Respuesta :
Answer:
a) D_u . ∀ f ( 2 , 1 ) = 3 , Q = 72 degrees
b) D_u . ∀ f ( 2 , 1 ) = 4*sqrt(2) , Q = 80 degrees
c) D_u . ∀ f ( 2 , 1 ) = - sqrt(2) , Q = -54.74 degrees
d) u = 1 / sqrt(34) *(-3i +5j) , D_u . ∀ f ( 2 , 1 ) = +/- sqrt (34) , Q = 80.27 degrees
Step-by-step explanation:
Given:
- The terrain is modeled as a surface:
f(x , y) = x*y + y^3 - x^2
At point P( 2 , 1 , -1 ) is the current position:
Find:
(a) Determine the slope you would encounter if you headed due West from your position. What angle of inclination does this correspond to?
(b) Determine the slope you would encounter if you headed due North-West from your position. What angle of inclination does this correspond to?
(c) Determine the slope you would encounter if you headed due South-West from your position. What angle of inclination does this correspond to?
(d) Determine the steepest slope you could encounter from your position and the direction of that slope (as a unit vector).
Solution:
- We will compute the gradient of the vector ∀ f @ point P( 2 , 1 , -1 ) as follows:
∀ f ( x , y ) = (y - 2x ) i + ( x + 3y^2) j
- Evaluate at point P ( 2 , 1 ):
∀ f ( 2 , 1 ) = (1 - 2*2 ) i + ( 2 + 3*1^2) j
∀ f ( 2 , 1 ) = -3 i + 5 j
- We will use the result of ∀ f @ point P( 2 , 1 , -1 ) for the all the parts.
a)
- The direction due west can be written as a unit vector u_w = - i .
- Now compute the directional derivative in the direction of u_w = - i
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . - i
D_u . ∀ f ( 2 , 1 ) = 3
- Now compute the angle of inclination Q for the following direction:
Q = arctan(3) = 71.57 = 72 degrees
Hence, along the direction due west from position we ascend with an inclination of approximately 72 degrees.
b)
- The direction due north-west can be written as a unit vector u_w = - i + j .
- Now compute the directional derivative in the direction of u_w = - i + j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i + j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = 8 / sqrt(2) = 4*sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(4*sqrt(2)) = 79.98 = 80 degrees
Hence, along the direction due north-west from position we ascend with an inclination of approximately 80 degrees.
c)
- The direction due south-west can be written as a unit vector u_w = - i - j .
- Now compute the directional derivative in the direction of u_w = - i - j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i - j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = -2 / sqrt(2) = - sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(-sqrt(2)) = -54.74 degrees
Hence, along the direction due south-west from position we descend with an inclination of approximately 55 degrees.
d)
- We know that ∀ f ( 2 , 1 ) gradient points in the direction greatest increase, hence, So from P the direction of greatest increase is ∇f(P) = -3i +5j. The unit vector pointing in this direction is:
u = 1 / sqrt(34) *(-3i +5j)
- so we have:
D_u . ∀ f ( 2 , 1 ) = u . ∀ f ( 2 , 1 ) = (-3i +5j) . (-3i +5j) / sqrt(34)
= +/- sqrt (34)
- Hence, the steepest ascent or decent is of :
Q = arctan ( sqrt(34)) = 80.27 degrees
