936. What is the greatest rate at which the function $u(M)=\frac{10}{x^{2}+y^{2}+z^{2}+1}$ can increase as the point $M(x, y, z)$ passes through the point $M_{0}(-1 ; 2 ;-2)$? In what direction should the point $M$ move as it passes through the point $M_{1}(2 ; 0 ; 1)$, so that the function $u(M)$ decreases at the greatest rate?
Solution. The greatest in absolute value the rate of change (increase or decrease) of the function and (M) when the point $M$ passes through the point $P$ numerically equals the modulus of the gradient of the function at point $P$. In this case, the function will increase or decrease at the greatest rate, depending on whether the point $M$ when passing through point $P$ moves in the direction of the gradient of the function at point $P$ or in the directly opposite direction.

Guided by these principles, we find the partial derivatives of the function and and by formula (b) - its gradient at any point:

$$
\operatorname{grad} u=-\frac{20}{\left(x^{2}+y^{2}+\tilde{z}^{2}+1\right)^{2}}(x \ddot{i}+y \ddot{j}+z k)
$$

Next, we find: 1) $\operatorname{grad} u\left(M_{6}\right)=\frac{1}{5}(\bar{i}-2 j+2 \bar{k})$; its modulus, numerically equal to the desired greatest rate of increase of the function $u(M)$ when $M$ passes through $M_{0}$, will be $\left|\operatorname{grad} u\left(M_{0}\right)\right|==$ $=\sqrt{\left(\frac{1}{5}\right)^{2}+\left(-\frac{2}{5}\right)^{2}+\left(\frac{2}{5}\right)^{2}}=\frac{3}{5}$.
2) $\operatorname{grad} u\left(M_{1}\right)=-\frac{10}{9} i-\frac{5}{9} k ;$ the desired vector, having the directly opposite direction, will be $-\operatorname{grad} u\left(M_{1}\right)=\frac{10}{9} \ddot{i}+\frac{5}{9} k$. For the function $u(M)$ to decrease at the greatest rate when passing through the point $M_{1}$, the point $M$ must move in the direction of the vector $-\operatorname{grad} u\left(M_{1}\right)$.