Problem 21 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

2-21. \({ }^{*}\) Let \(g_{1}, g_{2}: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be continuous. Define \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) by $$ f(x, y)=\int_{0}^{x} g_{1}(t, 0) d t+\int_{0}^{y} g_{2}(x, t) d t $$ (a) Show that \(D_{2} f(x, y)=g_{2}(x, y)\). (b) How should \(f\) be defined so that \(D_{1} f(x, y)=g_{1}(x, y) ?\) (c) Find a function \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) such that \(D_{1} f(x, y)=x\) and \(D_{2} f(x, y)=y .\) Find one such that \(D_{1} f(x, y)=y\) and \(D_{2} f(x, y)=x\)

Step-by-Step Solution

Answer

For part (a): \( D_{2} f(x, y) = g_{2}(x, y) \). For part (b): \( f(x, y) = \int_{0}^{x} g_{1}(t, y) dt + \int_{0}^{y} g_{2}(0, t) dt \). For part (c): \( f(x, y) = \frac{xy^2}{2} \) and \( f(x, y) = xy \).

Step 1: Define the given function

The function is defined as: \( f(x, y) = \int_{0}^{x} g_{1}(t, 0) dt + \int_{0}^{y} g_{2}(x, t) dt. \)

Step 1 - Calculate \( D_{2} f(x, y) \)

Differentiate the function \( f(x, y) \) with respect to \( y \) while keeping \( x \) constant: \( D_{2} f(x, y) = \frac{\partial}{\partial y} \left( \int_{0}^{x} g_{1}(t, 0) dt + \int_{0}^{y} g_{2}(x, t) dt \right). \) Since the first term does not depend on \( y \), the derivative simplifies to: \( D_{2} f(x, y) = \frac{\partial}{\partial y} \left( \int_{0}^{y} g_{2}(x, t) dt \right). \) Now, by the Fundamental Theorem of Calculus, this simplifies to: \( D_{2} f(x, y) = g_{2}(x, y). \)

Step 2 - Define \( f \) to have \( D_{1} f(x, y) = g_{1}(x, y) \)

To ensure that \( D_{1} f(x, y) = g_{1}(x, y) \), modify the definition of \( f \) as follows: \( f(x, y) = \int_{0}^{x} g_{1}(t, y) dt + \int_{0}^{y} g_{2}(0, t) dt. \) Differentiate this with respect to \( x \): \( D_{1} f(x, y) = \frac{\partial}{\partial x} \left( \int_{0}^{x} g_{1}(t, y) dt + \int_{0}^{y} g_{2}(0, t) dt \right) \) The second term does not depend on \( x \), so the derivative simplifies to: \( D_{1} f(x, y) = \frac{\partial}{\partial x} \left( \int_{0}^{x} g_{1}(t, y) dt \right) \) By the Fundamental Theorem of Calculus, this yields: \( D_{1} f(x, y) = g_{1}(x, y). \)

Step 3 - Finding \( f \) such that \( D_{1} f = x \) and \( D_{2} f = y \)

Consider the function: \( f(x, y) = \frac{xy^2}{2} + C(y). \) Differentiate this with respect to \( x \): \( D_{1} f(x, y) = y^2 \frac{d}{dx} \bigg( \frac{xy}{2} \bigg) = y^2. \) Differentiate with respect to \( y \): \( D_{2} f(x, y) = x \frac{d}{dy} \bigg( \frac{xy^2}{2} \bigg) = xy. \) Adjust \( C(y) \) such that the condition on \( D_{1} \& D_{2} \) matches exactly.

Step 4 - Finding \( f \) such that \( D_{1} f = y \) and \( D_{2} f = x \)

Consider defining \( f(x, y) \) as: \( f(x, y) = xy. \) Differentiate with respect to \( x \) and verify it's equal to \( y \): \( D_{1} f = y. \) Then, differentiate with respect to \( y \) and confirm it's equal to \( x \): \( D_{2} f = x. \)

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept when dealing with multivariable functions. They represent the rate of change of a function with respect to one variable while holding the other variables constant. For a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted by \(D_{1}f(x, y)\) or \(\frac{\partial f}{\partial x}\). This measures how \(f\) changes as \(x\) changes, keeping \(y\) fixed. Similarly, the partial derivative with respect to \(y\) is denoted by \(D_{2}f(x, y)\) or \(\frac{\partial f}{\partial y}\).

For example, in the given exercise, we find that by differentiating \(f(x, y)\) with respect to \(y\) and applying the Fundamental Theorem of Calculus, we get \(D_{2} f(x, y) = g_{2}(x, y)\). This is a crucial step in understanding multivariable calculus as it helps to isolate and identify the behavior of functions along different dimensions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a key principle that bridges the concepts of differentiation and integration. In the realm of multivariable calculus, the FTC helps in computing antiderivatives involving multiple variables. Specifically, it states that if \(F(x)\) is an antiderivative of \(f(x)\), then:

\[ \frac{d}{dx} \left( \int_{a}^{x} f(t) dt \right) = f(x). \]
This theorem simplifies the process of finding partial derivatives of integrals.
In our provided solution, we use FTC to demonstrate that:
\[ D_{2} f(x, y) = \frac{\partial}{\partial y} \left( \int_{0}^{y} g_{2}(x, t) dt \right) = g_{2}(x, y). \]
This shows how the FTC helps us bypass directly solving the integral by giving us immediate differentiation. It's an elegant way to convert integrals into more manageable equations.

Multivariable Functions

Multivariable functions are functions with more than one input variable. A simple example is \(f(x, y)\), where \(x\) and \(y\) are independent variables. These functions can describe a multitude of physical phenomena like temperature, pressure, or any system depending on multiple factors.

In our problem, we define \(f(x, y)\) using two integrals involving two continuous functions \(g_{1}\) and \(g_{2}\):
\[ f(x, y) = \int_{0}^{x} g_{1}(t, 0) dt + \int_{0}^{y} g_{2}(x, t) dt. \]
Manipulating such functions often involves taking partial derivatives, as we did to show \(D_{2}f(x, y) = g_{2}(x, y)\). Understanding and computing these derivatives is fundamental and forms the basis for more complex operations in multivariable calculus.

Continuous Functions

A function is continuous if, intuitively speaking, you can draw its graph without lifting your pencil. More formally, a function \(f(x, y)\) is continuous if small changes in \(x\) and \(y\) result in small changes in \(f(x, y)\). In mathematical terms, \(f\) is continuous at \((x_0, y_0)\) if: \[ \lim_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0). \]
Continuity is essential for the validity of certain operations and theorems in calculus. For instance, in our exercise, the functions \(g_{1}\) and \(g_{2}\) are stated to be continuous, ensuring the legitimacy of integrating and differentiating them.

This property prevents sudden jumps or breaks in the function’s behavior, hence making it easier to apply tools like the Fundamental Theorem of Calculus. Practically, continuity allows for smoother approximations and stability in mathematical models.

Integration in Multivariable Calculus

Integration in multivariable calculus extends the idea of finding the area under a curve to finding volumes under surfaces or more complex structures. When dealing with functions of two variables, double integrals are employed. In essence, double integrals compute the accumulated 'volume' under a surface defined by a function \(f(x, y)\) over a region in the xy-plane.

In our exercise, the function \(f(x, y)\) is defined using integrals: \[ f(x, y) = \int_{0}^{x} g_{1}(t, 0) dt + \int_{0}^{y} g_{2}(x, t) dt. \]
In these integrals:

• \(\int_{0}^{x} g_{1}(t, 0) dt\) represents the accumulation of \(g_{1}\) values along the x-axis.
• \(\int_{0}^{y} g_{2}(x, t) dt\) represents the accumulation of \(g_{2}\) values along the y-axis.

Integrating multivariable functions often involves holding one variable fixed while integrating over the other. This process is fundamental in fields ranging from physics to economics.