Question

-2j is -3. what is the derivative of ƒ in the direction of -i - 2j? give reasons for your answer.

Answer 1

The derivative of ƒ in the direction of -i - 2j can be found by taking the dot product of the gradient of ƒ and the given direction vector.

Step-by-step explanation:

The derivative of ƒ in the direction of -i - 2j can be found by taking the dot product of the gradient of ƒ and the given direction vector.

First, we find the derivative of ƒ by taking its partial derivatives with respect to x and y. Let's say ƒ(x, y) = z.

The gradient of ƒ is given by ∇ƒ = (∂z/∂x)i + (∂z/∂y)j.

Now, taking the dot product of ∇ƒ and -i - 2j, we have:

-∇ƒ · (-i - 2j)

= -(-∂z/∂x)i · i - (-∂z/∂y)j · i - 2(-∂z/∂x)i · j - 2(-∂z/∂y)j · j

= ∂z/∂x + 2∂z/∂y

So, the derivative of ƒ in the direction of -i - 2j is ∂z/∂x + 2∂z/∂y.