①  Using Pythagoras, since  -i and -j each have length -1, vector v has a length of -√2 (negative because it's pointing in the opposite direction to the x=y axis). 

②  Vector p (in the x-direction)  has a gradient equal to the partial derivative ∂z/∂x.  We already know that ∂z/∂x = 2x. Since x = -0.5 at the point where the gradient is taken, we get ∂z/∂x = -1.

Since the tangent has a constant gradient, we can sat that ∂z/∂x = Δz/ Δx and we know that Δx = -1. Hence Δz = -1 x -1 = 1.

Similarly, Vector q (in the y-direction)  has a gradient equal to the partial derivative ∂z/∂y.  We already know that ∂z/∂y = 4y. Since y = -0.5 at the point where the gradient is taken, we get ∂z/∂y = -2. Hence Δz = ∂z/∂y x Δy =  -2 x -1 = 2.

Adding these two Δz terms together, we get the total z-component of vector u:

Δz = 1 + 2 = 3.

③  The gradient of vector u = Δz / length of vector v  =  -3/√2

(which is, of course, the same as the value we obtained from focusing on tangent in question 4A-4).