Calculating Scalar Field Derivatives: A Step-by-Step Guide

Hey guys! Today, we're diving into a cool problem in calculus: figuring out the derivative of a scalar field. Specifically, we'll be looking at how to find the derivative of a function u = u(x, y, z) at a specific point M, along the direction of the normal n to a surface S. Sounds complex? Don't sweat it; we'll break it down step by step. This is a common problem in multivariable calculus, and understanding it is super important for all sorts of applications, from physics to engineering. Let's get started!

Understanding the Problem: Scalar Fields and Derivatives

Alright, so what exactly are we dealing with? We've got a scalar field, which is essentially a function that assigns a single numerical value (a scalar) to each point in space. In our case, the scalar field is represented by the function u = u(x, y, z) = 8 / (y - 4x²). This function tells us the value of u at any point (x, y, z). Think of it like a temperature map, where each point in space has a specific temperature associated with it.

Now, we want to find the derivative of this function. But not just any derivative – we want the derivative in a specific direction. This is where the concept of the directional derivative comes into play. The directional derivative tells us how the function u changes as we move from a point M along a particular direction. In our problem, the direction is defined by the normal vector n to the surface S. The surface S is given by the equation x² - 4z² - y = 0. The normal vector is essentially a vector that's perpendicular to the surface at a given point. Imagine a ball resting on the surface; the normal vector points directly upwards (or downwards) from the point of contact. Since, the normal vector n must form an acute angle with the positive direction of the z-axis (OZ). This condition imposes a constraint on the direction of the normal vector.

The point M(1, 0, 1/2) is the specific location where we want to calculate the derivative. So, we're not just looking at any point; we're focusing on this particular spot in space. The key is to find how u changes as we move away from M along the direction of the normal to the surface S at that point. The problem is to calculate the directional derivative, a concept deeply rooted in multivariable calculus. Understanding this process is crucial for anyone delving into fields that require spatial analysis and the study of functions in multiple dimensions. The directional derivative is a powerful tool for analyzing the behavior of scalar fields and understanding how they change in different directions. Let's dive in and make this concept our own!

Step-by-Step Solution: Finding the Derivative

Okay, let's get down to the nitty-gritty and solve this problem. We'll break it into several manageable steps to make sure we get it right.

1. Calculate the Gradient of the Scalar Field

The gradient of a scalar field is a vector that points in the direction of the greatest rate of increase of the function. It's a crucial tool for finding the directional derivative. The gradient of u is denoted by ∇u and is calculated as follows:

∇u = (∂u/∂x, ∂u/∂y, ∂u/∂z)

First, we'll find the partial derivatives of u with respect to x, y, and z:

• ∂u/∂x = ∂/∂x (8 / (y - 4x²)) = 8 * (-1) * (-8x) / (y - 4x²)² = 64x / (y - 4x²)²
• ∂u/∂y = ∂/∂y (8 / (y - 4x²)) = 8 * (-1) / (y - 4x²)² = -8 / (y - 4x²)²
• ∂u/∂z = ∂/∂z (8 / (y - 4x²)) = 0 (since u does not depend on z)

So, the gradient of u is:

∇u = (64x / (y - 4x²)², -8 / (y - 4x²)², 0)

2. Find the Gradient at Point M

Now, we need to evaluate the gradient at the specific point M(1, 0, 1/2). We substitute the coordinates of M (x = 1, y = 0, z = 1/2) into the gradient:

∇u(M) = (64(1) / (0 - 4(1)²)², -8 / (0 - 4(1)²)², 0) = (64/16, -8/16, 0) = (4, -1/2, 0)

3. Find the Normal Vector to the Surface S

The surface S is given by the equation x² - 4z² - y = 0. To find the normal vector to this surface, we can use the gradient of the function f(x, y, z) = x² - 4z² - y. The gradient of f is normal to the surface S. So, let's find that gradient:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Calculating the partial derivatives:

• ∂f/∂x = 2x
• ∂f/∂y = -1
• ∂f/∂z = -8z

Therefore, ∇f = (2x, -1, -8z). Now, we evaluate this gradient at the point M(1, 0, 1/2):

∇f(M) = (2(1), -1, -8(1/2)) = (2, -1, -4)

So, the normal vector to the surface S at point M is (2, -1, -4).

4. Normalize the Normal Vector

To use the normal vector in our calculations, we need to normalize it. This means we need to make it a unit vector (a vector with a magnitude of 1). We do this by dividing each component of the vector by its magnitude.

First, find the magnitude of the normal vector:

||(2, -1, -4)|| = √(2² + (-1)² + (-4)²) = √(4 + 1 + 16) = √21

Now, normalize the vector:

n = (2/√21, -1/√21, -4/√21)

5. Check the Angle Condition

The problem states that the normal vector n must form an acute angle with the positive direction of the z-axis. This means the z-component of the normal vector should be positive. However, in our calculated normal vector, the z-component is negative (-4/√21). Thus, we need to reverse the direction of the normal vector to satisfy this condition. So,

n = (-2/√21, 1/√21, 4/√21)

6. Calculate the Directional Derivative

The directional derivative of u in the direction of n at point M is given by the dot product of the gradient of u at M and the normalized normal vector n:

Du/Dn = ∇u(M) · n

Plugging in our values:

Du/Dn = (4, -1/2, 0) · (-2/√21, 1/√21, 4/√21) = (4 * -2/√21) + (-1/2 * 1/√21) + (0 * 4/√21) = -8/√21 - 1/(2√21) = -17/(2√21)

7. Simplify and Final Answer

We can rationalize the denominator to simplify the result:

-17/(2√21) = -17√21 / (2 * 21) = -17√21 / 42

So, the directional derivative of u at point M along the direction of the normal vector n is -17√21 / 42. This is the rate of change of the scalar field u at the point M in the specified direction. The negative value indicates that the function u is decreasing in this direction. The process of finding the directional derivative is fundamental in understanding the behavior of scalar fields. It is used extensively in physics, engineering, and various scientific disciplines to analyze and model phenomena that vary in space. This value represents the instantaneous rate of change of u at point M in the direction of n.

Conclusion: Key Takeaways

Alright guys, we've just successfully found the derivative of a scalar field at a specific point along a given direction! Here are the main takeaways from this problem:

• Scalar Fields: These functions assign a single value to each point in space.
• Gradient: The gradient points in the direction of the greatest rate of increase.
• Normal Vector: It's perpendicular to the surface.
• Directional Derivative: It tells you how the function changes in a specific direction. This is calculated using the dot product of the gradient and the normalized normal vector.

By understanding these concepts and practicing these steps, you'll be well on your way to mastering multivariable calculus. Remember to practice and apply these concepts to different problems to solidify your understanding. Keep up the great work, and happy calculating!

I hope this explanation was helpful. If you've got any questions or want to work through another problem, just ask! Good luck, and keep exploring the fascinating world of calculus. This approach ensures you grasp the core concepts and can apply them confidently. The skills gained here are crucial for tackling complex problems and are applicable in many areas of science and engineering. Keep practicing, and you will become a pro in no time! Understanding these core elements will help you build a solid foundation in multivariable calculus. Well done, guys!