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Chapter 14: Problem 45

Show that (0,0) is a critical point of \(f(x, y)=x^{2}+k x y+y^{2}\) no matter what value the constant \(k\) has. (Hint: Consider two cases: \(k=0 \text { and } k \neq 0 .)\)

Short Answer

Expert verified
The point \((0, 0)\) is a critical point of \(f(x, y)\) for any \(k\).

Step by step solution

01

Understand Critical Points

Critical points of a function occur when the gradient of the function is zero or undefined. For a function \(f(x, y)\), this means solving for \(abla f = 0\), where \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).
02

Calculate Partial Derivatives

Given the function \(f(x, y)=x^{2}+kxy+y^{2}\), calculate the partial derivatives: 1. \(\frac{\partial f}{\partial x} = 2x + ky\). 2. \(\frac{\partial f}{\partial y} = kx + 2y\).
03

Set Partial Derivatives to Zero

Find where both partial derivatives are zero: 1. \(2x + ky = 0\). 2. \(kx + 2y = 0\).
04

Analyze Case \(k = 0\)

Substitute \(k = 0\) into the equations:- \(2x = 0 \Rightarrow x = 0\).- \(2y = 0 \Rightarrow y = 0\).Thus, \((0, 0)\) is a critical point.
05

Analyze Case \(k \neq 0\)

Solve \(2x + ky = 0\) and \(kx + 2y = 0\):1. From \(2x + ky = 0\), express \(y\): \(y = -\frac{2}{k}x\).2. Substitute \(y = -\frac{2}{k}x\) into \(kx + 2y = 0\), which simplifies to \(kx + 2(-\frac{2}{k}x) = 0\), or \(kx - \frac{4}{k}x = 0\).3. Simplifying gives \(x(k^2 - 4) = 0\), implying \(x = 0\) since \(keq 0\).4. Substituting \(x = 0\) into \(y = -\frac{2}{k}x\) yields \(y = 0\).Thus, \((0, 0)\) is a critical point regardless of \(k eq 0\).
06

Conclusion

Based on the analyses, \((0, 0)\) is a critical point for the function \(f(x, y)\) for any value of \(k\).

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Key Concepts

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

Partial Derivatives
Let's start with partial derivatives, which are essential when dealing with functions of multiple variables, like our function \( f(x, y) = x^2 + kxy + y^2 \). Imagine you are walking along a hill; partial derivatives tell us the slope of the hill in the direction of the axis. For our function:
  • \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as \( x \) varies, holding \( y \) constant.
  • \( \frac{\partial f}{\partial y} \) tells us the change in \( f \) as \( y \) changes, while \( x \) is fixed.
For our specific function, the calculations give us:
  • \( \frac{\partial f}{\partial x} = 2x + ky \)
  • \( \frac{\partial f}{\partial y} = kx + 2y \)
These derivatives help us determine where the function's rate of change is zero, which is a crucial part of finding critical points.
Gradient of a Function
The gradient of a function is like a beacon that points in the direction of the greatest rate of increase of that function. For a function \( f(x, y) \), the gradient is represented as:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
For our function \( f(x, y) = x^2 + kxy + y^2 \), the gradient becomes:
  • \( abla f = (2x + ky, kx + 2y) \)
Setting the gradient to zero is akin to finding those quiet spots on our hill where the slopes in both the \( x \) and \( y \) directions flatten out. These spots are our critical points.
Solving Equations
To find critical points, we solve the system of equations where the gradient equals zero. Our system is:
  • \( 2x + ky = 0 \)
  • \( kx + 2y = 0 \)
Here, solving means finding \( x \) and \( y \) that satisfy both equations simultaneously. For this:- When \( k = 0 \), the equations boil down to:
  • \( 2x = 0 \rightarrow x = 0 \)
  • \( 2y = 0 \rightarrow y = 0 \)
- For \( k eq 0 \), we handle the equations together:
  • Express \( y \) from the first equation: \( y = -\frac{2}{k}x \)
  • Substitute into the second: \( kx + 2(-\frac{2}{k}x) = 0 \rightarrow x(k^2 - 4) = 0 \)
  • Thus, \( x = 0 \) since \( k eq 0 \)
  • Then \( y = -\frac{2}{k}(0) \rightarrow y = 0 \)
This analysis shows (0,0) fits both cases as a critical point.
Analysis of Cases
In mathematical problem-solving, analyzing different cases is key to finding universal truths or confirming whether results hold under various conditions. Here, we analyzed two cases:
  • Case \( k = 0 \): leads directly to both \( x \) and \( y \) being zero, as shown earlier.
  • Case \( k eq 0 \): By solving the modified systems where \( k \) could be any non-zero constant, we still found that both \( x \) and \( y \) equate to zero.
This kind of case analysis is fundamental, ensuring that critical points are not artifacts of a particular assumption or oversight. It reassures us that indeed, under any circumstance for \( k \), the point \( (0,0) \) remains a critical feature of the function. Understanding these analyses equips you to tackle more complex situations and ensures solutions hold true over different scenarios.

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Most popular questions from this chapter

Show that each function satisfies a Laplace equation. \(f(x, y)=e^{-2 y} \cos 2 x\)

(A) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. $$f(x, y)=\sin ^{-1}(y-x)$$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\). e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{aligned} &f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16, \quad-3 \leq x \leq 3,\\\ &-6 \leq y \leq 6 \end{aligned}$$

Find the limits by rewriting the fractions first. $$\lim _{(x, y) \rightarrow(2,2)} \frac{x-y}{x^{4}-y^{4}}$$

When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}},\quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right),\quad(3)$$ with all sums running from \(k=1\) to \(k=n\). Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\), 3\. handle data analytically. We demonstrated these ideas with a variety of applications in Section 1.4.
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