다변수 미분 전체 문제

The basic assumption is that the three quantites $x$, $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0$, which can be solved to

• determine $x$ as a function of $y$ and $z$ (say $x=f(y,z)$) and can alternatively be solved to
• determine $y$ as a function of $x$ and $z$ (say $y=g(x,z)$) and can alternatively be solved to
• determine $z$ as a function of $x$ and $y$ (say $z=h(x,y)$).

As an example, if $F(x,y,z) = xyz-1$, then

• $F(x,y,z)=xyz-1=0$ implies that $x=\frac{1}{yz}=f(y,z)$ and
• $F(x,y,z)=xyz-1=0$ implies that $y=\frac{1}{xz}=g(x,z)$ and
• $F(x,y,z)=xyz-1=0$ implies that $z=\frac{1}{xy}=h(x,y)$

In general, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

$$F\big(f(y,z),y,z\big)=0 \tag{$*$}$$

for all $y$ and $z$. Set $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$. Applying the chain rule to $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ (with $y$ and $z$ independent variables) gives

$$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z) &= \frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big) \end{aligned}$$

The equation $(*)$ says that $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ for all $y$ and $z$. So differentiating the equation $(*)$ with respect to $z$ gives

$$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z)&=\frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)=0 \\ &\implies \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)} {\frac{\partial F}{\partial x}\big(f(y,z),y,z\big)} \end{aligned}$$

for all $y$ and $z$. Similarly, differentiating $F\big(x,g(x,z),z\big)=0$ with respect to $x$ and $F\big(x,y,h(x,y)\big)=0$ with respect to $y$ gives

$$\begin{aligned} \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,g(x,z),z\big)} {\frac{\partial F}{\partial y}\big(x,g(x,z),z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,h(x,y)\big)} {\frac{\partial F}{\partial z}\big(x,y,h(x,y)\big)} \end{aligned}$$

If $(x,y,z)$ is any point satisfying $F(x,y,z)=0$ (so that $x=f(y,z)$ and $y=g(x,z)$ and $z=h(x,y)$), then

$$\begin{aligned} \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\qquad \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f}{\partial z}(y,z)\ \frac{\partial g}{\partial x}(x,z)\ \frac{\partial h}{\partial y}(x,y) &=-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \\ &=-1 \end{aligned}$$

Denote by $S$ the surface $G(x,y,z)=0$ and by $C$ the parametrized curve $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$. To start, we'll find the tangent plane to $S$ at $\mathbf{r}_0$ and the tangent line to $C$ at $\mathbf{r}_0$.

• The tangent vector to $C$ at $\mathbf{r}_0$ is $\left< x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right>$, so the parametric equations for the tangent line to $C$ at $\mathbf{r}_0$ are

$$x-x_0 = t x'(t_0)\qquad y-y_0 = t y'(t_0)\qquad z-z_0 = t z'(t_0) \tag{$E_1$}$$

• The gradient $\left<\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right>$ is a normal vector to the surface $S$ at $(x_0,y_0,z_0)$. So the tangent plane to the surface $S$ at $(x_0,y_0,z_0)$ is

$$\left<\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right> \cdot \left< x-x_0\,,\, y-y_0\,,\,z-z_0\right> = 0$$

or

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}$$

Next, we'll show that the tangent vector $\left< x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right>$ to $C$ at $\mathbf{r}_0$ and the normal vector $\left<\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right>$ to $S$ at $\mathbf{r}_0$ are perpendicular to each other. To do so, we observe that, for every $t$, the point $\big(x(t),y(t),z(t)\big)$ lies on the surface $G(x,y,z)=0$ and so obeys

$$\begin{aligned} G\big(x(t),y(t),z(t)\big) =0 \end{aligned}$$

Differentiating this equation with respect to $t$ gives, by the chain rule,

$$\begin{aligned} 0&= \frac{\mathrm{d}}{\mathrm{d}t}G\big(x(t),y(t),z(t)\big) \\ &=\frac{\partial G}{\partial x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\frac{\partial G}{\partial y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\frac{\partial G}{\partial z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{aligned}$$

Then setting $t=t_0$ gives

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$}$$

Finally, we are in a position to show that if $(x,y,z)$ is any point on the tangent line to $C$ at $\mathbf{r}_0$, then $(x,y,z)$ is also on the tangent plane to $S$ at $\mathbf{r}_0$. As $(x,y,z)$ is on the tangent line to $C$ at $\mathbf{r}_0$ then there is a $t$ such that, by $(E_1)$,

$$\begin{aligned} &\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{x-x_0\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{y-y_0\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{z-z_0\} \\ &=\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,x'(t_0)\big\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,y'(t_0)\big\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,z'(t_0)\big\}\\ &= t\left[\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) \right] =0 \end{aligned}$$

by $(E_3)$. That is, $(x,y,z)$ obeys the equation, $(E_2)$, of the tangent plane to $S$ at $\mathbf{r}_0$ and so is on that tangent plane. So the tangent line to $C$ at $\mathbf{r}_0$ is contained in the tangent plane to $S$ at $\mathbf{r}_0$.