다변수 미분 전체 문제

partial

In general, this is false. Consider $f(x,y)=12-(1-10x)^2-(1-10y)^2$.

• $\lim\limits_{(x,y)\to(0,0)}f(x,y)=12-1-1=10$ (the function is continuous)
• $f(0.1,0.1)=12-(1-1)^2-(1-1)^2=12$
• $f(0.2,0.2)=12-(1-2)^2-(1-2)^2=10$

We often (somewhat lazily) interpret the limit “$\lim\limits_{(x,y)\to(0,0)}f(x,y)=10$" to mean that, as $(x,y)$ gets closer and closer to the origin, $f(x,y)$ gets closer and closer to 10. This isn't exactly what the definition means, though. The definition tells us that, we can guarantee that $f(x,y)$ be very close to 10 by choosing $(x,y)$ very close to $(0,0)$.

The function $f(x,y)$ can also be very close to 10 for some $(x,y)$'s that are not close to $(0,0)$. Moreover, we don't know how close to $(0,0)$ we have to be in order for $f(x,y)$ to be “very close" to 10.

(a) The temperature gradient at $(3,2,1)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } T(3,2,1) = \left< 4x\,,\,2y\,,\,-2z\right>\Big|_{(x,y,z)=(3,2,1)} = \left< 12\,,\,4\,,\,-2\right> \end{aligned}$$

She wishes to fly in a direction that is perpendicular to ${ \boldsymbol{\nabla} } T(3,2,1)$. That is, she wishes to fly in a direction $\left< a\,,\,b\,,\,c\right>$ that obeys

$$0 = \left< 12\,,\,4\,,\,-2\right>\cdot \left< a\,,\,b\,,\,c\right> = 12a+4b-2c$$

Any nonzero $\left< a\,,\,b\,,\,c\right>$ that obeys $12a+4b-2c=0$ is an allowed direction. Four allowed unit vectors are $\pm\frac{\left< 0\,,\,1\,,\,2\right>}{\sqrt{5}}$ and $\pm\frac{\left< 1\,,\,-3\,,\,0\right>}{\sqrt{10}}$.

(b) No they need not be the same. Four different explicit directions were given in part (a).

(c) To cool down as quickly as possible, she should move in the direction opposite to the temperature gradient. A unit vector in that direction is $-\frac{\left< 6\,,\,2\,,\,-1\right>}{\sqrt{41}}$.