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a. Find the linear approximation to the function \(f\) at the given point. b. Use part (a) to estimate the given function value. $$f(x, y)=-x^{2}+2 y^{2} ;(3,-1) ; \text { estimate } f(3.1,-1.04).$$

Short Answer

Expert verified
Question: Estimate the value of the function \(f(x,y) = -x^2 + 2y^2\) at the point \((3.1, -1.04)\) using linear approximation at the point \((3, -1)\). Answer: The estimated value of \(f(3.1, -1.04)\) using linear approximation is \(-7.44\).

Step by step solution

01

1. Calculate the partial derivatives

We need to find the partial derivatives of the function \(f(x,y)\) with respect to \(x\) and \(y\). To do that, we will differentiate the function with respect to one variable while treating the other variable as a constant. $$f_x(x,y) = \frac{\partial f}{\partial x} = \frac{\partial (-x^2 + 2y^2)}{\partial x} = -2x$$ $$f_y(x,y) = \frac{\partial f}{\partial y} = \frac{\partial (-x^2 + 2y^2)}{\partial y} = 4y$$
02

2. Evaluate the partial derivatives at the given point

Now we need to evaluate \(f_x\) and \(f_y\) at the point \((3, -1)\). So, we have: $$f_x(3, -1) = -2(3) = -6$$ $$f_y(3, -1) = 4(-1) = -4$$
03

3. Evaluate the function f at the given point

To find the constant term for the linear approximation, we need to evaluate the function \(f(x,y)\) at the given point (3, -1): $$f(3, -1) = -(3^2) + 2(-1)^2 = -9 + 2 = -7$$
04

4. Find the linear approximation

Now we have all the components to write the linear approximation \(L(x,y)\): $$L(x,y) = f(3, -1) + f_x(3, -1)(x-3) + f_y(3, -1)(y+1)$$ Replacing the values found in previous steps, we get: $$L(x,y) = -7 -6(x-3) -4(y+1)$$
05

5. Use the linear approximation to estimate the function value

In order to estimate the value of the function at the point \((3.1, -1.04)\), we will substitute these values into the linear approximation \(L(x,y)\): $$L(3.1, -1.04) = -7 -6(3.1-3) -4(-1.04+1)$$ Evaluating this expression, we get: $$L(3.1, -1.04) = -7 -6(0.1) -4(-0.04) = -7 -0.6 +0.16 = -7.44$$ The estimated value of \(f(3.1, -1.04)\) using linear approximation is \(-7.44\).

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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 are used to understand how a function changes as only one of its variables changes, while others remain constant. Imagine you have a surface that represents a function of two variables, like our function, \(-x^2 + 2y^2\). To explore the surface, you would take a cross-section along one direction, say the x-direction, to see how the elevation (or value of the function) changes as x changes. This is what a partial derivative does.
  • The partial derivative with respect to \(x\) is calculated by holding \(y\) constant and differentiating with respect to \(x\). For our exercise, this gives us \(f_x(x,y) = -2x\).
  • Similarly, the partial derivative with respect to \(y\) is obtained by keeping x constant and differentiating with respect to \(y\), yielding \(f_y(x,y) = 4y\).
These derivatives are crucial for understanding the behavior of the function and form the foundation for the linear approximation process.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions with more than one variable. This branch of mathematics is especially important because many real-world phenomena depend on multiple variables. In our exercise, the function \(f(x, y) = -x^2 + 2y^2\) involves two variables, x, and y.To work with such functions, we use tools like partial derivatives to explore changes in each variable independently, but multivariable calculus is not just about derivatives. It also examines how variables interact with each other, often using concepts like gradients and limits to analyze functions in space.
  • By examining the surface or graph of a multivariable function, we can visualize data and understand relationships between variables.
  • The linear approximation is just one of many applications, providing a way to estimate function values at nearby points reliably.
Multivariable calculus is essential for modeling and solving problems in physics, engineering, economics, and more.
Function Estimation
Function estimation involves approximating the output of a function based on known data. In our exercise, we use linear approximation — a technique derived from calculus — to estimate the value of our function at a new point close to a given one.Linear approximation simplifies a complex function into a linear equation or plane that closely matches the function near the point of interest.This involves crucial steps such as:
  • Determining the constant term by evaluating the function at a specific point.
  • Finding partial derivatives to understand how the function changes as each variable is altered slightly.
  • Using these derivatives in a linear formula, \(L(x, y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\), where \((x_0, y_0)\) is the given point.
Finally, substituting the point of interest into the linear approximation gives an estimate of the function’s value. This technique is beneficial when small changes occur in input, providing reliable predictions with minimal calculation.

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