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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. $$f(x, y)=-x^{2}+2 y^{2} ;(3,-1) ; \text { estimate } f(3.1,-1.04)$$

Short Answer

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Question: Use linear approximation to estimate the value of the function \(f(3.1, -1.04)\) if \(f(x, y) = -x^2 + 2y^2\) and the given point is \((3,-1)\). Answer: The estimated value of \(f(3.1, -1.04)\) using linear approximation is approximately \(-7.44\).

Step by step solution

01

Calculate the partial derivatives

First, compute the partial derivatives of the function with respect to both x and y. These derivatives will be used to build the linear approximation function. $$\frac{\partial f}{\partial x} = -2x$$ $$\frac{\partial f}{\partial y} = 4y$$
02

Evaluate the partial derivatives at the given point

Now, evaluate the partial derivatives at the given point, \((3,-1)\). $$\frac{\partial f}{\partial x}(3,-1) = -2(3) = -6$$ $$\frac{\partial f}{\partial y}(3,-1) = 4(-1) = -4$$
03

Define the linear approximation function

Use the computed values to define the linear approximation function, \(L(x,y)\), with \(f(x_0, y_0)\) being the functional value and \(f_{x_0}\) and \(f_{y_0}\) as evaluated partial derivatives in (Step 2). $$L(x,y) = f(3,-1) + \frac{\partial f}{\partial x}(3,-1)(x-3) + \frac{\partial f}{\partial y}(3,-1)(y+1)$$
04

Plug in the given values

Now plug in the given point \((3,-1)\) into the original function \(f(x,y)\) and compute the functional value \(f(3,-1)\): $$f(3, -1) = -3^2 + 2(-1)^2 = -9 + 2 = -7$$
05

Compute the linear approximation function

Now, by plugging in values from Steps 2 and 4, we can get the linear approximation function, \(L(x,y)\): $$L(x,y) = -7 -6(x-3) - 4(y+1)$$
06

Estimate the given function value

Finally, use the linear approximation function to estimate the given function value at point \((3.1,-1.04)\): $$L(3.1, -1.04) = -7 -6(3.1-3) - 4(-1.04+1)$$ $$L(3.1, -1.04) = -7 - 0.6 + 0.16$$ $$L(3.1, -1.04) \approx -7.44$$ So the estimated value of \(f(3.1,-1.04)\) using the linear approximation is approximately \(-7.44\).

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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, a partial derivative is a derivative that focuses on one variable at a time, keeping all other variables constant. This concept is essential when finding the rate of change in functions with more than one variable. For example, in the exercise provided, the function is defined as \(f(x, y)=-x^{2}+2y^{2}\). To find the linear approximation, it is crucial to calculate the partial derivatives with respect to \(x\) and \(y\).
  • The partial derivative with respect to \(x\) is found by differentiating the function treating \(y\) as a constant. Hence, \(\frac{\partial f}{\partial x} = -2x\).
  • Similarly, the partial derivative with respect to \(y\) is obtained by treating \(x\) as a constant, giving \(\frac{\partial f}{\partial y} = 4y\).
Evaluating these partial derivatives at a point, such as \((3,-1)\), provides the slope of the tangent plane at that point. This slope helps construct the linear approximation to predict the behavior of the function around the point.
Function Estimation
Function estimation involves using known data to approximate unknown values of a function. Linear approximation is one of the simplest techniques for function estimation. It uses the concept of the tangent plane in multivariable calculus to estimate the function.
In the context of the provided exercise, once the partial derivatives are known and evaluated at the point \((3,-1)\), they are used to construct the linear approximation function, \(L(x,y)\). This is done using the formula:\[ L(x,y) = f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y-y_0) \]Once \(L(x,y)\) is formulated, it serves to estimate the value of \(f(3.1,-1.04)\) by plugging these values into \(L(x,y)\).
This estimation is particularly useful when \(x\) and \(y\) are close to \((3,-1)\), thus making linear approximation a practical technique for short-range predictions.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to multiple dimensions, analyzing functions that depend on several variables. This field includes the study of partial derivatives, gradients, and more complex operations.
Linear approximation in multivariable calculus involves approximating a surface with a plane at a given point, similar to how single-variable calculus approximates a curve with a tangent line. The objective is to use the tangent plane, which is determined by partial derivatives, to simplify complex calculations by providing a linear estimate around a specific point.
The entire process begins with evaluating the function's value at a point, followed by finding the slopes in each direction using partial derivatives. This will guide the creation of the linear approximation function. By understanding this process, students can appreciate how linear approximation effectively estimates values for functions of multiple variables, showcasing the powerful techniques of multivariable calculus in practical applications.

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