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Chapter 2: Problem 4

A function \(f(x, y)\), defined for all \((x, y)\), is such that \(\frac{\partial f}{\partial y}=0\). Show that there is a function \(g(x)\) such that $$ f(x, y) \equiv g(x) . $$

Short Answer

Expert verified
Question: Given that the partial derivative of a function f(x, y) with respect to y is zero, show that there exists a function g(x) such that f(x, y) = g(x). Answer: We have demonstrated through integration and analyzing the resulting expression that when the partial derivative of a function f(x, y) with respect to y is zero, there exists a function g(x) such that f(x, y) = g(x).

Step by step solution

01

Write down the given information

We are given that the partial derivative of \(f(x, y)\) with respect to \(y\) is zero: $$ \frac{\partial f}{\partial y} = 0 $$
02

Integrate with respect to \(y\)

Now, we will integrate both sides of the equation with respect to \(y\): $$ \int \frac{\partial f}{\partial y} dy = \int 0 dy $$
03

Evaluate the integrals

Evaluating both sides of the equation, we get: $$ \int \frac{\partial f}{\partial y} dy = f(x, y) + C_1(x) $$ For the right-hand side, we have: $$ \int 0 dy = C_2(y) $$
04

Combine the two sides of the equation

Combine the results from Steps 3 to get: $$ f(x, y) + C_1(x) = C_2(y) $$ Rearranging the equation, we can isolate the term \(f(x, y)\): $$ f(x, y) = C_2(y) - C_1(x) $$
05

Analyze the result

Recall that our goal is to show there is a function \(g(x)\) such that \(f(x, y) \equiv g(x)\). Notice that if we define \(g(x)\) as \(g(x) = -C_1(x)\), we can write the expression for \(f(x, y)\) as: $$ f(x, y) = C_2(y) + g(x) $$ Since \(C_2(y)\) only depends on \(y\) and \(g(x)\) only depends on \(x\), the result obtained above shows that there exists a function \(g(x)\) such that \(f(x, y)\) can be written as a combination of a function dependent only on \(x\) and a function dependent only on \(y\). Therefore, we have shown that there exists a function \(g(x)\) such that \(f(x, y) \equiv g(x)\).

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Key Concepts

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

Multivariable Calculus
Multivariable calculus, also known as multivariate calculus, extends single variable calculus concepts to higher dimensions. In this discipline, functions have more than one input, often denoted by variables such as \(x, y\), \(x, y, z\), and so on, representing coordinates in multidimensional spaces. The beauty of multivariable calculus lies in its ability to handle curves, surfaces, and shapes in two-dimensional (2D) planes and three-dimensional (3D) spaces.

The powerful tools of multivariable calculus include partial derivatives and multiple integrals. Partial derivatives help us understand how a function changes as each variable is varied, holding the others constant. Multiple integrals are used to calculate the content (such as area or volume) under or within a surface.
Partial Differentiation
Partial differentiation is a fundamental concept in multivariable calculus, where we deal with functions of more than one variable. The process focuses on the rate of change of a function with respect to one variable at a time, treating all other variables as constants. For instance, \(\frac{\partial f}{\partial x}\) signifies the rate at which the function \(f\) changes with \(x\), keeping \(y\) constant.

It's a useful technique when analyzing systems where multiple factors influence the outcome. For example, in physics, \(\frac{\partial}{\partial t}\) could represent how temperature changes over time in a specific location. Partial differentiation can guide us in optimizing processes, understanding physical phenomena, and solving complex equations where variables are interdependent.
Integration of Partial Derivatives
Integration is the inverse process of differentiation, and the integration of partial derivatives allows us to reconstruct a multivariable function from its rates of change. If we are given \(\frac{\partial f}{\partial x}\) or \(\frac{\partial f}{\partial y}\), by integrating these partial derivatives with respect to their respective variables, we can potentially find the original function \(f(x, y)\).

When \(\frac{\partial f}{\partial y} = 0\), as in our exercise, it implies that \(f\) does not change as \(y\) changes and is therefore not a function of \(y\). Upon integrating with respect to \(y\), any constant of integration would depend only on \(x\), leading us to conclude \(f(x, y)\) is indeed some function \(g(x)\). Calculating these integrals correctly helps physicists, engineers, and mathematicians to solve real-world problems involving heat transfer, waves, and growth processes.

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Most popular questions from this chapter

Let \(u(x, y)\) and \(v(x, y)\) be harmonic in a domain \(D\) and have no critical points in \(D\). Show that if \(u\) and \(v\) are functionally dependent, then they are "linearly" dependent: \(u=a v+b\) for suitable constants \(a\) and \(b\). [Hint: Assume a relation of form \(u=f(v)\) and take the Laplacian of both sides.]

For each of the following curves (represented by intersecting surfaces), find the equations of the tangent line at the point indicated, verifying that the point is on the curve: a) \(2 x+y-z=6, x+2 y+2 z=7\) at \((3,1,1)\); b) \(x^{2}+y^{2}+z^{2}=9, x^{2}+y^{2}-8 z^{2}=0\) at \((2,2,1)\); c) \(x^{2}+y^{2}=1, x+y+z=0\) at \((1,0,-1)\); d) \(x^{2}+y^{2}+z^{2}=9, x^{2}+2 y^{2}+3 z^{2}=9\) at \((3,0,0)\). Why does the procedure break down in (d)? Show that solution is impossible.

Obtain the Jacobian matrix for each of the following mappings: a) \(y_{1}=5 x_{1}+2 x_{2}, y_{2}=2 x_{1}+3 x_{2}\). b) \(y_{1}=2 x_{1}^{2}+x_{2}^{2}, y_{2}=3 x_{1} x_{2}\). c) \(y_{1}=x_{1} x_{2} x_{3}, y_{2}=x_{1}^{2} x_{3}\) d) \(u=x \cos y, v=x \sin y, w=x^{2}\). e) \(w=x^{2} y z\). f) \(w=x^{2}+y^{2}-z^{2}\). g) \(x=t^{2}, y=t^{3}, z=t^{4}\).

. Show that the following sets of functions are functionally dependent: a) \(f=\frac{y}{x}, g=\frac{x-y}{x+y}\); b) \(f=x^{2}+2 x y+y^{2}+2 x+2 y, g=e^{x} e^{y}\); c) \(f=x^{2} y-x y^{2}+x y z, g=x y+x-y+z, h=x^{2}+y^{2}+z^{2}-2 y z+2 x z\); d) \(f=u+v-x, g=x-y+u, h=u-2 v+5 x-3 y\).

A certain function \(z=f(x, y)\) is known to have the value \(f(1,2)=3\) and derivatives \(f_{x}(1,2)=2, f_{y}(1,2)=5\). Make "reasonable" estimates of \(f(1.1,1.8), f(1.2,1.8)\), and \(f(1,3,1.8)\).
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