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The gradient is like a compass pointing in the direction of the steepest increase of a function. It is a vector that consists of the partial derivatives of the function with respect to each of the variables. So, if you have a function of two variables, like in our case, the gradient is a 2D vector. For the function \( f(x, y) = x^3 - x^2y + xy^2 + y^3 \), the gradient \( abla f \) is calculated by finding:

• The partial derivative with respect to \( x \): \( f_x = 3x^2 - 2xy + y^2 \).
• The partial derivative with respect to \( y \): \( f_y = -x^2 + 2xy + 3y^2 \).

Therefore, the gradient is \( abla f = (f_x, f_y) = (3x^2 - 2xy + y^2, -x^2 + 2xy + 3y^2) \). The gradient at a specific point gives not only the direction but also the rate of the fastest increase from that point. When evaluated at \( P(1, -1) \), it becomes \( abla f(1, -1) = (6, 0) \), which means the increase is in the x-direction.

Partial derivatives help us understand how a function changes as we tweak one variable at a time, keeping the others constant. For a function with several variables, each partial derivative shows the rate of change of the function as you change one variable, which essentially "slices" the function.Take the function \( f(x, y) = x^3 - x^2y + xy^2 + y^3 \). The two partial derivatives are:

• \( f_x = 3x^2 - 2xy + y^2 \) shows how \( f \) changes as \( x \) changes, keeping \( y \) constant.
• \( f_y = -x^2 + 2xy + 3y^2 \) shows how \( f \) changes as \( y \) changes, keeping \( x \) constant.

These derivatives are essential because they form the components of the gradient vector, directing us toward how the function behaves in different directions when one of the variables is changed.

A unit vector is a vector with a magnitude of 1. It is used to indicate direction, without contributing to the vector's magnitude or length.To find a unit vector in the direction of \( \mathbf{v} = 2 \mathbf{i} + 3 \mathbf{j} \), you first need to calculate the magnitude of \( \mathbf{v} \). This is done using the formula:\[ |\mathbf{v}| = \sqrt{2^2 + 3^2} = \sqrt{13} \]Then, you divide \( \mathbf{v} \) by its magnitude to convert it into a unit vector \( \mathbf{u} \):\[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \left( \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) \]This unit vector \( \mathbf{u} \) indicates the direction we are interested in, scaled to a size of 1, making it ideal for computing things like directional derivatives.

The dot product is a way of multiplying two vectors to find a scalar (a single number). It's useful for finding the projection of one vector onto another and is central to calculating the directional derivative.The dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In the context of directional derivatives, you find the dot product of the gradient at a point and the unit vector in the direction you're interested in. For example, with the gradient \( abla f(1, -1) = (6, 0) \) and the unit vector \( \mathbf{u} = \left( \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \right) \), the directional derivative is:\[ D_{\mathbf{u}} f(P) = abla f(1, -1) \cdot \mathbf{u} = 6 \times \frac{2}{\sqrt{13}} + 0 \times \frac{3}{\sqrt{13}} = \frac{12}{\sqrt{13}} \]This result shows the rate of change of the function \( f \) at the point in the direction of \( \mathbf{u} \).