Problem 25 - Calculus Early Transcendentals

Question

a. Find the linear approximation for the following functions at the given point. b. Use part (a) to estimate the given function value. $$f(x, y)=x y+x-y ;(2,3) ; \text { estimate } f(2.1,2.99)$$

Step-by-Step Solution

Answer

The estimated value of the function at point (2.1, 2.99) using the linear approximation method is approximately 5.39.

Step 1: Calculate partial derivatives

To find the linear approximation of a function at a point, we need to calculate its partial derivatives with respect to both x and y. In this case, the function is \(f(x, y) = xy + x - y\). Calculate the partial derivatives: $$\frac{\partial f}{\partial x} = \frac{\partial(xy + x - y)}{\partial x} = y + 1$$ $$\frac{\partial f}{\partial y} = \frac{\partial(xy + x - y)}{\partial y} = x - 1$$

Step 2: Evaluate partial derivatives at the given point

Now, we need to evaluate the partial derivatives at the given point (2, 3): $$\frac{\partial f}{\partial x}(2,3) = 3 + 1 = 4$$ $$\frac{\partial f}{\partial y}(2,3) = 2 - 1 = 1$$

Step 3: Write down the linear approximation formula

Next, let's write down the linear approximation formula: $$L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(a, b)(x - a) + \frac{\partial f}{\partial y}(a, b)(y - b)$$ Here, (a, b) is the given point (2, 3).

Step 4: Substitute the values into the linear approximation formula

Now, we have all the values we need to calculate the linear approximation L(x, y) at the given point (2, 3). Let's substitute these values into the formula: $$L(x, y) = f(2, 3) + \frac{\partial f}{\partial x}(2, 3)(x - 2) + \frac{\partial f}{\partial y}(2, 3)(y - 3)$$ $$L(x, y) = (2\cdot3 + 2 - 3) + 4(x - 2) + 1(y - 3) = 5 + 4(x - 2) + (y - 3)$$

Step 5: Estimate the function value using the linear approximation

Finally, we need to estimate the function value f(2.1, 2.99) using the linear approximation. Simply substitute these values into L(x, y): $$L(2.1, 2.99) = 5 + 4(2.1 - 2) + (2.99 - 3) = 5 + 4\cdot0.1 - 0.01 = 5 + 0.4 - 0.01 = 5.39$$ Thus, after completing these steps, we estimate that the function value \(f(2.1, 2.99) \approx 5.39\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They help us understand how a function changes concerning each variable separately. When you have a function of multiple variables, such as \( f(x, y) = xy + x - y \), partial derivatives are used to compute the rate of change of the function with respect to each variable.
In the context of linear approximation, partial derivatives are crucial for constructing the tangent plane at a specific point. For the function \( f(x, y) \) given in the problem, the partial derivatives are:

• \( \frac{\partial f}{\partial x} = y + 1 \)
• \( \frac{\partial f}{\partial y} = x - 1 \)

Once these derivatives are calculated, we evaluate them at a specific point, in this case, \((2, 3)\). This gives us the basis for creating a linear approximation, which essentially "flattens" the function to a plane at that point, simplifying computation of nearby values.

Function Estimation

Function estimation involves predicting the output value of a function for a given input, especially when the exact computation might be complex. Linear approximation provides a convenient method for estimating function values.
This technique uses the tangent plane created by partial derivatives to estimate the function's value close to a specific point. The idea is that linear functions, like planes, are easier to work with because they involve simple arithmetic operations.
In this exercise, we estimated \( f(2.1, 2.99) \) using the linear approximation formula, \( L(x, y) \). By calculating the linear function using the derivatives we found and substituting the values, \( L(2.1, 2.99) \) gives us 5.39, an approximate value for \( f(x, y) \) near the point \((2, 3)\). This shows how linear approximation can save calculation time while providing a reasonable estimate.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of two or more variables. It encompasses a range of methods for analyzing and estimating changes in these functions. This field of mathematics is particularly useful in understanding systems where multiple factors interact at once.
In this problem, multivariable calculus helps us delve into the behavior of functions like \( f(x, y) = xy + x - y \) in different dimensions. One of the primary tools used is the gradient, which consists of partial derivatives with respect to each variable.
The application of multivariable calculus in linear approximation involves creating a tangent plane that makes estimating values around a known point easier. This highlights the power of using derivatives and the linearization technique, drawing from foundational calculus principles to solve real-world problems with complex variables.