Chapter 14: Problem 25
In Exercises \(25-30,\) find the linearization \(L(x, y)\) of the function at each point. $$ f(x, y)=x^{2}+y^{2}+1 \text { at } \quad \text { a. }(0,0), \quad \text { b. }(1,1) $$
Short Answer
Expert verified
At (0,0), \(L(x, y) = 1\); at (1,1), \(L(x, y) = 2x + 2y - 1\).
Step by step solution
01
Understand the Linearization Formula
The linearization of a function \(f(x, y)\) at a point \((x_0, y_0)\) is given by the formula: \(L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\). Here, \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.
02
Calculate Partial Derivatives
First, find \(f_x\) by differentiating \(f(x, y) = x^2 + y^2 + 1\) with respect to \(x\), which gives \(f_x = 2x\). Next, find \(f_y\) by differentiating with respect to \(y\), resulting in \(f_y = 2y\).
03
Evaluate at Point (0,0)
Evaluate the function and its derivatives at the point \((0, 0)\):- \(f(0, 0) = 0^2 + 0^2 + 1 = 1\)- \(f_x(0, 0) = 2(0) = 0\)- \(f_y(0, 0) = 2(0) = 0\)Thus, the linearization at \((0,0)\) is: \(L(x, y) = 1 + 0(x - 0) + 0(y - 0) = 1\).
04
Evaluate at Point (1,1)
Evaluate the function and its derivatives at the point \((1, 1)\):- \(f(1, 1) = 1^2 + 1^2 + 1 = 3\)- \(f_x(1, 1) = 2(1) = 2\)- \(f_y(1, 1) = 2(1) = 2\)Thus, the linearization at \((1,1)\) is: \(L(x, y) = 3 + 2(x - 1) + 2(y - 1) = 2x + 2y - 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linearization
Linearization is a technique used in calculus to approximate the value of a function near a point using its first derivatives. It's like finding a simple line that matches the curve of your function at the neighborhood of a specific point. This can be incredibly helpful when dealing with complex functions that are tricky to analyze or visualize as they are.
- Linearization uses the first-order Taylor series expansion.
- Gives an approximate value with the same tangent at the point.
- It is generally easier to work with than the original function.
Partial Derivatives
Partial derivatives are pivotal when dealing with functions of more than one variable. They represent how a function changes as each variable is varied individually, keeping others constant. This concept bolsters our understanding of how multivariable functions behave in a multidimensional space.For the function \[f(x, y) = x^2 + y^2 + 1\], we compute the partial derivatives with respect to each variable:
- The partial derivative with respect to \(x\) is denoted as \(f_x = \frac{\partial}{\partial x}(x^2 + y^2 + 1) = 2x\). Here, \(y\) is treated as a constant, focusing on changes along the \(x\)-direction.
- The partial derivative with respect to \(y\) is \(f_y = \frac{\partial}{\partial y}(x^2 + y^2 + 1) = 2y\) — understanding changes along the \(y\)-axis with \(x\) held steady.
Multivariable Functions
Multivariable functions are an extension of single-variable functions, mapping inputs from more than one dimension into a single output. These functions are foundational in understanding systems where multiple factors or measurements influence an outcome. Consider our function \[f(x, y) = x^2 + y^2 + 1\], which accepts two variables. Visualizing this can be imagined as a surface in 3D space, where each pair \((x, y)\) corresponds to a point on that surface.
- They are particularly significant in fields like physics, engineering, and economics.
- Require tools like partial derivatives and linearization for analysis.
- Help model phenomena where dependencies aren't optional and multiple variables play a role.