Chapter 10: Problem 42
Find a linear approximation to each func\mathrm{tion } \(f(x, y)\) at the indicated point. \(\mathbf{f}(x, y)=\left[\begin{array}{c}(x+y)^{2} \\ x y\end{array}\right]\) at \((-1,1)\)
Short Answer
Expert verified
The linear approximation is \( \left[ \begin{array}{c} 0 \\ x - y + 1 \end{array} \right] \).
Step by step solution
01
Identify functions in vector form
We have two functions derived from the vector-valued function: \ \(f_1(x, y) = (x + y)^2\) and \(f_2(x, y) = xy\). We need approximations for each at the point \((-1, 1)\).
02
Find partial derivatives for f1
The partial derivative of \(f_1\) with respect to \(x\) is \(\frac{\partial}{\partial x}(x+y)^2 = 2(x+y)\). \ The partial derivative of \(f_1\) with respect to \(y\) is \(\frac{\partial}{\partial y}(x+y)^2 = 2(x+y)\).
03
Evaluate partial derivatives of f1 at the point (-1, 1)
Substitute \((-1, 1)\) into the partials: \ \(\frac{\partial f_1}{\partial x}\bigg|_{(-1, 1)} = 2(-1+1) = 0\) \ \(\frac{\partial f_1}{\partial y}\bigg|_{(-1, 1)} = 2(-1+1) = 0\).
04
Linear equation for f1
The linear approximation for \(f_1\) is given by: \ \(L_1(x, y) = f_1(-1, 1) + \frac{\partial f_1}{\partial x}(-1, 1)(x + 1) + \frac{\partial f_1}{\partial y}(-1, 1)(y - 1)\). \ Since the derivative values are zero, \(L_1(x, y) = f_1(-1, 1) = (-1+1)^2 = 0\).
05
Find partial derivatives for f2
The partial derivative of \(f_2\) with respect to \(x\) is simply \(\frac{\partial}{\partial x}(xy) = y\). \ The partial derivative of \(f_2\) with respect to \(y\) is \(\frac{\partial}{\partial y}(xy) = x\).
06
Evaluate partial derivatives of f2 at the point (-1, 1)
Substitute \((-1, 1)\) into the partials: \ \(\frac{\partial f_2}{\partial x}\bigg|_{(-1, 1)} = 1\) \ \(\frac{\partial f_2}{\partial y}\bigg|_{(-1, 1)} = -1\).
07
Linear equation for f2
The linear approximation for \(f_2\) is given by: \ \(L_2(x, y) = f_2(-1, 1) + \frac{\partial f_2}{\partial x}(-1, 1)(x + 1) + \frac{\partial f_2}{\partial y}(-1, 1)(y - 1)\). \ Calculate \(f_2(-1, 1) = (-1)(1) = -1\), so \ \(L_2(x, y) = -1 + 1(x + 1) - 1(y - 1)\).
08
Simplify the linear equation for f2
Simplify \(L_2\): \ \(L_2(x, y) = -1 + x + 1 - y + 1 = x - y + 1\).
09
Combine approximations
Combine the approximations for \(f_1\) and \(f_2\) into a vector: \ \(L(x, y) = \left[ \begin{array}{c} 0 \ x - y + 1 \end{array} \right]\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector-Valued Function
Vector-valued functions are mathematical expressions that produce vectors as output, instead of just a single scalar value. This means that when you input certain values, the output is a vector. Understanding vector-valued functions is crucial for grasping complex systems as they often describe phenomena in physics and engineering.
In our example, the vector-valued function is given by:
These components represent how inputs \(x\) and \(y\) are transformed into outputs. We need to approximate each component at the point \((-1, 1)\).
The linear approximation becomes very useful in simplifying complex vector outputs near a particular point, making the solutions more manageable and interpretable.
In our example, the vector-valued function is given by:
- \[ \mathbf{f}(x, y) = \left[ (x + y)^2, \ xy \right] \]
These components represent how inputs \(x\) and \(y\) are transformed into outputs. We need to approximate each component at the point \((-1, 1)\).
The linear approximation becomes very useful in simplifying complex vector outputs near a particular point, making the solutions more manageable and interpretable.
Partial Derivatives
Partial derivatives are fundamental in calculus and are essential when dealing with functions of multiple variables. They show how a function changes as one of the variables change, keeping the other variables constant.
Let's break down the task of finding partial derivatives for our functions:
Let's break down the task of finding partial derivatives for our functions:
- For \(f_1(x, y) = (x + y)^2\), taking the partial derivative with respect to \(x\) and \(y\) yields: - \(\frac{\partial f_1}{\partial x} = 2(x+y)\) - \(\frac{\partial f_1}{\partial y} = 2(x+y)\)
- For \(f_2(x, y) = xy\), the partial derivatives are: - \(\frac{\partial f_2}{\partial x} = y\) - \(\frac{\partial f_2}{\partial y} = x\)
Evaluation at a Point
Evaluating a function or its derivatives at a particular point is an essential skill in calculus, helping to calculate specific values that can aid in real-life applications. For linear approximation, this evaluation allows you to produce a local linear model of the function.
In our task, we evaluated the partial derivatives at point \((-1, 1)\):
Finally, we combined the linear approximations of \(f_1\) and \(f_2\) at \((-1, 1)\) yielding \(L(x, y) = [0, x - y + 1]\). This linear model is a simple but powerful tool to predict behavior closely around the point of interest, making complex functions easier to handle.
In our task, we evaluated the partial derivatives at point \((-1, 1)\):
- - For \(f_1\), both derivatives, \(\frac{\partial f_1}{\partial x}\) and \(\frac{\partial f_1}{\partial y}\), contribute as zero at \((-1, 1)\), showing that near this point, the slope is flat.
- - For \(f_2\), evaluating the derivatives: - \(\frac{\partial f_2}{\partial x}\) becomes \(1\) - \(\frac{\partial f_2}{\partial y}\) turns into \(-1\)
Finally, we combined the linear approximations of \(f_1\) and \(f_2\) at \((-1, 1)\) yielding \(L(x, y) = [0, x - y + 1]\). This linear model is a simple but powerful tool to predict behavior closely around the point of interest, making complex functions easier to handle.