Understanding the Directional Derivative for f(x, y) = 0

The question asks us to find the directional derivative of the function $f(x, y) = 0$ at the point $c = (0, 0)$ in the direction of the vector $u = (a, b)$. Let's break down what this means and how to calculate it.

What is a Directional Derivative?

The directional derivative of a function $f(x, y)$ at a point $(x_0, y_0)$ in the direction of a unit vector $\mathbf{\hat{u}} = (u_1, u_2)$ measures the rate of change of the function at that point as you move in the direction of $\mathbf{\hat{u}}$. It is calculated using the dot product of the gradient of the function at the point and the unit direction vector:

Directional Derivative $= \nabla f(x_0, y_0) \cdot \mathbf{\hat{u}}$

where $\nabla f(x, y)$ is the gradient vector, given by $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$.

Calculating the Directional Derivative

We are given the function $f(x, y) = 0$. This is a very simple function – its value is always zero, regardless of the input values of $x$ and $y$.

Step 1: Find the Partial Derivatives

We need to calculate the partial derivatives of $f(x, y)$ with respect to $x$ and $y$.

• The partial derivative with respect to $x$, denoted as $\frac{\partial f}{\partial x}$, treats $y$ as a constant and differentiates with respect to $x$. Since $f(x, y) = 0$, which is a constant, its derivative with respect to $x$ is 0.

  $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(0) = 0$

• The partial derivative with respect to $y$, denoted as $\frac{\partial f}{\partial y}$, treats $x$ as a constant and differentiates with respect to $y$. Since $f(x, y) = 0$, which is a constant, its derivative with respect to $y$ is also 0.

  $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(0) = 0$

Step 2: Find the Gradient Vector

The gradient vector $\nabla f(x, y)$ is formed by these partial derivatives:

$\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (0, 0)$

Step 3: Evaluate the Gradient at the Given Point

We need to find the gradient at the point $c = (0, 0)$.

$\nabla f(0, 0) = (0, 0)$

The gradient of $f(x, y) = 0$ is the zero vector $(0, 0)$ at any point $(x, y)$, including $(0, 0)$.

Step 4: Determine the Unit Direction Vector

The direction vector is given as $u = (a, b)$. To use the formula for the directional derivative, we need the unit vector in this direction, $\mathbf{\hat{u}}$. The unit vector is found by dividing the vector by its magnitude:

$\mathbf{\hat{u}} = \frac{u}{|u|} = \frac{(a, b)}{\sqrt{a^2 + b^2}}$, provided $u \neq (0, 0)$.

Let $\mathbf{\hat{u}} = (u_1, u_2)$. The exact values of $u_1$ and $u_2$ depend on $a$ and $b$, but the magnitude $\sqrt{u_1^2 + u_2^2}$ will be 1 (if $u \neq (0, 0)$).

If $u = (0, 0)$, the direction is undefined. However, the standard definition of the directional derivative often results in 0 in such cases or it might be considered undefined depending on context. Let's proceed with the calculation using the gradient.

Step 5: Calculate the Directional Derivative using the Dot Product

Now we compute the dot product of the gradient at $(0, 0)$ and the unit direction vector $\mathbf{\hat{u}} = (u_1, u_2)$.

Directional Derivative $= \nabla f(0, 0) \cdot \mathbf{\hat{u}} = (0, 0) \cdot (u_1, u_2)$

The dot product is calculated as $(0 \times u_1) + (0 \times u_2) = 0 + 0 = 0$.

Alternatively, using the original direction vector $u = (a, b)$ instead of the unit vector (some definitions use the non-unit vector, leading to the same directional derivative value when the gradient is zero), the calculation is still:

Directional Derivative $= \nabla f(0, 0) \cdot u = (0, 0) \cdot (a, b) = (0 \times a) + (0 \times b) = 0 + 0 = 0$.

In either case, because the gradient vector of $f(x, y) = 0$ is the zero vector at $(0, 0)$, the directional derivative in any direction $u = (a, b)$ is 0.

Conclusion

The directional derivative of $f(x, y) = 0$ at $c = (0, 0)$ along the direction $u = (a, b)$ is 0.

Revision Table: Key Concepts

Review the fundamental concepts used in this problem:

Additional Information: Constant Functions and Derivatives

Understanding why the directional derivative is 0 for $f(x, y) = 0$ is straightforward if you think about the function's behavior. The function $f(x, y) = 0$ represents a flat plane at $z=0$ in 3D space. Since the function's value is constant everywhere, there is no change in height (the value of $f$) no matter which direction you move from any point, including $(0, 0)$.

• Constant Function: A function $f$ is constant if $f(x, y) = C$ for some constant $C$ and for all $(x, y)$ in its domain. In this problem, $C=0$.
• Derivatives of Constants: The derivative of any constant with respect to any variable is always zero. This applies to both single-variable and partial derivatives.

  For $f(x, y) = C$, $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.

• Gradient of a Constant Function: Since all partial derivatives are zero, the gradient vector of a constant function is always the zero vector: $\nabla f = (0, 0)$.
• Directional Derivative with Zero Gradient: If the gradient $\nabla f$ is the zero vector at a point, the directional derivative at that point in any direction $\mathbf{\hat{u}}$ is $\nabla f \cdot \mathbf{\hat{u}} = (0, 0) \cdot \mathbf{\hat{u}} = 0$. This confirms that for a constant function, the rate of change is zero in all directions.

This problem demonstrates a fundamental concept: the rate of change of a constant function is always zero.