Problem 258 - Calculus Volume 3

Question

The \(x\) and \(y\) components of a fluid moving in two dimensions are given by the following functions: \(u(x, y)=2 y\) and \(v(x, y)=-2 x ; \quad x \geq 0 ; y \geq 0 .\) The speed of the fluid at the point \((x, y)\) is \(s(x, y)=\sqrt{u(x, y)^{2}+v(x, y)^{2}}\). Find \(\frac{\partial s}{\partial x}\) and \(\frac{\partial s}{\partial y}\) using the chain rule.

Step-by-Step Solution

Answer

\(\frac{\partial s}{\partial x} = \frac{4x}{\sqrt{x^2 + y^2}}\) and \(\frac{\partial s}{\partial y} = \frac{4y}{\sqrt{x^2 + y^2}}\).

Step 1: Express the Speed Function

The speed of the fluid is given by the equation \( s(x,y) = \sqrt{u(x, y)^{2} + v(x, y)^{2}} \). Substituting the expressions for \( u(x, y) \) and \( v(x, y) \), we get \( s(x,y) = \sqrt{(2y)^{2} + (-2x)^{2}} \), which simplifies to \( s(x,y) = \sqrt{4y^{2} + 4x^{2}} \). Further simplification gives \( s(x,y) = \sqrt{4(y^{2} + x^{2})} = 2\sqrt{x^{2} + y^{2}} \).

Step 2: Calculate Partial Derivative with respect to x

To find \( \frac{\partial s}{\partial x} \), apply the chain rule. Let \( z = x^2 + y^2 \). Then \( s(x, y) = 2\sqrt{z} \). The derivative \( \frac{ds}{dz} = \frac{1}{\sqrt{z}} \). Substitute \( z = x^2 + y^2 \), so \( \frac{dz}{dx} = 2x \). Thus, \( \frac{\partial s}{\partial x} = \frac{ds}{dz} \cdot \frac{dz}{dx} = 2 \cdot \frac{1}{\sqrt{x^2 + y^2}} \cdot 2x = \frac{4x}{\sqrt{x^2 + y^2}} \).

Step 3: Calculate Partial Derivative with respect to y

To find \( \frac{\partial s}{\partial y} \), apply similar steps using the chain rule. With \( z = x^2 + y^2 \), and \( s(z) = 2 \sqrt{z} \), the derivative with respect to \( z \) is \( \frac{1}{\sqrt{z}} \). For \( z \), \( \frac{dz}{dy} = 2y \). Thus, \( \frac{\partial s}{\partial y} = \frac{ds}{dz} \cdot \frac{dz}{dy} = 2 \cdot \frac{1}{\sqrt{x^2 + y^2}} \cdot 2y = \frac{4y}{\sqrt{x^2 + y^2}} \).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus that involve taking the derivative of a function with respect to one variable while keeping others constant. In this exercise, we looked at finding the partial derivatives of the speed function \( s(x, y) = 2\sqrt{x^2 + y^2} \) with respect to both \( x \) and \( y \).

Here's a simple breakdown to understand it better:

• **Keeping Variables Constant**: When finding \( \frac{\partial s}{\partial x} \), treat \( y \) as a constant and differentiate only with respect to \( x \). Conversely, treat \( x \) as constant when finding \( \frac{\partial s}{\partial y} \).
• **Why Use Partial Derivatives?**: These derivatives are crucial in finding how a function changes in multi-dimensional spaces, like in this fluid dynamics problem where movement happens in a plane.

By following the steps given, you learn how to apply this concept using an example function, showing changes in speed with respect to the variables independently.

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of a composition of two or more functions. You apply it when a function relies on another function of a variable. In our exercise, the speed \( s(x,y) \) is expressed as a function of \( z = x^2 + y^2 \), which helps break down into smaller problems.

Here’s how you can think about using the chain rule more clearly:

• **Identify Inner and Outer Functions**: For \( s(x, y) = 2\sqrt{x^2 + y^2} \), recognize it as \( s(z) = 2\sqrt{z} \) where \( z = x^2 + y^2 \).
• **Differentiate Step-by-Step**: First, find the derivative of the outer function in terms of the inner \( z \), then multiply by the derivative of \( z \) itself with respect to \( x \) or \( y \).
• **Combine Results**: This gives a complete student understanding through a modular approach, handling complex derivatives by dividing them into simpler parts.

Using the chain rule allows you to tackle more sophisticated problems in multi-variable calculus confidently by linking related functions together.

Vector Fields

Vector fields are mathematical constructs used in physics and engineering to represent quantities with both magnitude and direction across a region. In 2D, vector fields can describe things like fluid flows or electromagnetic fields. Here, our exercise involves fluid movement represented by the vector field with components \(u(x, y) = 2y\) and \(v(x, y) = -2x\).

Key insights into vector fields:

• **Basics of Vector Fields**: Each point \((x, y)\) has a vector \(\mathbf{F}(x, y) = \langle u(x, y), v(x, y) \rangle\) showing the direction and speed of movement at that location.
• **Application in Real World**: Understanding such components aids in modeling how a fluid flows—critical for fields such as meteorology, oceanography, and aerodynamics.
• **Graphical Representation**: Visualizing vector fields helps you see patterns of flow or force across a region. This often makes abstract concepts more tangible.

Grasping vector fields helps in comprehending how dynamic systems behave, influencing not just abstract mathematics but practical applications in scientific and engineering disciplines.