91Ó°ÊÓ

It has been given that the temperature is $\mathrm{T}={\mathrm{x}}^2-\mathrm{xy}+{\mathrm{z}}^2$.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

(a) To calculate the gradient of u, first calculate the partial derivatives with respect to each coordinate.

$$\begin{gathered} \frac{∂\mathrm{T}}{∂\mathrm{x}}=2\mathrm{x}-\mathrm{y}, \\ \frac{∂\mathrm{T}}{∂\mathrm{y}}=-\mathrm{x}\mathrm{and} \\ \frac{∂\mathrm{T}}{∂\mathrm{z}}=2\mathrm{z} \end{gathered}$$

Hence, the equation.

$∇\mathrm{T}=(2\mathrm{x}-\mathrm{y})\mathrm{i}-\mathrm{xj}+2\mathrm{zk}$

At the point the gradient becomes as described below.

$∇{\mathrm{T}}_{\left(2,1,-1\right)}=3\mathrm{i}-2\mathrm{j}-2\mathrm{k}$

So, the direction of the heat flow is

$\mathrm{u}=-3\mathrm{i}+2\mathrm{j}+2\mathrm{k}$

(b) The rate of change of the temperature in a given direction is the directional derivative of the temperature in that direction,

To calculate the directional derivative of T in a specific direction, a unit vector in that direction is needed.

Let the unit vector in the given direction be r.

$$\begin{gathered} \mathrm{r}=\frac{\mathrm{j}-\mathrm{k}}{\left|\mathrm{j}-\mathrm{k}\right|} \\ =\frac{\mathrm{j}-\mathrm{k}}{\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}} \\ =\frac{1}{\sqrt{2}}\left(\mathrm{j}-\mathrm{k}\right) \end{gathered}$$

The directional derivative of T in that direction is the dot product between gradient T and the unit vector in that direction.

$$\begin{gathered} {\overline{){\mathrm{D}}_{\mathrm{r}}\left(\mathrm{T}\right)}}_{\left(2,1-1\right)}=∇\mathrm{T}.\mathrm{r} \\ =\left(3\mathrm{i}-2\mathrm{j}-2\mathrm{k}\right).\left(\frac{1}{\sqrt{2}}\left(\mathrm{j}-\mathrm{k}\right)\right) \\ =0 \end{gathered}$$

Hence, the solutions are given below.

$$\begin{gathered} a)u=-3i+2j+2k \\ b){\overline{)D_r\left(T\right)}}_{\left(2,1,-1\right)}=0 \end{gathered}$$