Physics, 13.08.2019 01:30 brodycruce

Suppose that over a certain region of space the electrical potential v is given by the following equation. v(x, y, z) = 3x2 − 3xy + xyz (a) find the rate of change of the potential at p(3, 6, 5) in the direction of the vector v = i + j − k. (b) in which direction does v change most rapidly at p? (c) what is the maximum rate of change at p?

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Answer from: bikerhomie

(a) 18

(b) in the direction of the gradient vector $\nabla v(3,6,5) = \langle30,6,18\rangle$

explanation:

the function

$v(x, y, z) = 3x^2-3xy + xyz$

is a differentiable function of x,y,z since it is a polynomial. therefore, the directional derivative at the point $(x_0,y_0,z_0)$ in the direction of a unit vector $\bf u$ is the dot product

${\bf d_u} v(x_0,y_0,z_0) = \langle v_x(x_0,y_0,z_0), v_y(x_0,y_0,z_0), v_z(x_0,y_0,z_0)\rangle \cdot {\bf u}$

(the derivative is the rate of change, so this is the rate of change in the direction of the unit vector $\bf u$ at the point $(x_0,y_0,z_0)$ .)

we need a unit vector in the direction of vector ${\bf v}={\bf i}+{\bf j}-{\bf k}$ . the magnitude of $\bf v$ is

$|{\bf v}| = |{\bf i}+{\bf j}-{\bf k}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$

so a unit vector in the direction of $\bf v$ is

${\bf u} = \frac{1}{\sqrt{3}}{\bf v} = \frac{1}{\sqrt{3}}({\bf i}+{\bf j}-{\bf k})=\frac{1}{\sqrt{3}}\langle1,1,-1\rangle.$

now we have to calculate the partial derivatives of v at (3,6,5). as we have $v(x, y, z)= 3x^2-3xy+xyz$ , we calculate partial derivative with respect to x of v by treating y and z as constants and differentiating with respect to x:

$\begin{aligned}v_x(x_0,y_0,z_0) & = 6x - 3y + yz \\v_x(3,6,5)& = 6(3)-3(6)+6(5) = 30 \end{aligned}$

similarly, we calculate $v_y(3,6,5)$

$\begin{aligned}v_y(x_0,y_0,z_0) & = - 3x + xz \\v_y(3,6,5)& = -3(3)+3(5) = 6 \end{aligned}$

and we calculate $v_z(3,6,5)$

$\begin{aligned}v_z(x_0,y_0,z_0) & = xy\\v_z(3,6,5)& =3(6) = 18 \end{aligned}$

(a)

we have

$\begin{aligned}{\bf d_u} v(3,6,5) & = \langle v_x(3,6,5), v_y(3,6,5), v_z(3,6,5)\rangle \cdot {\bf u} \\& = \langle30,6,18\rangle \cdot \frac{1}{\sqrt{3}}\langle1,1,-1\rangle \\& = \frac{1}{\sqrt{3}}(30(1)+6(1)+ 18(-1)) \\& = \frac{18}{\sqrt{3}}\end{aligned}$

(b)

we want a unit vector $u$ that maximizes the value of ${\bf d_u} v(3,6,5)$ . so here, we fix the point (x,y,z) and find $\bf u$ . since we have the gradient vector defined as $\nabla v(x,y,z)=\langle v_x, v_y, v_z\rangle$ we can rewrite the formula for the directional derivative as

${\bf d_u} v(x,y,z) = \nabla v(x,y,z) \cdot {\bf u}.$

this is a dot product of two vectors. so if $\theta$ is the angle between these two vectors (where $0 \le \theta \le \pi$ ), then we have

$\nabla v(x,y,z) \cdot {\bf u} = | \nabla v(x,y,z)||{\bf u}|\cos\theta$

but clearly |u| = 1 since it is a unit vector, so we have

${\bf d_u}v(x,y,z)=|\nabla v(x,y,z)|\cos\theta.$

the maximum of this occurs when $\cos\theta$ has its maximum, and for $0 \le \theta \le \pi$ , the maximum of $\cos\theta$ occurs when $\theta = 0$ . this means the unit vector $\bf u$ has to be in the same direction as the gradient vector $\nabla v(x,y,z)$ .

and since $\cos 0 = 1$ , the maximum value of the directional derivative at a point is $\nabla v(x,y,z)|$ .

we calculated before that $\nabla v(3,6,5) = \langle30,6,18\rangle$ so v increases fastest in the direction that gradient vector.

(c)

as discussed before, maximum value of the directional derivative at a point is $\nabla v(x,y,z)|$ . so the maximum rate of change of v at point (3,6,5) is

$|\nabla v(3,6,5)|=|\langle30,6,18\rangle|=\sqrt{30^2+6^2+18^2}=\sqrt{1260}$

or since 1260 = 36*35, we have

$\sqrt{1260} = \sqrt{36\cdot35} = \sqrt{36}\cdot\sqrt{35} = 6\sqrt{35}$

as our final answer.

Answer from: Quest

What is the things about?

Answer from: Quest

hello there!

your answer is a “generator”

a generator does not create energy but instead it transforms it.

it’s basically a motor working in reverse.

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