ at $(x,y)=(-1,1)$. So we find them first.

$$\begin{aligned} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9} \\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{aligned}$$

The linear approximation to $f(x,y)$ about $(-1,1)$ is

$$\begin{aligned} f(x,y)\approx=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \end{aligned}$$

In particular

$$\begin{aligned} f(-0.9,1.1)\approx \frac{1}{3} -\frac{2}{9}\,(0.1) -\frac{1}{9}\,(0.1) =\frac{27}{90}=0.3 \end{aligned}$$

Denote by $x$ and $y$ the lengths of sides with $x=3\pm 0.1$ and $y=4\pm 0.2$. Then the length of the hypotenuse is $f(x,y)=\sqrt{x^2+y^2}$. Note that

$$\begin{aligned} f(x,y)&=\sqrt{x^2+y^2} & f(3,4)&=5 \\ f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}\qquad & f_x(3,4)&=\frac{3}{5} \\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}} & f_y(3,4)&=\frac{4}{5} \end{aligned}$$

By the linear approximation

$$\begin{aligned} f(x,y)\approx f(3,4) + f_x(3,4)\,(x-3) + f_y(3,4)\,(y-4) =5 +\frac{3}{5}\,(x-3) +\frac{4}{5}\,(y-4) \end{aligned}$$

So the approximate maximum error in calculating the length of the hypotenuse is

$$\begin{aligned} \frac{3}{5}\,(0.1) +\frac{4}{5}\,(0.2) =\frac{1.1}{5} = 0.22 \end{aligned}$$

The function $R(R_1,R_2)$ is defined implictly by

$$\frac{1}{R(R_1,R_2)} =\frac{1}{R_1}+\frac{1}{R_2} \tag{$*$}$$

In particular

$$\frac{1}{R(2,8)} = \frac{1}{2}+\frac{1}{8} = \frac{5}{8} \implies R(2,8) = \frac{8}{5}$$

We wish to use the linear approximation

$$R(R_1,R_2) \approx R(2,8) + \frac{\partial R}{\partial R_1}(2,8)\,(R_1-2) + \frac{\partial R}{\partial R_2}(2,8)\,(R_2-8)$$

To do so, we need the partial derivatives $\frac{\partial R}{\partial R_1}(2,8)$ and $\frac{\partial R}{\partial R_2}(2,8)$. To find them, we differentiate $(*)$ with respect to $R_1$ and $R_2$:

$$\begin{aligned} -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_1}(R_1,R_2) &= -\frac{1}{R_1^2} \\ -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_2}(R_1,R_2) &= -\frac{1}{R_2^2} \\ \end{aligned}$$

Setting $R_1=2$ and $R_2=8$ gives

$$\begin{aligned} -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_1}(2,8) &= -\frac{1}{4} &\quad\implies\quad & &\frac{\partial R}{\partial R_1}(2,8) &=\frac{16}{25} \\ -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_2}(2,8) &= -\frac{1}{64} &\quad\implies\quad & &\frac{\partial R}{\partial R_2}(2,8) &=\frac{1}{25} \end{aligned}$$

So the specified change in $R$ is

$$\begin{aligned} R(1.9,8.1)-R(2,8) \approx \frac{16}{25} (-0.1) + \frac{1}{25}(0.1) =-\frac{15}{250} = -0.06 \end{aligned}$$

The linear approximation to $P(s,r)$ at $(2,2)$ is

$$\begin{aligned} P(s,r)\approx P(2,2) +P_s(2,2)\,(s-2) + P_r(2,2)\,(r-2) \end{aligned}$$

As

$$\begin{aligned} P(2,2) &= (2)(2)\big[4(2)^2-(2)^2-2\big] = 40 \qquad\text{(which we don't actually need) } \\ P_s(2,2) &= \Big[12s^2r-r^3-2r\Big]_{s=r=2} = 84 \\ P_r(2,2) &= \Big[4s^3-3sr^2-2s\Big]_{s=r=2} = 4 \end{aligned}$$

the linear approximation is

$$\begin{aligned} P(s,r)\approx 40 +84\,(s-2) + 4\,(r-2) \end{aligned}$$

Under method 1, the maximum error in $P$ will have magnitude at most (approximately)

$$\begin{aligned} 84(0.01) + 4(0.1) = 1.24 \end{aligned}$$

Under method 2, the maximum error in $P$ will have magnitude at most (approximately)

$$\begin{aligned} 84(0.02) + 4(0.02) = 1.76 \end{aligned}$$

Method 1 is better.

The first order partial derivatives of $f$ are

$$\begin{aligned} f_x(x,y) & = -\frac{4xy}{{(x^2+y^2)}^2}\quad & f_x(-1,2) & = \frac{8}{25} \\ f_y(x,y) & = \frac{2}{x^2+y^2}-\frac{4y^2}{{(x^2+y^2)}^2}\quad & f_y(-1,2) & = \frac{2}{5}-\frac{16}{25} =-\frac{6}{25} \\ \end{aligned}$$