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Chapter 2: Problem 22

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.

Short Answer

Expert verified
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.

Step by step solution

01

- 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.
02

- 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.
03

- 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.
04

- 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.

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Key Concepts

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

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.
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.
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:
  • **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
Understanding these concepts is crucial for solving problems that involve varying input, such as in physics, engineering, and economics.

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.

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Most popular questions from this chapter

2-9. Two functions \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) are equal up to \(\boldsymbol{n}\) th order at \(a\) if $$ \lim _{h \rightarrow 0} \frac{f(a+h)-g(a+h)}{h^{n}}=0 $$ (a) Show that \(f\) is differentiable at \(a\) if and only if there is a function \(g\) of the form \(g(x)=a_{0}+a_{1}(x-a)\) such that \(f\) and \(g\) are equal up to first order at \(a\). (b) If \(f^{\prime}(a), \ldots, f^{(n)}(a)\) exist, show that \(f\) and the function \(g\) defined by $$ g(x)=\sum_{i=0}^{n} \frac{f^{(i)}(a)}{i !}(x-a)^{i} $$ are equal up to \(n\)th order at \(a\). Hint: The limit $$ \lim _{x \rightarrow a} \frac{f(x)-\sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i !}(x-a)^{i}}{(x-a)^{n}} $$ may be evaluated by L'Hospital's rule.

2-29. Let \(f: \mathbf{R}^{n} \rightarrow\) R. For \(x \in \mathbf{R}^{n}\), the limit $$ \lim _{t \rightarrow 0} \frac{f(a+t x)-f(a)}{t} $$ if it exists, is denoted \(D_{x} f(a)\), and called the directional derivative of \(f\) at \(a\), in the direction \(x\). (a) Show that \(D_{e_{i}} f(a)=D_{i} f(a)\). (b) Show that \(D_{t x} f(a)=t D_{x} f(a)\). (c) If \(f\) is differentiable at \(a\), show that \(D_{x} f(a)=D f(a)(x)\) and therefore \(D_{x+y} f(a)=D_{x} f(a)+D_{y} f(a)\).

2-32. (a) Let \(f: \mathbf{R} \rightarrow \mathbf{R}\) be defined by $$ f(x)= \begin{cases}x^{2} \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0\end{cases} $$ Show that \(f\) is differentiable at 0 but \(f^{\prime}\) is not continuous at 0 . (b) Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be defined by $$ f(x, y)= \begin{cases}\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}} & (x, y) \neq 0 \\ 0 & (x, y)=0\end{cases} $$ Show that \(f\) is differentiable at \((0,0)\) but \(D_{i} f\) is not continuous at \((0,0)\)

2-4. Let \(g\) be a continuous real-valued function on the unit circle \(\left\\{x \in \mathbf{R}^{2}:|x|=1\right\\}\) such that \(g(0,1)=g(1,0)=0\) and \(g(-x)=\) \(-g(x) . \quad\) Define \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) by $$ f(x)- \begin{cases}|x| \cdot g\left(\frac{x}{|x|}\right) & x \neq 0 \\ 0 & x=0\end{cases} $$ (a) If \(x \in \mathbf{R}^{2}\) and \(h: \mathbf{R} \rightarrow \mathbf{R}\) is defined by \(h(t)=f(t x)\), show that \(h\) is differentiable. (b) Show that \(f\) is not differentiable at \((0,0)\) unless \(g=0\) Hint: First show that \(D f(0,0)\) would have to be 0 by considering \((h, k)\) with \(k=0\) and then with \(h=0\)

2-34. A function \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) is homogeneous of degree \(m\) if \(f(t x)=\) \(t^{m} f(x)\) for all \(x .\) If \(f\) is also differentiable, show that $$ \sum_{i=1}^{n} x^{i} D_{i} f(x)=m f(x) $$ Hint: If \(g(t)=f(t x)\), find \(g^{\prime}(1)\)
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