Chapter 10: Problem 45
Find a linear approximation to $$\mathbf{f}(x, y)=\left[\begin{array}{c} (x-y)^{2} \\ 2 x^{2} y \end{array}\right]$$ at \((2,-3)\). Use your result to find an approximation for \(f(1.9,-3.1)\), and compare the approximation with the value of \(f(1.9,-3.1)\) that you get when you use a calculator.
Short Answer
Step by step solution
Find the Partial Derivatives
Evaluate Partial Derivatives at Given Point
Formulate Linear Approximation
Calculate Linear Approximation at (1.9, -3.1)
Compare with Actual Function Value Using Calculator
Analyze the Difference
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
For a vector function like \(\mathbf{f}(x, y) = \left[\begin{array}{c} (x-y)^{2} \ 2 x^{2} y \end{array}\right]\), we focus on finding the derivative of each individual element with respect to one variable at a time. Consider the first component \((x-y)^2\). The partial derivative with respect to \(x\), denoted as \(f_x\), would be \(2(x-y)\), because the term involves \(x\) linearly. The partial derivative with respect to \(y\), \(f_y\), is \(-2(x-y)\), due to the influence of \(y\) being subtracted.
For the second component, \(2x^2y\), the partial with respect to \(x\) results in \(4xy\) because it considers the quadratic term in \(x\). Meanwhile, \(g_y = 2x^2\) reflects the linearity in \(y\). Through these derivatives, we gauge how each part of the function varies independently with changes in \(x\) or \(y\), forming the basis for linear approximation.
Vector Functions
Consider our vector function example, \(\mathbf{f}(x, y) = \left[\begin{array}{c} (x-y)^{2} \ 2 x^{2} y \end{array}\right]\). Here, for every input pair \((x, y)\), we get a vector with two entries. Each entry represents different functional outputs based on the given inputs.
This type of function is invaluable in fields requiring calculations across multiple dimensions, such as physics for force vectors, or computer graphics for transformations. Understanding how to work with and analyze vector functions, especially in the context of partial derivatives and approximations, enables us to solve complex real-world problems by breaking them down into multiple, comprehensible parts.
Function Evaluation
For the function \(\mathbf{f}(x, y) = \left[\begin{array}{c} (x-y)^{2} \ 2 x^{2} y \end{array}\right]\), evaluating at a point like \((2, -3)\) involves substituting \(x\) with 2 and \(y\) with -3.
- The first component: \((2 - (-3))^2 = 25\)
- The second component: \(2 \cdot 2^2 \cdot (-3) = -24\)
Non-linear Functions
In our context, \(\mathbf{f}(x, y) = \left[\begin{array}{c} (x-y)^{2} \ 2 x^{2} y \end{array}\right]\) is a clear example of a non-linear function. The presence of exponentiation in terms like \((x-y)^2\) and \(x^2\) confirms the non-linearity.
Non-linear functions often require more sophisticated tools for analysis, such as differentiation and linear approximation. Approximating a non-linear function with a linear one near a point simplifies understanding complex behaviors locally. This is the essence of linear approximation, which acts as a powerful tool in calculus to handle non-linear functions efficiently, especially in fields like data modeling, physics, and engineering.