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

explanation:

Answer from: Quest

Honestly "global warming" may be a slight possiblity, however they called it "climate change" because its not real, in fact we had one of our coldest days here not too long ago, and the "issues" with this subject is a blindfold for actual problems.

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Suppose that over a certain region of space the electrical potential v is given by the following equ...

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