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The equation$a^2+b^2+c^2=1$ is given.

In mathematics and physics, scalar field or scalar-valued function referred to a scalarvalueto everypointin aspace鈥� possiblyphysical space. The scalar may either be a (dimensionless)mathematical numberor aphysical quantity.

Let$蠒=蠒\left(x,y,z\right)$be a scalar field.

$\frac{d蠒}{ds}=\frac{鈭�蠒}{鈭�x}\frac{dx}{ds}+\frac{鈭�蠒}{鈭�y}\frac{dy}{ds}+\frac{鈭�蠒}{鈭�z}\frac{dz}{ds}..........\left(1\right)$

It is known that $\frac{d蠒}{ds}$is the rate of change of $蠒$at that point along a given direction.

Consider an arbitrary point $\left(x_0,y_0,z_0\right)$.

Let $u=ai+bj+ck$be a unit vector in a given direction.

Since u is a unit vector, therefore$a^2+b^2+c^2=1......\left(2\right)$

Start at a point$\left(x_0,y_0,z_0\right)$and move a distance s in the direction of u to the point $\left(x,y,z\right)$.

The vector joining the points is $us=\left(ai+bj+ck\right)s$.

$$\begin{gathered} x=x_0+as \\ y=y_0+as \\ z=z_0+as \\ \frac{dx}{ds}=a......\left(3\right) \\ \frac{dy}{ds}=b.......\left(4\right) \\ \frac{dz}{ds}=c.......\left(5\right) \end{gathered}$$

Put the values of equation 3, 4, 5 in equation 1.

$\frac{d蠒}{ds}=\frac{鈭�蠒}{鈭�x}a+\frac{鈭�蠒}{鈭�y}b+\frac{鈭�蠒}{鈭�z}c.....\left(6\right)$

Use equation (6) to find the maximum of $\frac{d蠒}{ds}$.

Apply Lagrange multiplier method.

Let, a function be F=(x,y,z).

$$\begin{gathered} F=\frac{d蠒}{ds}+位\left(a^2+b^2+c^2\right) \\ =\frac{鈭�蠒}{鈭�x}a+\frac{鈭�蠒}{鈭�y}+b\frac{鈭�蠒}{鈭�z}c+位\left(a^2+b^2+c^2\right) \end{gathered}$$

Solve the following equations to find the value of a,b,c.

$$\begin{gathered} \frac{鈭�F}{鈭�a}=0 \\ \frac{鈭�蠒}{鈭�x}+2a位=0 \\ a=-\frac{1}{2位}\frac{鈭�蠒}{鈭�x} \\ \frac{鈭�F}{鈭�b}=0 \\ \frac{鈭�蠒}{鈭�y}+2b位=0 \\ b=-\frac{1}{2位}\frac{鈭�蠒}{鈭�y} \end{gathered}$$

localid="1659157825632" $$\begin{gathered} \frac{鈭�F}{鈭�c}=0 \\ \frac{鈭�蠒}{鈭�z}+2c位=0 \\ c=-\frac{1}{2位}\frac{鈭�蠒}{鈭�z} \end{gathered}$$

Put the values of a,b,c in equation (2).

$a^2+b^2+c^2=1$

$$\begin{gathered} {\left(-\frac{1}{2位}\frac{鈭�蠒}{鈭�x}\right)}^2+{\left(-\frac{1}{2位}\frac{鈭�蠒}{鈭�y}\right)}^2+{\left(-\frac{1}{2位}\frac{鈭�蠒}{鈭�z}\right)}^2=1 \\ 位=卤\frac{1}{2}\sqrt{{\left(\frac{鈭�蠒}{鈭�x}\right)}^2+{\left(\frac{鈭�蠒}{鈭�y}\right)}^2+{\left(\frac{鈭�蠒}{鈭�z}\right)}^2} \\ =卤\frac{1}{2}\left|鈭�蠒\right| \end{gathered}$$

Put the value of$位ina,b,c$.

For$位=+\frac{1}{2}\left|鈭�蠒\right|$

The values of a,b,c are as follows:

$$\begin{gathered} a=-\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�x} \\ b=-\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�y} \\ c=-\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�z} \end{gathered}$$

For$位=-\frac{1}{2}\left|鈭�蠒\right|$

The values of a,b,c are as follows:

$$\begin{gathered} a=\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�x} \\ b=\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�y} \\ c=\frac{1}{\left|鈭�蠒\right|}\frac{鈭�蠒}{鈭�z} \end{gathered}$$

Put the value of a, b, c in equation (1).

For $位=+\frac{1}{2}\left|鈭�蠒\right|$

$$\begin{gathered} \frac{d蠒}{ds}=\frac{鈭�蠒}{鈭�x}a+\frac{鈭�蠒}{鈭�y}+b+\frac{鈭�蠒}{鈭�z}c \\ =-\frac{1}{\left|鈭�蠒\right|}\left({\left(\frac{鈭�蠒}{鈭�x}\right)}^2+{\left(\frac{鈭�蠒}{鈭�y}\right)}^2{\left(\frac{鈭�蠒}{鈭�z}\right)}^2\right) \\ =-\frac{1}{\left|鈭�蠒\right|}{\left|鈭�蠒\right|}^2 \\ =-\left|鈭�蠒\right| \end{gathered}$$

For $位=-\frac{1}{2}\left|鈭�蠒\right|$

$$\begin{gathered} \frac{d蠒}{ds}=\frac{鈭�蠒}{鈭�x}a+\frac{鈭�蠒}{鈭�y}+b+\frac{鈭�蠒}{鈭�z}c \\ =\frac{1}{\left|鈭�蠒\right|}\left({\left(\frac{鈭�蠒}{鈭�x}\right)}^2+{\left(\frac{鈭�蠒}{鈭�y}\right)}^2{\left(\frac{鈭�蠒}{鈭�z}\right)}^2\right) \\ =\frac{1}{\left|鈭�蠒\right|}{\left|鈭�蠒\right|}^2 \\ =\left|鈭�蠒\right| \end{gathered}$$

Hence, the solutions are mentioned below.

$\frac{d蠒}{ds}=\left|鈭�蠒\right|$is the maximum value.

$\frac{d蠒}{ds}=-\left|鈭�蠒\right|$is the minimum value.