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Fluid flow The \(x\) - and \(y\) -components of a fluid moving in two dimensions are given by the following functions \(u\) and \(v .\) The speed of the fluid at \((x, y)\) is \(s(x, y)=\sqrt{u(x, y)^{2}+v(x, y)^{2}} .\) Use the Chain Rule to find \(\partial s / \partial x\) and \(\partial s / \partial y\). $$u(x, y)=2 y \text { and } v(x, y)=-2 x ; x \geq 0 \text { and } y \geq 0$$

Short Answer

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Question: Find the partial derivatives of the speed function \(s(x, y) = \sqrt{u(x, y)^2 + v(x, y)^2}\), where \(u(x, y) = 2y\) and \(v(x, y) = -2x\), with respect to \(x\) and \(y\). Answer: The partial derivatives of the speed function \(s(x, y)\) are given by: $$\frac{\partial s}{\partial x} = \frac{4x}{\sqrt{(2y)^2 + (-2x)^2}}$$ $$\frac{\partial s}{\partial y} = \frac{4y}{\sqrt{(2y)^2 + (-2x)^2}}$$

Step by step solution

01

Write the expression for the speed function

We are given the speed function: $$s(x, y) = \sqrt{u(x, y)^2 + v(x, y)^2}$$ where $$u(x, y) = 2y \text{ and } v(x, y) = -2x$$
02

Apply the Chain Rule to find the partial derivative with respect to x

The Chain Rule for partial derivatives states: $$\frac{\partial s}{\partial x} = \frac{\partial s}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial s}{\partial v}\frac{\partial v}{\partial x}$$ To find \(\frac{\partial s}{\partial x}\), we need to calculate the following partial derivatives: \(\frac{\partial s}{\partial u}\), \(\frac{\partial u}{\partial x}\), \(\frac{\partial s}{\partial v}\), and \(\frac{\partial v}{\partial x}\). 1. \(\frac{\partial s}{\partial u} = \frac{2u(x, y)}{2\sqrt{u(x, y)^2 + v(x, y)^2}} = \frac{u(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}}\) 2. \(\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(2y) = 0\) 3. \(\frac{\partial s}{\partial v} = \frac{2v(x, y)}{2\sqrt{u(x, y)^2 + v(x, y)^2}} = \frac{v(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}}\) 4. \(\frac{\partial v}{\partial x} = \frac{\partial}{\partial x}(-2x) = -2\) Now, substituting these values into the Chain Rule formula: $$\frac{\partial s}{\partial x} = \frac{u(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}} \cdot 0 + \frac{v(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}} \cdot (-2)$$ $$\frac{\partial s}{\partial x} = -\frac{2v(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}}$$ $$\frac{\partial s}{\partial x} = \frac{4x}{\sqrt{(2y)^2 + (-2x)^2}}$$
03

Apply the Chain Rule to find the partial derivative with respect to y

Similarly, applying the Chain Rule with respect to \(y\): $$\frac{\partial s}{\partial y} = \frac{\partial s}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial s}{\partial v}\frac{\partial v}{\partial y}$$ We need to find the following partial derivatives: \(\frac{\partial u}{\partial y}\) and \(\frac{\partial v}{\partial y}\). 1. \(\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(2y) = 2\) 2. \(\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(-2x) = 0\) Now, substituting these values into the Chain Rule formula: $$\frac{\partial s}{\partial y} = \frac{u(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}} \cdot 2 + \frac{v(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}} \cdot 0$$ $$\frac{\partial s}{\partial y} = \frac{2u(x, y)}{\sqrt{u(x, y)^2 + v(x, y)^2}}$$ $$\frac{\partial s}{\partial y} = \frac{4y}{\sqrt{(2y)^2 + (-2x)^2}}$$
04

Write down the final expressions for the partial derivatives

The partial derivatives of the speed function \(s(x, y)\) with respect to \(x\) and \(y\) are: $$\frac{\partial s}{\partial x} = \frac{4x}{\sqrt{(2y)^2 + (-2x)^2}}$$ $$\frac{\partial s}{\partial y} = \frac{4y}{\sqrt{(2y)^2 + (-2x)^2}}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
In multivariable calculus, the chain rule is an essential tool that allows us to compute the derivative of a composite function. For functions of a single variable, the chain rule is straightforward, taking the derivative of the outer function and multiplying it by the derivative of the inner function. However, when dealing with functions of multiple variables, as we often do in fluid dynamics, the process becomes a bit more complex.

The chain rule for partial derivatives can be visualized as following the pathways through which one variable affects another. For instance, in the context of the example exercise provided, the speed of fluid flow, denoted by s(x, y), is affected by changes in the variables x and y through the functions u(x, y) and v(x, y). When finding the partial derivative of s(x, y) with regards to x, we consider how x affects u and v, and in turn, how these functions affect the speed.

Therefore, the chain rule enables us to decompose the change in s with respect to x into a sum of terms, each representing a distinct 'pathway' of dependence. In practice, this involves calculating and then multiplying partial derivatives along each pathway until we reach the variable for which we are calculating the change. The final step is summing these products to obtain the total partial derivative of the composite function. This method ensures that all the ways in which the input variables could affect the output variable are fully accounted for.
Fluid Dynamics
The study of fluid dynamics is concerned with how fluids (liquids and gases) move and the forces acting upon them. It has applications in a wide range of fields, from engineering systems like pipelines and air conditioning units to natural phenomena such as weather patterns and ocean currents. In our specific exercise, we are looking at a fluid moving in two dimensions, where u(x, y) and v(x, y) represent the fluid's components of velocity in the x and y directions, respectively.

Mathematically describing the behavior of fluids requires applying principles of conservation of mass, momentum, and energy. These principles often manifest as complex partial differential equations. The speed of the fluid at a particular point is not only a measure of how fast it is moving but also a vectorial property combining the motion in both the x and y directions. Multivariable calculus, particularly partial derivatives, helps us analyze changes in fluid speed in response to changes in position. This becomes crucial in optimizing and understanding fluid behavior in various applications, such as aerodynamics and hydrodynamic systems.
Multivariable Calculus
The role of multivariable calculus is pivotal in solving problems involving more than one variable, which is often the case in real-world scenarios. This branch of mathematics generalizes the concepts of single-variable calculus to higher dimensions, allowing for the analysis of functions that depend on several variables. Techniques like partial differentiation, as exhibited in the given exercise, enable us to isolate the effect of varying one variable at a time, while keeping the others constant.

Understanding how to operate with partial derivatives is crucial when working with physical phenomena such as fluid flows, where each component of the flow can be influenced by multiple factors. For example, the partial derivatives of the speed function s(x, y) with respect to x and y provide insights into how changes in the fluid's position affect its speed, separately in each direction. Thus, multivariable calculus offers the tools necessary to dissect complex systems into more manageable sub-parts, study the relations between them, and synthesize this information to comprehend the overall system behavior fully.

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Most popular questions from this chapter

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) The gravitational potential associated with two objects of mass \(M\) and \(m\) is \(\varphi=-G M m / r,\) where \(G\) is the gravitational constant. If one of the objects is at the origin and the other object is at \(P(x, y, z),\) then \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between the objects. The gravitational field at \(P\) is given by \(\mathbf{F}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. Show that the force has a magnitude \(|\mathbf{F}|=G M m / r^{2}\) Explain why this relationship is called an inverse square law.

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Second Derivative Test Suppose the conditions of the Second Derivative Test are satisfied on an open disk containing the point \((a, b) .\) Use the test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b) < 0 < f_{\text {vy }}(a, b)\) or \(f_{y y}(a, b) < 0 < f_{x x}(a, b),\) then \(f\) has a saddle point at \((a, b)\)

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