Example 3. Find the derivative of the scalar field $u=x z^{2}+2 y z$ at the point $M_{0}(1,0,2)$ along the circle

$$
\left\{\begin{array}{l}
x=1+\cos t \\
y=\sin t-1 \\
z=2
\end{array}\right.
$$
Solution. The vector equation of the circle has the form

$$
\mathbf{r}(t)=(1+\cos t) \mathbf{i}+(\sin t-1) \mathbf{j}+2 \mathbf{k} .
$$

We find the vector $T$, tangent to it at any point $M$. We have

$$
\left.T=\frac{d r}{d t}=-\sin t \mathbf{i}+\cos t \mathbf{j}\right]
$$

The given point $M_{0}(1,0,2)$ lies in the $\boldsymbol{O} \boldsymbol{z}$ plane in the first octant, and corresponds to the parameter value $t=\frac{\pi}{2}$. At this point we will have

$$
\left.\tau\right|_{M_{0}}=-\sin \frac{\pi}{2} \mathbf{i}+\cos \frac{\pi}{2} \mathbf{j}=-\mathbf{i} .
$$

From this we obtain that the direction cosines of the tangent to the circle are $\cos \alpha=-1, \cos \beta=0, \cos \gamma=0$. The values of the partial derivatives of the given scalar field at the point $M_{0}(1,0,2)$ are

$$
\left.\frac{\partial u}{\partial x}\right|_{M_{0}}=\left.z^{2}\right|_{M_{0}}=4,\left.\quad \frac{\partial u}{\partial y}\right|_{M_{0}}=\left.2 z\right|_{M_{0}}=4,\left.\quad \frac{\partial u}{\partial z}\right|_{M_{0}}=\left.(2 x z+2 y)\right|_{M_{0}}=4
$$

Therefore, the sought derivative is

$$
\left.\frac{\partial u}{\partial l}\right|_{M_{0}}=\left.\frac{\partial u}{\partial r}\right|_{M_{0}}=4 \cdot(-1)+4 \cdot 0+4 \cdot 0=-4
$$

## Problems for Independent Solution

In the following problems, find the derivative of the given functions at the point $M_{0}\left(x_{0}, y_{0}, z_{0}\right)$ in the direction of the point $M_{1}\left(x_{1}, y_{1}, z_{1}\right)$.

58. $u=\sqrt{x^{2}+y^{2}+z^{2}}, \quad M_{0}(1,1,1), M_{1}(3,2,1)$
59. $u=x^{2} y+x z^{2}-2, \quad M_{0}(1,1,-1), M_{1}(2,-1,3)$.
60. $u=x e^{y}+y e^{x}-z^{2}, \quad M_{0}(3,0,2), M_{1}(4,1,3)$
61. $u=\frac{x}{y}-\frac{y}{x}, \quad M_{0}(1,1), M_{1}(4,5)$.
62. Find the derivative of the scalar field $u=\ln \left(x^{2}+y^{2}\right)$ at the point $M_{0}(1,2)$ of the parabola $y^{2}=4 x$ in the direction of this curve.
63. Find the derivative of the scalar field $u=\operatorname{arctg} \frac{y}{x}$ at the point $M_{0}(2,-2)$ of the circle $x^{2}+y^{2}-4 x=0$ along the arc of this circle.
64. Find the derivative of the scalar field $u=x^{2}+y^{2}$ at the point $M_{0}\left(x_{0}, y_{0}\right)$ of the circle $x^{2}+y^{2}=\boldsymbol{R}^{2}$ in the direction of this circle.
65. Find the derivative of the scalar field $u=2 x y+y^{2}$ at the point $(\sqrt{2}, 1)$ of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ in the direction of the external normal to the ellipse at this point.
66. Find the derivative of the scalar field $u=x^{2}-y^{2}$ at the point $(5,4)$ of the hyperbola $x^{2}-y^{2}=9$ in the direction of this curve.
67. Find the derivative of the scalar field $u=\ln (x y+y z+x z)$ at the point $M_{0}(0,1,1)$ in the direction of the circle $x=\cos t, y=\sin t, z=1$.
68. Find the derivative of the scalar field $u=x^{2}+y^{2}+z^{2}$ at the point $M_{0}$ corresponding to the value of the parameter $t=\frac{\pi}{2}$ in the direction of the helical line $x=R \cos t$, $y=R \sin t, z=a t$.

## §9. Gradient of a Scalar Field

Let us have a scalar field defined by the scalar function

$$
u=f(x, y, z)
$$

the function $f$ is assumed to be differentiable.

Definition. The gradient of a scalar field at a given point $M$ is a vector denoted by the symbol grad $u$ and defined by the formula

$$
\operatorname{grad} u=\frac{\partial u}{\partial x} \mathbf{i}+\frac{\partial u}{\partial y} \mathbf{j}+\frac{\partial u}{\partial z} \mathbf{k}
$$

We have

Using the formula (1) from §8 for the directional derivative,

$$
\frac{\partial u}{\partial l}=\left(\operatorname{grad} u, \mathbf{l}^{0}\right)
$$

where $\mathbf{l}^{0}$ is the unit vector in the direction of $\mathbf{l}$, i.e.

$$
\mathbf{l}^{0}=\frac{\mathbf{l}}{|\mathbf{l}|}=\mathbf{i} \cos \alpha+\mathbf{j} \cos \beta+\mathbf{k} \cos \gamma
$$

## Properties of the Gradient

1. The gradient is directed along the normal to the level surface (or level line, if the field is planar).
2. The gradient is directed in the direction of the increase of the field function.
3. The magnitude of the gradient is equal to the maximum directional derivative at the given point of the field:

$$
\left.\max \frac{\partial u}{\partial l}=|\operatorname{grad} u|=\sqrt{\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}+\left(\frac{\partial u}{\partial z}\right)^{2}}\right)
$$

These properties provide an invariant characteristic of the gradient. They indicate that the vector grad $u$ points in the direction and magnitude of the greatest change in the scalar field at the given point.