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These exercises are concerned with functions of two variables. Let \(f(x, y)=x+\sqrt[3]{x y} .\) Find $$ \begin{array}{llll}{\text { (a) } f\left(t, t^{2}\right)} & {\text { (b) } f\left(x, x^{2}\right)} & {\text { (c) } f\left(2 y^{2}, 4 y\right)} & {}\end{array} $$

Short Answer

Expert verified
(a) \(2t\), (b) \(2x\), (c) \(2y^2 + 2y\)."

Step by step solution

01

Substitute Values for Part (a)

To find \( f(t, t^2) \), substitute \( x = t \) and \( y = t^2 \) into the function. This gives: \[ f(t, t^2) = t + \sqrt[3]{t \cdot t^2} \] which simplifies to \[ f(t, t^2) = t + \sqrt[3]{t^3}. \] Since \( \sqrt[3]{t^3} = t \), the expression becomes: \[ f(t, t^2) = t + t = 2t. \]
02

Substitute Values for Part (b)

For \( f(x, x^2) \), substitute \( x = x \) and \( y = x^2 \) into the function. This leads to: \[ f(x, x^2) = x + \sqrt[3]{x \cdot x^2} \] which simplifies to \[ f(x, x^2) = x + \sqrt[3]{x^3}. \] Again, \( \sqrt[3]{x^3} = x \), hence the expression becomes: \[ f(x, x^2) = x + x = 2x. \]
03

Substitute Values for Part (c)

To find \( f(2y^2, 4y) \), substitute \( x = 2y^2 \) and \( y = 4y \) in the function. This gives: \[ f(2y^2, 4y) = 2y^2 + \sqrt[3]{(2y^2) \cdot 4y} \] which simplifies to \[ f(2y^2, 4y) = 2y^2 + \sqrt[3]{8y^3}. \] Since \( \sqrt[3]{8y^3} = 2y \), the expression simplifies to: \[ f(2y^2, 4y) = 2y^2 + 2y. \]

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Key Concepts

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

Substitution Method
The substitution method is a useful technique when analyzing functions of two variables like this one. It involves replacing the variables in the function with specific values or expressions, making calculations more manageable and understandable.

For instance, in our given problem, when looking for the results of functions such as \( f(t, t^2) \), \( f(x, x^2) \), or \( f(2y^2, 4y) \), we substitute:
  • For \( f(t, t^2) \), substitute \( x = t \) and \( y = t^2 \).
  • For \( f(x, x^2) \), substitute \( x = x \) and \( y = x^2 \).
  • For \( f(2y^2, 4y) \), substitute \( x = 2y^2 \) and \( y = 4y \).
By replacing the variables, you transform a complex function into a simpler one, making it easier to carry out further calculations. This method is essential because it allows us to solve the problem step by step, reducing potential errors and gaining deeper understanding of variable interactions.
Simplifying Expressions
Simplifying expressions is another core skill when working with functions of two variables. After you have made substitutions, it's important to simplify what you have. This process makes the expression less complicated and easier to work with.

When simplifying, look for opportunities to combine like terms or apply arithmetic rules. In the exercise, after substitution, expressions like \( \sqrt[3]{t \cdot t^2} \) can be simplified to \( \sqrt[3]{t^3} \). Similarly, \( \sqrt[3]{x \cdot x^2} \) simplifies to \( \sqrt[3]{x^3} \).

These simplifications pave the way for straightforward conclusions. For example:
  • \( \sqrt[3]{t^3} \) simplifies to \( t \), resulting in the expression \( 2t \) after combining terms.
  • \( \sqrt[3]{x^3} \) becomes \( x \), leading to \( 2x \).
  • \( \sqrt[3]{8y^3} \) results in \( 2y \), simplifying the expression to \( 2y^2 + 2y \).
Simplification is crucial in mathematics because it makes complex expressions accessible and allows for effective problem solving.
Cubic Roots
Understanding cubic roots is pivotal when simplifying expressions like the ones found in our exercise. The cubic root of a number or expression is what you multiply by itself twice to get the original number or expression. Mathematically, the cubic root of \( a \) is denoted as \( \sqrt[3]{a} \).

This skill is crucial when dealing with terms such as \( \sqrt[3]{t^3} \) and \( \sqrt[3]{x^3} \). Similarly, for numeric calculations, \( \sqrt[3]{8y^3} \) simplifies due to the fact that 8 is \( 2^3 \), making its cubic root 2, multiplied by \( y \) gives \( 2y \).

Grasping the concept of cubic roots aids in swiftly breaking down complex expressions into manageable parts. Key takeaways include:
  • Recognize patterns like \( \sqrt[3]{x^3} = x \) as a straightforward simplification.
  • Numerical cubic roots simplify complex arithmetic within expressions, as with \( 8y^3 \) reducing to \( 2y \).
Mastering cubic roots is essential for students working with polynomial expressions and solving equations involving cubic terms efficiently.

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Most popular questions from this chapter

Show that the function satisfies Laplace's equation $$\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0$$ $$\begin{array}{l}{\text { (a) } z=x^{2}-y^{2}+2 x y} \\ {\text { (b) } z=e^{x} \sin y+e^{y} \cos x} \\ {\text { (c) } z=\ln \left(x^{2}+y^{2}\right)+2 \tan ^{-1}(y / x)}\end{array}$$

A common problem in experimental work is to obtain a mathematical relationship \(y=f(x)\) between two variables \(x\) and \(y\) by "fitting" a curve to points in the plane that correspond to experimentally determined values of \(x\) and \(y,\) say $$ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) $$ The curve \(y=f(x)\) is called a mathematical model of the data. The general form of the function \(f\) is commonly determined by some underlying physical principle, but sometimes it is just determined by the pattern of the data. We are concerned with fitting a straight line \(y=m x+b\) to data. Usually, the data will not lie on a line (possibly due to experimental error or variations in experimental conditions), so the problem is to find a line that fits the data "best" according to some criterion. One criterion for selecting the line of best fit is to choose \(m\) and \(b\) to minimize the function $$ g(m, b)=\sum_{i=1}^{n}\left(m x_{i}+b-y_{i}\right)^{2} $$ This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometrically, \(\left|m x_{i}+b-y_{i}\right|\) is the vertical distance between the data point \(\left(x_{i}, y_{i}\right)\) and the line \(y=m x+b\) These vertical distances are called the residuals of the data points, so the effect of minimizing \(g(m, b)\) is to minimize the sum of the squares of the residuals. In these exercises, we will derive a formula for the regression line. The purpose of this exercise is to find the values of \(m\) and \(b\) that produce the regression line. (a) To minimize \(g(m, b),\) we start by finding values of \(m\) and \(b\) such that \(\partial g / \partial m=0\) and \(\partial g / \partial b=0 .\) Show that these equations are satisfied if \(m\) and \(b\) satisfy the conditions $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right) m+\left(\sum_{i=1}^{n} x_{i}\right) b=\sum_{i=1}^{n} x_{i} y_{i} $$ \(\left(\sum_{i=1}^{n} x_{i}\right) m+n b=\sum_{i=1}^{n} y_{i}\) (b) Let \(\bar{x}=\left(x_{1}+x_{2}+\cdots+x_{n}\right) / n\) denote the arithmetic average of \(x_{1}, x_{2}, \ldots, x_{n} .\) Use the fact that $$ \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq 0 $$ to show that $$ n\left(\sum_{i=1}^{n} x_{i}^{2}\right)-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geq 0 $$ with equality if and only if all the \(x_{i}\) 's are the same. (c) Assuming that not all the \(x_{i}\) 's are the same, prove that the equations in part (a) have the unique solution $$ \begin{aligned} m=& \frac{n \sum_{i=1}^{n} x_{i} y_{i}-\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}}{n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2}} \\ b=& \frac{1}{n}\left(\sum_{i=1}^{n} y_{i}-m \sum_{i=1}^{n} x_{i}\right) \end{aligned} $$ [Note: We have shown that \(g\) has a critical point at these values of \(m\) and \(b\). In the next exercise we will show that \(g\) has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line \(y=m x+b\) is the regression line for these values of \(m \text { and } b .]\)

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