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Linear approximation a. Find the linear approximation to the function \(f\) at the given point. b. Use part (a) to estimate the given function value. $$\begin{aligned} &f(x, y, z)=\ln (1+x+y+2 z) ;(0,0,0) ; \text { estimate }\\\ &f(0.1,-0.2,0.2) \end{aligned}$$

Short Answer

Expert verified
Answer: The estimated value of the function at the point \((0.1, -0.2, 0.2)\) is \(0.3\).

Step by step solution

01

Find Partial Derivatives

First, we will find the partial derivatives of the given function with respect to \(x, y, z\). We have: $$\frac{\partial f}{\partial x} = \frac{1}{1+x+y+2z}$$ $$\frac{\partial f}{\partial y} = \frac{1}{1+x+y+2z}$$ $$\frac{\partial f}{\partial z} = \frac{2}{1+x+y+2z}$$
02

Evaluate Partial Derivatives at the Given Point

Next, we evaluate the partial derivatives at the given point \((0,0,0)\). Doing so, we get: $$\frac{\partial f}{\partial x}(0,0,0) = \frac{1}{1+0+0+2\cdot0} = 1$$ $$\frac{\partial f}{\partial y}(0,0,0) = \frac{1}{1+0+0+2\cdot0} = 1$$ $$\frac{\partial f}{\partial z}(0,0,0) = \frac{2}{1+0+0+2\cdot0} = 2$$
03

Use the Linear Approximation Formula

Now, we have all the necessary information to find the linear approximation of the function at the given point. The formula for the linear approximation is: $$f(a+\Delta x, b+\Delta y, c+\Delta z) \approx f(a, b, c) + \frac{\partial f}{\partial x}(a, b, c)\Delta x + \frac{\partial f}{\partial y}(a, b, c)\Delta y + \frac{\partial f}{\partial z}(a, b, c)\Delta z$$ where \((a, b, c) = (0,0,0)\) and \(\Delta x = 0.1, \Delta y = -0.2, \Delta z = 0.2\). Thus, we have: $$f(0.1,-0.2,0.2) \approx f(0,0,0) + \left( \frac{\partial f}{\partial x}(0,0,0) \cdot 0.1\right) + \left(\frac{\partial f}{\partial y}(0,0,0) \cdot (-0.2)\right) + \left(\frac{\partial f}{\partial z}(0,0,0) \cdot 0.2\right)$$
04

Solve for the Linear Approximation

Finally, we solve for the linear approximation using the values calculated above: $$f(0.1,-0.2,0.2) \approx \ln(1+0+0+2\cdot 0) + 1\cdot0.1 + 1\cdot(-0.2) + 2\cdot 0.2$$ $$f(0.1,-0.2,0.2) \approx \ln(1) + 0.1 - 0.2 + 0.4$$ $$f(0.1,-0.2,0.2) \approx 0 + 0.3$$ Hence, the linear approximation to the function \(f(x, y, z)\) at the point \((0.1,-0.2,0.2)\) is \(0.3\).

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Key Concepts

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

Partial Derivatives
In multivariable calculus, partial derivatives play a crucial role in analyzing the behavior of functions with several variables. A partial derivative represents the rate at which a function changes as one variable changes, holding other variables constant. When we have a function like \( f(x, y, z) = \ln(1 + x + y + 2z) \) and we need to find its linear approximation, the first step is to compute these derivatives for each variable individually.

In the given exercise, we found the partial derivatives with respect to \( x \), \( y \) and \( z \), which are denoted as \( \frac{\partial f}{\partial x} \) , \( \frac{\partial f}{\partial y} \) , and \( \frac{\partial f}{\partial z} \), respectively. These are the gradients that indicate the slope of the function in the direction of each variable at a specific point. By evaluating them at the point \( (0,0,0) \), we can see how the function \( f \) changes near this point when each coordinate is slightly modified while the others remain unchanged.
Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus to higher dimensions. It allows us to study functions that have more than one input, which is critical in fields like physics, engineering, and economics. Concepts such as partial derivatives, gradients, and multiple integrals help us to understand how these multivariable functions behave.

For example, when we talk about a function \( f(x, y, z) \) as in the exercise, we're dealing with a three-dimensional space. This means that every point can be represented by three coordinates, and the function's output can be visualized as a surface within this space. The beauty of multivariable calculus is in its ability to analyze the curvature, slopes, and other geometrical properties of this surface. When dealing with real-world applications, such geometric interpretations can be incredibly insightful for predicting and optimizing outcomes related to the function.
Linearization
Linearization is the process of approximating a function near a given point using a linear function, also known as the tangent line (or plane, in higher dimensions). This approximation is extremely useful for complex functions where calculating the exact value is difficult or unnecessary.

In simple terms, what we are doing is creating the best straight-line approximation that touches our function at a particular point. This line or plane will be a close estimate to the function's value near that point. The linear approximation, as performed in the exercise, uses the value of the function and its partial derivatives at a specific point to calculate an estimate for a point close by. The general formula applied in the example exercise \( f(a + \Delta x, b + \Delta y, c + \Delta z) \approx f(a, b, c) + \frac{\partial f}{\partial x}(a, b, c)\Delta x + \frac{\partial f}{\partial y}(a, b, c)\Delta y + \frac{\partial f}{\partial z}(a, b, c)\Delta z \) is fundamental in linearization and shows how the concept of slopes from single-variable calculus extends to functions of several variables. It is a necessary approximation method for making predictions in various scientific and engineering tasks.

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Most popular questions from this chapter

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