Question

The temperature in the big hall is approximated by the function

T(x, y, z) = x

2 − 2xyz + z

2 + 5;

0 ≤ x ≤ 2, 0 ≤ y ≤ 3 and 0 ≤ z ≤ 2.

If a person located at (1, 1, 1), in which direction he should walk

to cool off as rapidly as possibly.

Answer 1

$$T(x, y, z)=x^2-2xyz+z^2+5$$

$$\dfrac{\partial T}{\partial x}=2x-2yz$$

$$\dfrac{\partial T}{\partial y}=-2xz$$

$$\dfrac{\partial T}{\partial z}=-2xy+2z$$

The gradient of $T$ is

$$\nabla T=\dfrac{\partial T}{\partial x}i+\dfrac{\partial T}{\partial y}j+\dfrac{\partial T}{\partial z}k$$

$$=(2x-2yz)i+(-2xz)j+(-2xy+2z)k$$

At the point $(1,1,1)$ the gradient vector is

$$\nabla T(1,1,1)=(2(1)-2(1)(1))i+(-2(1)(1))j$$

$$+(-2(1)(1)+2(1))k=-2j$$

The temperature increases fastest in the direction of the gradient vector $\nabla T(1,1,1)=-2j.$

A person should walk in direction opposite to the gradient vector $\nabla T(1,1,1),$ or equivalently, in the direction of $2j$ or the unit vector $j$ to cool off as rapidly as possibly.

The maximum rate of decrease is the length of the gradient vector:

$$|\nabla T(1,1,1)|=|-2j|=2$$

Therefore the maximum rate of decrease of temperature is $2\ \dfrac{unit\ of\ temperature}{unit \ of\ length}.$