Respuesta :
Answer:
See the step by step solution.
Step-by-step 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
[tex]\nabla f = (\frac[\partial f][\partial x], \frac[\partial f][\partial y],\frac[\partial f][\partial z])[/tex]. (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 [tex]P=(x_0,y_0,z_0)[/tex] by replacing x,y,z in the formulas with [tex]x_0,y_0,z_0[/tex] respectively.
a. The given function is [tex] e^{xyz}-3[/tex] and the given point is P=(0,3,-2).
The partial derivatives of f are: [tex]\frac[\partial f][\partial x]= yze^{xyz}, \frac[\partial f][\partial y]= xze^{xyz}, \frac[\partial f][\partial z]= xye^{xyz}[/tex].
Then, the gradient function of f is [tex]\nabla f(x,y,z) =(yze^{xyz},xze^{xyz},xye^{xyz})[/tex]. So, in this case, we want to know [tex]\nabla f(0,3,-2)[/tex] which is [tex]\nabla f(0,3,-2)=(-6,0,0)[/tex]
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 [tex]\vec{v}(-6,0,0)/6 = (-1,0,0)[/tex] 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 [tex]\vec{v}=(-1,0,0)[/tex]. So, the rate of change of f in that direction is (-6)*(-1)+0*0+0*0=6
d. Using the vector [tex]\vec{v} = (\frac[7][9],\frac[4][9],\frac[4][9])=[/tex] the rate of change of is given by (-6)*7/9+0*4/9+0*4/9 = -42/9.
Answer:
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)
Step-by-step 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)
