Question

A) What is the rate of change of \( f(x, y)=7 x y+y^{2} \) at the point \( (4,1 \) in the direction \( \vec{v}=6 \vec{i}-\vec{j} \) ? NOTE: Enter an exact answer or round to three decimal places. \[ f

A) 41
B) -11
C) 30
D) 16

Answer 1

The rate of change of the function f(x, y)=7xy+y² at the point (4,1) in the direction of the vector äv = 6êéi - êéj is found by computing the directional derivative. The directional derivative, approximately 1.974, is calculated as the dot product of the gradient of f at (4,1) and the normalized direction vector v.

Step-by-step explanation:

The question asks for the rate of change of the function f(x, y)=7xy+y² at the point (4,1) in the direction of the vector äv = 6êéi - êéj. To find this rate of change, we calculate the directional derivative of f at the given point in the direction of the vector v. The directional derivative is found by taking the dot product of the gradient of f and the unit vector in the direction of v.

First, we find the gradient of f at (4,1):

∇f(x, y) = ∇(7xy + y²) = (7y, 7x + 2y)

∇f(4, 1) = (7 · 1, 7 · 4 + 2 · 1) = (7, 30)

Next, we normalize the direction vector v:

ëvé = ë6êéi - êéjé = ë6, -1éá

ôüvé = ë6, -1é / √(6² + (-1)²) = ë6, -1é / √(37)

Finally, we compute the directional derivative as the dot product of the gradient and the normalized direction vector:

D_vf(4, 1) = ∇f(4, 1) · ôüvé

=(7, 30) · ë6/√(37), -1/√(37)é = 7·(6/√(37)) + 30·(-1/√(37))

= 42/√(37) - 30/√(37)

=(42 - 30)/√(37)

= 12/√(37)

Therefore, the rate of change of f at the point (4,1) in the direction of äv is 12/√(37), which is approximately 1.974.