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Chapter 14: Problem 3

In Exercises \(1-4,\) find the specific function values. $$ f(x, y, z)=\frac{x-y}{v^{2}+z^{2}} $$ $$ \begin{array}{ll}{\text { a. } f(3,-1,2)} & {\text { b. } f\left(1, \frac{1}{2},-\frac{1}{4}\right)} \\ {\text { c. } f\left(0,-\frac{1}{3}, 0\right)} & {\text { d. } f(2,2,100)}\end{array} $$

Short Answer

Expert verified
a. \(\frac{4}{13}\), b. \(\frac{8}{17}\), c. Undefined, d. 0.

Step by step solution

01

Understand the Function

The given function is \( f(x, y, z) = \frac{x-y}{v^2 + z^2} \). We need to substitute the specific values given for \(x\), \(y\), and \(z\) into the function to find \(f(x, y, z)\). Note that \(v\) should actually be \(x\), since it's a typo. Correct function expression should be \( f(x, y, z) = \frac{x-y}{x^2 + z^2} \).
02

Compute \( f(3, -1, 2) \)

Substitute into the function: \( f(3, -1, 2) = \frac{3 - (-1)}{3^2 + 2^2} = \frac{3 + 1}{9 + 4} = \frac{4}{13} \).
03

Compute \( f\left(1, \frac{1}{2}, -\frac{1}{4}\right) \)

Substitute into the function: \( f\left(1, \frac{1}{2}, -\frac{1}{4}\right) = \frac{1 - \frac{1}{2}}{1^2 + \left(-\frac{1}{4}\right)^2} = \frac{\frac{1}{2}}{1 + \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{17}{16}} = \frac{1}{2} \times \frac{16}{17} = \frac{8}{17} \).
04

Compute \( f\left(0, -\frac{1}{3}, 0\right) \)

Substitute into the function: \( f\left(0, -\frac{1}{3}, 0\right) = \frac{0 - \left(-\frac{1}{3}\right)}{0^2 + 0^2} = \frac{\frac{1}{3}}{0} \). Since division by zero is undefined, \( f\left(0, -\frac{1}{3}, 0\right) \) is undefined.
05

Compute \( f(2, 2, 100) \)

Substitute into the function: \( f(2, 2, 100) = \frac{2 - 2}{2^2 + 100^2} = \frac{0}{4 + 10000} = \frac{0}{10004} = 0 \).

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Key Concepts

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

Function Evaluation
In multivariable calculus, evaluating a function means that you substitute given values for each variable and calculate the result. For example, in the function\( f(x, y, z) = \frac{x-y}{x^2 + z^2} \), if you're asked to find \( f(3, -1, 2) \), you simply plug in \( x=3 \), \( y=-1 \), and \( z=2 \) into the function. Here's how you go about it:
  • Step 1: Replace \( x \), \( y \), and \( z \) with the given numbers.
  • Step 2: Perform any arithmetic operations like subtraction in the numerator and addition in the denominator.
  • Step 3: Divide the result of the operations to find the function's output.
In our example, you get \( f(3, -1, 2) = \frac{3 - (-1)}{3^2 + 2^2} = \frac{4}{13} \).

Function evaluation is essential as it checks what output the function gives for specific inputs. It's crucial in fields like physics and engineering where functions describe real-world phenomena.
Substitution in Functions
Substitution is a technique used to find the value of a function at specific points. You replace each variable in the function's expression with a value to compute a numerical outcome. For some functions, this process can be straightforward, but it may require careful steps, especially if involved with fractions or negative numbers.Let's take the example of \( f\left(1, \frac{1}{2}, -\frac{1}{4}\right) \). To solve this, you substitute \( x=1 \), \( y=\frac{1}{2} \), and \( z=-\frac{1}{4} \) into the equation:
  • Begin substituting: \( \frac{1 - \frac{1}{2}}{1^2 + \left(-\frac{1}{4}\right)^2} \)
  • Simplify the numerator and denominator: the numerator becomes \( \frac{1}{2} \), and the denominator \( 1 + \frac{1}{16} \).
  • Final computation: \( \frac{\frac{1}{2}}{\frac{17}{16}} = \frac{8}{17} \)
This substitution process allows you to break down complex functions into simpler arithmetic calculations.
Division by Zero
Division by zero is a concept where the denominator in a fraction equals zero. This is a special situation in mathematics because division by zero is undefined. In the exercise, when computing \( f\left(0, -\frac{1}{3}, 0\right) \), the denominator becomes \( 0^2 + 0^2 = 0 \), which results in a division of the form \( \frac{\frac{1}{3}}{0} \).Here's why this is undefined:
  • Imagine dividing a number by smaller and smaller numbers approaching zero, the result grows larger without bound.
  • This creates an inconsistency because dividing by zero doesn鈥檛 lead to a unique result.
  • Therefore, expressions where division by zero occurs have no meaning in real number arithmetic.
In practical terms, spotting a zero in the denominator should raise a flag to check for possible errors or undefined values. Recognizing these situations is important for sound mathematical practices, avoiding incorrect conclusions.

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Most popular questions from this chapter

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) $$ \begin{array}{l}{f(x, y, z)=x y+2 y z-3 x z \text { at } P_{0}(1,1,0)} \\ {R :|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z| \leq 0.01}\end{array} $$

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2} $$

In Exercises \(61-64,\) find an equation for the level surface of the function through the given point. $$ f(x, y, z)=\sqrt{x-y}-\ln z, \quad(3,-1,1) $$

Let $$f(x, y)=\left\\{\begin{array}{ll}{x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}},} & {\text { if }(x, y) \neq 0} \\ {0,} & {\text { if }(x, y)=0}\end{array}\right.$$ a. Show that \(\frac{\partial f}{\partial y}(x, 0)=x\) for all \(x,\) and \(\frac{\partial f}{\partial x}(0, y)=-y\) for all \(y\) b. Show that \(\frac{\partial^{2} f}{\partial y \partial x}(0,0) \neq \frac{\partial^{2} f}{\partial x \partial y}(0,0)\) The graph of \(f\) is shown on page 788 . The three-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}=0$$ is satisfied by steady-state temperature distributions \(T=f(x, y, z)\) in space, by gravitational potentials, and by electrostatic potentials. The two-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0$$ obtained by dropping the \(\partial^{2} f / \partial z^{2}\) term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) may be treated as a thin slice of the solid (b) perpendicular to the \(z\) -axis.

Establish the fact, widely used in hydrodynamics, that if \(f(x, y, z)=0,\) then \begin{equation}\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial z}{\partial x}\right)_{y}=-1.\end{equation} (Hint: Express all the derivatives in terms of the formal partial derivatives \(\partial f / \partial x, \partial f / \partial y,\) and \(\partial f / \partial z . )\)
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