To solve these questions, we need to calculate the partial derivatives of the function $z = f(x, y) = xy + y^3 - x^2$ to determine the slopes in various directions and the steepest slope.

Step 1: Compute Partial Derivatives

We start by calculating the partial derivatives $f_x$ and $f_y$ at the point $(x, y) = (2, 1)$.

Given: $f(x, y) = xy + y^3 - x^2$

1. Partial derivative with respect to $x$: $f_x = \frac{\partial}{\partial x}(xy + y^3 - x^2) = y - 2x$ Substitute $(x, y) = (2, 1)$: $f_x(2, 1) = 1 - 2 \cdot 2 = 1 - 4 = -3$

2. Partial derivative with respect to $y$: $f_y = \frac{\partial}{\partial y}(xy + y^3 - x^2) = x + 3y^2$ Substitute $(x, y) = (2, 1)$: $f_y(2, 1) = 2 + 3 \cdot 1^2 = 2 + 3 = 5$

So, the gradient vector at the point $(2, 1, -1)$ is: $\nabla f(2, 1) = \langle f_x(2, 1), f_y(2, 1) \rangle = \langle -3, 5 \rangle$

Part (a): Slope Heading Due West

When heading due West, the direction vector is $\langle -1, 0 \rangle$.

The slope in a given direction is the directional derivative, which is calculated as: $D_u f = \nabla f \cdot u = \langle -3, 5 \rangle \cdot \langle -1, 0 \rangle$ $D_u f = (-3)(-1) + (5)(0) = 3$

Thus, the slope heading due West is $\boxed{3}$.

Part (b): Slope Heading Due North-West

When heading due North-West, the direction vector is $\langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$.

The slope in this direction is: $D_u f = \nabla f \cdot u = \langle -3, 5 \rangle \cdot \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$ $D_u f = \left(-3 \cdot -\frac{1}{\sqrt{2}}\right) + \left(5 \cdot \frac{1}{\sqrt{2}}\right) = \frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$

Thus, the slope heading North-West is $\boxed{4\sqrt{2}}$.

Part (c): Slope Heading Due South-East

When heading due South-East, the direction vector is $\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \rangle$.

The slope in this direction is: $D_u f = \nabla f \cdot u = \langle -3, 5 \rangle \cdot \langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \rangle$ $D_u f = \left(-3 \cdot \frac{1}{\sqrt{2}}\right) + \left(5 \cdot -\frac{1}{\sqrt{2}}\right) = -\frac{3}{\sqrt{2}} - \frac{5}{\sqrt{2}} = -\frac{8}{\sqrt{2}} = -4\sqrt{2}$

Thus, the slope heading South-East is $\boxed{-4\sqrt{2}}$.

Part (d): Steepest Slope

The steepest slope is in the direction of the gradient vector $\nabla f(2, 1) = \langle -3, 5 \rangle$. The magnitude of this vector gives the steepest slope.

$|\nabla f(2, 1)| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$

Therefore, the steepest slope is $\boxed{\sqrt{34}}$.

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Would you like a further breakdown of any part, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How does the gradient vector indicate the direction of the steepest ascent?
2. What would the slope be if you headed due South from your position?
3. Can you calculate the slope for any arbitrary direction on this terrain?
4. How would the results change if you were at a different point on this terrain, such as $(1, 1)$?
5. What does a negative slope indicate about the terrain in that direction?

Tip: In multivariable calculus, the gradient vector not only gives the steepest ascent but also provides valuable insight into the behavior of the function in any given direction.