Question

Consider the function ​f(x,y,z)equalse Superscript xyz minus 3​, the point P(0 comma 3 comma negative 2 )​, and the unit vector Bold uequalsleft angle seven ninths comma four ninths comma four ninths right angle .

a. Compute the gradient of f and evaluate it at P.
b. Find the unit vector in the direction of maximum increase of f at P.
c. Find the rate of change of the function in the direction of maximum increase at P.
d. Find the directional derivative at P in the direction of the given vector.

Answer 1

a) ∇f(0, 3, -2) = (-6*e⁻³, 0, 0)

b) u = -1 i + 0 j + 0 k = (-1, 0 , 0)

c) Df_u(0, 3, -2) = 6*e⁻³

d) Df_B(0, 3, -2) = (- 14/3)*(e⁻³)

Explanation:

a) We apply the formula

∇f = e∧(xyz-3)(yz) i + e∧(xyz-3)(xz) j + e∧(xyz-3)(xy) k

∇f(0, 3, -2) = e∧(-3)(-6) i + e∧(-3)(0) j + e∧(-3)(0) k = -6*e⁻³i + 0 j + 0 k

⇒ ∇f(0, 3, -2) = (-6*e⁻³, 0, 0)

b) u = ∇f / ║∇f║

we get ║∇f║ as follows

║∇f║= √((-6*e⁻³)²+(0)²+(0)²) = 6*e⁻³

⇒ u = (-6*e⁻³/6*e⁻³) i + (0/6*e⁻³) j + (0/6*e⁻³) k

⇒ u = -1 i + 0 j + 0 k = (-1, 0 , 0)

c) Df_u(P) = ∇f(0, 3, -2).u

⇒ Df_u(0, 3, -2) = (-6*e⁻³, 0, 0)(-1, 0, 0) = (-6*e⁻³)(-1)+(0)(0)+(0)(0)

⇒ Df_u(0, 3, -2) = 6*e⁻³

d) Df_B(0, 3, -2) = (-6*e⁻³, 0, 0)(7/9, 4/9, 4/9) = (-6*e⁻³)(7/9)+(0)(4/9)+(0)(4/9)

⇒ Df_B(0, 3, -2) = (- 14/3)*(e⁻³)

Answer 2

See the step by step solution.

Explanation:

The main topics of this question are gradient and directional derivative of a multivariable function. Given a function f(x,y,z), its' gradient is defined by

`\\abla f = (\frac[\partial f][\partial x], \frac[\partial f][\partial y],\frac[\partial f][\partial z])`. (i.e each of the components correspond to the partial derivative of the function f with respect each the its' variables). The gradient is a vector function, which means it could be evaluated at a given point
`P=(x_0,y_0,z_0)` by replacing x,y,z in the formulas with
`x_0,y_0,z_0` respectively.

a. The given function is
`e^(xyz)-3` and the given point is P=(0,3,-2).

The partial derivatives of f are:
`\frac[\partial f][\partial x]= yze^(xyz), \frac[\partial f][\partial y]= xze^(xyz), \frac[\partial f][\partial z]= xye^(xyz)`.

Then, the gradient function of f is
`\\abla f(x,y,z) =(yze^(xyz),xze^(xyz),xye^(xyz))`. So, in this case, we want to know
`\\abla f(0,3,-2)` which is
`\\abla f(0,3,-2)=(-6,0,0)`

b. The gradient function evaluated at a given point gives the vector in which there is maximum increase. On the other hand, an unit vector is a vector, whose norm is equal to 1. Then, we just need to divide the previous answer by its' norm. So, the unit vector
`\vec{v}(-6,0,0)/6 = (-1,0,0)` is the correct answer.

For questions c and d: The directional derivative (or rate of change) of a function f in the direction of vector v is given by the dot product of v and the gradient of f.

c. From b) we got that the vector of maximum increase at P is
`\vec{v}=(-1,0,0)`. So, the rate of change of f in that direction is (-6)*(-1)+0*0+0*0=6

d. Using the vector
`\vec{v} = (\frac[7][9],\frac[4][9],\frac[4][9])=` the rate of change of is given by (-6)*7/9+0*4/9+0*4/9 = -42/9.