(4A-5)
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).