Chapter 14: Problem 32
Find the linearization \(L(x, y)\) of the function at each point. $$ f(x, y)=e^{2 y-x} \quad \text { at } \quad \text { a. }(0,0), \quad \text { b. }(1,2) $$
Short Answer
Expert verified
\(L(x, y) = 1-x+2y\) for (0,0) and \(L(x, y) = e^3(-1+x+2y)\) for (1,2).
Step by step solution
01
Understand Linearization
Linearization is used to approximate a function near a given point. The formula for the linearization of a function \(f(x, y)\) at the point \((a, b)\) is \(L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)\), where \(f_x\) and \(f_y\) are partial derivatives.
02
Compute Partial Derivatives
To find the partial derivatives, first compute \(f_x(x, y)\) and \(f_y(x, y)\). \( f_x(x, y) = \frac{\partial}{\partial x}(e^{2y-x}) = -e^{2y-x} \)\( f_y(x, y) = \frac{\partial}{\partial y}(e^{2y-x}) = 2e^{2y-x} \)
03
Evaluate at Point (0,0)
Evaluate the function and its partial derivatives at the point \((0,0)\).\( f(0, 0) = e^{2(0)-0} = e^0 = 1 \)\( f_x(0, 0) = -e^0 = -1 \)\( f_y(0, 0) = 2e^0 = 2 \)Substitute these into the linearization formula: \( L(x, y) = 1 - 1(x-0) + 2(y-0) = 1 - x + 2y \)
04
Evaluate at Point (1,2)
Evaluate the function and its partial derivatives at the point \((1,2)\).\( f(1, 2) = e^{2(2)-1} = e^3 \)\( f_x(1, 2) = -e^3 \)\( f_y(1, 2) = 2e^3 \)Substitute these into the linearization formula: \( L(x, y) = e^3 - e^3(x-1) + 2e^3(y-2) \) Simplify to: \( L(x, y) = e^3(1 - (x-1) + 2(y-2)) = e^3(1 -x + 1 + 2y - 4) = e^3(2 - x + 2y - 3) = e^3(-1-x+2y) \) Thus, \( L(x, y) = e^3(-1+x+2y) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, especially in the study of functions of multiple variables. They help us understand the rate of change of a function concerning one of its variables while keeping the other variables constant. For a function like \(f(x, y) = e^{2y - x}\), partial derivatives are instrumental in finding the linearization of the function.
- The partial derivative of \(f\) with respect to \(x\), denoted as \(f_x(x, y)\), represents the change in \(f\) as \(x\) changes, with \(y\) held constant. In this case, \(f_x(x, y) = -e^{2y-x}\).
- Similarly, the partial derivative with respect to \(y\), \(f_y(x, y)\), indicates the change in \(f\) as \(y\) changes, with \(x\) constant. Here, \(f_y(x, y) = 2e^{2y-x}\).
Function Approximation
Function approximation is a powerful tool in mathematics that allows us to estimate complex functions using simpler ones. Linearization is a technique that involves approximating a nonlinear function with a linear one near a given point. This can be immensely useful for solving problems where direct calculation or analysis of the original function is difficult.
When dealing with the function \(f(x, y) = e^{2y-x}\), the linearization at a point \((a, b)\) involves finding a linear function \(L(x, y)\) that closely resembles \(f(x, y)\) around \((a, b)\). The formula \(L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)\) helps achieve this by incorporating both the value of the function and the rates of change provided by the partial derivatives.
When dealing with the function \(f(x, y) = e^{2y-x}\), the linearization at a point \((a, b)\) involves finding a linear function \(L(x, y)\) that closely resembles \(f(x, y)\) around \((a, b)\). The formula \(L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)\) helps achieve this by incorporating both the value of the function and the rates of change provided by the partial derivatives.
- At \((0,0)\), the linear approximation yields \(L(x, y) = 1 - x + 2y\).
- At \((1,2)\), it simplifies to \(L(x, y) = e^3(-1 + x + 2y)\).
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions of more than one variable. It opens up a broader set of tools for understanding complex functions such as \(f(x, y) = e^{2y-x}\).
In multivariable calculus, functions have more than one input, which results in a need to account for multiple dimensions. This is where partial derivatives and linearization become crucial.
In multivariable calculus, functions have more than one input, which results in a need to account for multiple dimensions. This is where partial derivatives and linearization become crucial.
- Partial derivatives help us analyze how a function changes in each variable's direction independently.
- Linearization provides a strategy to approximate these functions in their local neighborhoods, converting complex surfaces and curves into almost flat planes at small scales.