If g(x, y) = x2 + y2 − 6x, find the gradient vector ∇g(2, 6) and use it to find the tangent line to the level curve g(x, y) = 28 at the point (2, 6).

The gradient of vector

   ∇g = i⁻(2 x -6 ) +  j⁻ (2y)

The gradient of vector at the point (2 ,6) is

( ∇g)₍₂,₆) =    (-2)  i⁻ +   (12) j⁻

Step-by-step explanation:

Explanation:-

Given  g (x, y) = x² +y² -6x ...(i)

The gradient of vector

     ∇g =( i⁻ δ/δx + j⁻ δ/δy+ k⁻δ/δZ)g

          = ( i⁻ δg/δx + j⁻ δg/δy+ k⁻δg/δZ)  (ii)

Differentiating equation (i) partially with respective to 'x' , treated 'y' as constant.

δg/δx = 2x - 6

Differentiating equation (i) partially with respective to 'x' , treated 'y' as constant.

δg/δy = 2y

Step(ii):-

The gradient of vector

   ∇g = i⁻(2 x -6 ) +  j⁻ (2y)

At the point ( 2,6)

( ∇g)₍₂,₆) = i⁻(2 (2) -6 ) +  j⁻ (2(6))

( ∇g)₍₂,₆) =    i⁻(-2) +  j⁻ (12)

2/3

step-by-step explanation:

the slope of a line is equal to the rise/run. in this case the line rises 2 units for every +3 units it moves

a

step-by-step explanation: