Exploring Directional Derivatives Function F(x, Y) At Point (1, 1)

Hey guys! Let's dive into an exciting problem in multivariable calculus – directional derivatives! We're going to explore a function ${ f(x, y) }$ at a specific point, ${ (1, 1) }$, and see how its rate of change behaves in different directions. It's like being a hiker on a mountain and figuring out how steep the path is in various directions. So, buckle up, and let's get started!

Understanding Directional Derivatives

Before we jump into the nitty-gritty, let's quickly recap what directional derivatives are all about. Imagine you're standing on a surface defined by a function ${ f(x, y) }$. The directional derivative tells you the rate of change of the function at a particular point in a specific direction. It's essentially the slope of the surface in that direction. Think of it like walking on a hill; the directional derivative tells you how steep the hill is in the direction you're walking.

Directional derivatives are a fundamental concept in multivariable calculus, providing a way to analyze the rate of change of a function along any direction. Unlike partial derivatives, which only consider changes along the coordinate axes, directional derivatives offer a more general perspective. The directional derivative is the rate at which the function changes at a given point in the direction of a specific vector. This vector, often a unit vector, defines the path along which we are measuring the change. Understanding directional derivatives allows us to analyze how a function behaves in various directions, crucial for optimization problems, physics applications, and more. For instance, consider a temperature distribution across a surface; the directional derivative can tell you the direction in which the temperature changes most rapidly. This concept extends the idea of a slope in single-variable calculus to surfaces in three dimensions, providing a powerful tool for analyzing and understanding multivariable functions.

The formula for the directional derivative of ${ f(x, y) }$ in the direction of a unit vector ${ \mathbf{u} = \langle a, b \rangle }$ is given by:

${ D_{\mathbf{u}} f(x, y) = \nabla f(x, y) \cdot \mathbf{u} = f_x(x, y)a + f_y(x, y)b }$

Where ${ \nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle }$ is the gradient of ${ f }$, and ${ f_x }$ and ${ f_y }$ are the partial derivatives of ${ f }$ with respect to ${ x }$ and ${ y }$, respectively.

The Problem at Hand

Now, let's tackle the specific problem. We have a function ${ f(x, y) }$, and we're looking at it at the point ${ (1, 1) }$. We're given two key pieces of information:

1. The directional derivative in the direction parallel to ${ \langle 2, 1 \rangle }$ is ${ \frac{4}{\sqrt{5}} }$.
2. [The directional derivative in another direction will be given, which is required to complete the problem].

Our goal is to extract as much information about the function ${ f(x, y) }$ as we can from these clues. This usually involves finding the gradient of the function at the point ${ (1, 1) }$, which tells us the direction of the steepest ascent and the rate of change in that direction.

Step-by-Step Solution

1. Normalize the Direction Vector

The first thing we need to do is normalize the direction vector ${ \langle 2, 1 \rangle }$. Normalizing a vector means turning it into a unit vector, which has a length of 1. This is important because the directional derivative formula uses a unit vector. To normalize, we divide the vector by its magnitude.

The magnitude of ${ \langle 2, 1 \rangle }$ is:

${ \| \langle 2, 1 \rangle \| = \sqrt{2^2 + 1^2} = \sqrt{5} }$

So, the unit vector ${ \mathbf{u} }$ in the direction of ${ \langle 2, 1 \rangle }$ is:

${ \mathbf{u} = \frac{\langle 2, 1 \rangle}{\sqrt{5}} = \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle }$

Normalizing the direction vector is a critical step in computing directional derivatives. The directional derivative formula relies on using a unit vector because it isolates the direction of change from the magnitude of the vector. A unit vector has a magnitude of 1, ensuring that the calculated derivative reflects only the rate of change in the specified direction, not the length of the vector itself. To normalize a vector, you divide each component of the vector by its magnitude. The normalization process is essential because it allows for consistent comparisons of rates of change across different directions. If we were to use a non-unit vector, the resulting value would be a scaled version of the actual directional derivative, making it difficult to interpret the true rate of change. Therefore, normalization ensures that the directional derivative accurately represents the rate of change of the function in the desired direction.

2. Use the Directional Derivative Formula

We know that the directional derivative in the direction of ${ \mathbf{u} }$ at ${ (1, 1) }$ is ${ \frac{4}{\sqrt{5}} }$. So, using the directional derivative formula, we have:

${ D_{\mathbf{u}} f(1, 1) = \nabla f(1, 1) \cdot \mathbf{u} = f_x(1, 1)\left(\frac{2}{\sqrt{5}}\right) + f_y(1, 1)\left(\frac{1}{\sqrt{5}}\right) = \frac{4}{\sqrt{5}} }$

This gives us our first equation:

${ 2f_x(1, 1) + f_y(1, 1) = 4 \quad (1) }$

Applying the directional derivative formula is the core step in relating the given information to the partial derivatives of the function. This formula, ${ D_{\mathbf{u}} f(x, y) = \nabla f(x, y) \cdot \mathbf{u} }$, connects the directional derivative to the gradient of the function and the unit direction vector. By substituting the known values—the directional derivative, the unit vector, and the point of evaluation—we establish an equation involving the partial derivatives ${ f_x }$ and ${ f_y }$. This equation is crucial because it provides a relationship between the rates of change of the function along the x and y axes. The directional derivative formula effectively decomposes the directional change into its component contributions along the coordinate axes. In our specific case, this application leads to the equation ${ 2f_x(1, 1) + f_y(1, 1) = 4 }$, which is a linear equation in terms of the partial derivatives at the point ${ (1, 1) }$. This is the first piece of a system of equations that we will need to solve to find the gradient.

3. Obtain a Second Equation

To find both ${ f_x(1, 1) }$ and ${ f_y(1, 1) }$, we need another piece of information – another directional derivative in a different direction. Let's assume (for the sake of continuing the solution) that we're given that the directional derivative in the direction parallel to ${ \langle 1, -1 \rangle }$ is ${ 0 }$. We'll go through the same process as before.

First, normalize the direction vector ${ \langle 1, -1 \rangle }$:

${ \| \langle 1, -1 \rangle \| = \sqrt{1^2 + (-1)^2} = \sqrt{2} }$

The unit vector ${ \mathbf{v} }$ is:

${ \mathbf{v} = \frac{\langle 1, -1 \rangle}{\sqrt{2}} = \left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right\rangle }$

Now, use the directional derivative formula:

${ D_{\mathbf{v}} f(1, 1) = \nabla f(1, 1) \cdot \mathbf{v} = f_x(1, 1)\left(\frac{1}{\sqrt{2}}\right) + f_y(1, 1)\left(-\frac{1}{\sqrt{2}}\right) = 0 }$

This simplifies to:

${ f_x(1, 1) - f_y(1, 1) = 0 \quad (2) }$

Obtaining a second equation is a crucial step because it allows us to solve for the individual partial derivatives. In this problem, we've assumed that the directional derivative in the direction of ${ \langle 1, -1 \rangle }$ is 0. Following the same process of normalizing the direction vector and applying the directional derivative formula, we derive a second equation involving ${ f_x(1, 1) }$ and ${ f_y(1, 1) }$. This equation, ${ f_x(1, 1) - f_y(1, 1) = 0 }$, provides a different relationship between the partial derivatives, making it possible to solve the system of equations. The importance of having two independent equations cannot be overstated; without the second equation, we would be left with an underdetermined system, and we could not uniquely determine the values of ${ f_x(1, 1) }$ and ${ f_y(1, 1) }$. By obtaining a second equation, we set the stage for solving for the components of the gradient vector.

4. Solve the System of Equations

We now have a system of two linear equations:\begin{align*}2f_x(1, 1) + f_y(1, 1) &= 4 \f_x(1, 1) - f_y(1, 1) &= 0 \end{align*}

We can solve this system using various methods, such as substitution or elimination. Let's use elimination. Add the two equations together: ${ (2f_x(1, 1) + f_y(1, 1)) + (f_x(1, 1) - f_y(1, 1)) = 4 + 0 }$

${ 3f_x(1, 1) = 4 }$

${ f_x(1, 1) = \frac{4}{3} }$

Now, substitute this value back into equation (2):

${ \frac{4}{3} - f_y(1, 1) = 0 }$

${ f_y(1, 1) = \frac{4}{3} }$

So, we've found that ${ f_x(1, 1) = \frac{4}{3} }$ and ${ f_y(1, 1) = \frac{4}{3} }$.

Solving the system of equations is the culmination of the process, where we leverage the relationships established between the partial derivatives to find their specific values. With two linear equations derived from the directional derivative information, we can employ standard algebraic techniques such as substitution, elimination, or matrix methods to determine the values of ${ f_x(1, 1) }$ and ${ f_y(1, 1) }$. In our example, using the elimination method, we add the two equations together to eliminate ${ f_y(1, 1) }$, allowing us to solve for ${ f_x(1, 1) }$. Subsequently, we substitute this value back into one of the original equations to find ${ f_y(1, 1) }$. The result of this step is the determination of the partial derivatives at the point of interest, which in our case, are ${ f_x(1, 1) = \frac{4}{3} }$ and ${ f_y(1, 1) = \frac{4}{3} }$. This is a crucial milestone, as these values form the components of the gradient vector.

5. Determine the Gradient Vector

Now that we have the partial derivatives, we can find the gradient vector at ${ (1, 1) }$:

${ \nabla f(1, 1) = \left\langle f_x(1, 1), f_y(1, 1) \right\rangle = \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle }$

Determining the gradient vector is the final step in synthesizing the information gathered to understand the behavior of the function at the given point. The gradient vector, ${ \nabla f(1, 1) }$, is a vector composed of the partial derivatives of the function evaluated at the point ${ (1, 1) }$. In our case, with ${ f_x(1, 1) = \frac{4}{3} }$ and ${ f_y(1, 1) = \frac{4}{3} }$, the gradient vector is ${ \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle }$. This vector points in the direction of the steepest ascent of the function at the point ${ (1, 1) }$, and its magnitude represents the maximum rate of change of the function at that point. The gradient vector is a powerful tool for analyzing multivariable functions, providing insight into the direction and magnitude of the greatest increase in the function's value. It is fundamental in optimization problems, where the goal is to find the maximum or minimum values of a function, and in various applications in physics and engineering.

Interpreting the Results

The gradient vector ${ \nabla f(1, 1) = \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle }$ tells us some important things about the function ${ f(x, y) }$ at the point ${ (1, 1) }$:

1. Direction of Steepest Ascent: The direction of the steepest increase in the function's value is in the direction of the vector ${ \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle }$, which is the same as the direction ${ \langle 1, 1 \rangle }$.

2. Maximum Rate of Change: The magnitude of the gradient vector gives us the maximum rate of change. The magnitude is:

   ${ \left\| \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle \right\| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3} }$

   So, the maximum rate of change of ${ f }$ at ${ (1, 1) }$ is ${ \frac{4\sqrt{2}}{3} }$.

Interpreting the results derived from the gradient vector provides crucial insights into the behavior of the function at the point of interest. The gradient vector, ${ \nabla f(1, 1) = \left\langle \frac{4}{3}, \frac{4}{3} \right\rangle }$, offers two primary pieces of information: the direction of the steepest ascent and the maximum rate of change. The direction of steepest ascent is given by the direction of the gradient vector itself, which in our case is the same as the direction ${ \langle 1, 1 \rangle }$. This means that if we were to move from the point ${ (1, 1) }$ in the direction of ${ \langle 1, 1 \rangle }$, we would experience the most rapid increase in the function's value. The maximum rate of change is quantified by the magnitude of the gradient vector, calculated as ${ \frac{4\sqrt{2}}{3} }$. This value represents the rate at which the function increases in the direction of steepest ascent. By understanding these interpretations, we gain a comprehensive understanding of how the function changes around the point ${ (1, 1) }$, which is vital for applications in optimization, physics, and other fields.

Conclusion

And there you have it! By using the concept of directional derivatives and some clever algebra, we were able to determine the gradient of the function ${ f(x, y) }$ at the point ${ (1, 1) }$. This gradient gives us valuable information about the function's behavior – the direction of steepest ascent and the maximum rate of change. These types of problems really show how powerful multivariable calculus can be in analyzing functions in multiple dimensions.

I hope you found this explanation helpful, guys! Keep exploring the world of calculus, and you'll discover even more fascinating concepts and applications!