Problem 22
Question
2-22. \(^{*}\) If \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) and \(D_{2} f=0\), show that \(f\) is independent of the second variable. If \(D_{1} f=D_{2} f=0\), show that \(f\) is constant.
Step-by-Step Solution
Verified Answer
If \(D_2 f = 0\), then f is independent of the second variable. If \(D_1 f = D_2 f = 0\), then f is constant.
1Step 1 - Understanding Partial Derivatives
Given that the partial derivative of the function with respect to the second variable, denoted as \(D_2 f\), is 0, the function does not change with respect to this variable. This implies that the function f is independent of the second variable.
2Step 2 - Using the Partial Derivative Information
Since \(D_2 f = 0\), the function can be written in the form \[f(x,y) = g(x)\]. Here, \(g(x)\) is some function that only depends on the first variable x.
3Step 3 - Checking Independence of Both Variables
Next, we are given that both partial derivatives, \(D_1 f\) and \(D_2 f\), are 0. This means \[\frac{\frac{\text{d}f}{\text{d}x}=0 \text{and}\frac{\text{d}f}{\text{d}y}=0\]. Since the partial derivatives with respect to both x and y are 0, it implies that f is constant.
4Step 4 - Concluding the Result
If \(D_1 f = D_2 f = 0\), then the original function \(f \text{(x,y)}}\) does not depend on either variable. Therefore, \[f(x,y) = c\], where c is a constant value.
Key Concepts
independence of variablesconstant functionmultivariable calculus
independence of variables
In multivariable calculus, a function can depend on several variables. For a function f(x, y), if the partial derivative with respect to the second variable y is zero, this means the function does not change when y changes. Formally, this is written as \(D_{2} f = 0\). When the partial derivative with respect to a variable is zero, it implies that the function is 'independent' of that variable. So, for our given function, \(f(x, y)\) can be simplified to \(g(x)\), indicating that it only changes with respect to x and is unaffected by y. This simplification helps because it reduces the number of variables we need to consider when analyzing the function.
Remember, independence of variables tells us how each variable impacts the value of the function. It's a crucial concept when dissecting functions in multivariable calculus.
Remember, independence of variables tells us how each variable impacts the value of the function. It's a crucial concept when dissecting functions in multivariable calculus.
constant function
A constant function is a fundamental concept in calculus. Such a function has the same value regardless of the values of its independent variables. In mathematical terms, if the partial derivatives with respect to all variables are zero, \(D_{1} f = 0\) and \(D_{2} f = 0\), then the function hovers at a constant value. For example, consider \(f(x, y) = c\), where c is a constant. No matter the values of x and y, the function outputs c.
To check if a function is constant, we look at its partial derivatives. If all partial derivatives are zero, the function must be constant. This is because there are no changes happening in any direction within the function. So, for our exercise, if both \(D_{1} f\) and \(D_{2} f\) are 0, then \(f(x,y)\) doesn't change as either x or y changes, resulting in a constant function.
To check if a function is constant, we look at its partial derivatives. If all partial derivatives are zero, the function must be constant. This is because there are no changes happening in any direction within the function. So, for our exercise, if both \(D_{1} f\) and \(D_{2} f\) are 0, then \(f(x,y)\) doesn't change as either x or y changes, resulting in a constant function.
multivariable calculus
Multivariable calculus extends the principles of calculus to functions of multiple variables. This branch of mathematics allows analyzing functions like \(f(x, y)\) which depend on more than one variable, providing tools to understand how changes in variables affect overall function behavior.
Core concepts include:
In our exercise, we use partial derivatives to analyze changes in individual variables, making it easier to determine if the function is constant or independent of certain variables.
Core concepts include:
- **Partial Derivatives:** Measures how the function changes as only one variable changes while others are held constant
- **Gradients:** Vector containing all the partial derivatives of a function, representing the rate and direction of change
- **Double Integrals:** Allows computing the volume under a surface in a two-variable function
In our exercise, we use partial derivatives to analyze changes in individual variables, making it easier to determine if the function is constant or independent of certain variables.
Other exercises in this chapter
Problem 20
2-20. Find the partial derivatives of \(f\) in terms of the derivatives of \(g\) and \(h\) if (a) \(f(x, y)=g(x) h(y) .\) (b) \(f(x, y)=g(x)^{h(y)}\). (c) \(f(x
View solution Problem 21
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)=\
View solution Problem 23
2-22. \(^{*}\) If \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) and \(D_{2} f=0\), show that \(f\) is independent of the second variable. If \(D_{1} f=D_{2} f=0\
View solution Problem 24
2-24. Define \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) by $$ f(x, y)= \begin{cases}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}} & (x, y) \neq 0 \\\ 0 & (x, y)=0\end{c
View solution