Problem 30 - 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. $$f(x, y)=(x+y) /(x-y) ;(3,2) ; \text { estimate } f(2.95,2.05)$$

Step-by-Step Solution

Answer

Answer: The estimated value at the point (2.95, 2.05) is approximately $$4.9$$.

Step 1: Find the Partial Derivatives of the Function

To find the linear approximation, we need the partial derivatives of the function with respect to x and y. We will find these derivatives by applying the quotient rule. Let's first find the partial derivative with respect to x, $$\frac{\partial f}{\partial x}$$: $$\frac{\partial f}{\partial x} = \frac{(1)(x-y)-(x+y)(1)}{(x-y)^2} = \frac{-2y}{(x-y)^2}$$ Now, let's find the partial derivative with respect to y, $$\frac{\partial f}{\partial y}$$: $$\frac{\partial f}{\partial y} = \frac{(x+y)(-1)-(1)(x-y)}{(x-y)^2} = \frac{-2x}{(x-y)^2}$$

Step 2: Evaluate Partial Derivatives at the Given Point

Now that we have the partial derivatives, we will evaluate them at the given point (3, 2): $$\frac{\partial f}{\partial x}(3,2) = \frac{-2(2)}{(3-2)^2} = -4$$ $$\frac{\partial f}{\partial y}(3,2) = \frac{-2(3)}{(3-2)^2} = -6$$

Step 3: Determine the Linear Approximation Equation

With the obtained partial derivative values, we can now write the linear approximation (also known as the tangent plane) at the point (3, 2): $$L(x, y) = f(3, 2) + \frac{\partial f}{\partial x}(3,2)(x-3) + \frac{\partial f}{\partial y}(3,2)(y-2)$$ We also need to find the value of the function at the given point, f(3, 2): $$f(3, 2) = \frac{3 + 2}{3 - 2} = \frac{5}{1} = 5$$ Now, substitute the function value and the partial derivatives into the equation: $$L(x, y) = 5 - 4(x-3) - 6(y-2)$$

Step 4: Estimate the Function Value at (2.95, 2.05)

Finally, we will use the linear approximation equation to estimate the function value at the point (2.95, 2.05): $$L(2.95, 2.05) \approx 5 - 4(2.95-3) - 6(2.05-2) = 5 - 4(-0.05) - 6(0.05) = 5 + 0.2 - 0.3 = 4.9$$ Hence, we can estimate that the function value at the point (2.95, 2.05) is approximately $$4.9$$.

Key Concepts

Partial Derivatives

In calculus, partial derivatives represent the rate at which a function changes as one of its variables changes while all other variables are held constant. For a function of two variables, like \( f(x, y) \), you can take the partial derivative with respect to \( x \) and \( y \).
To do this, you treat every other variable as if it were a constant.With our function \( f(x, y) = (x+y)/(x-y) \), we first have to calculate the partial derivatives with respect to each variable. Specifically, \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). These represent how \( f \) changes when either \( x \) or \( y \) changes.

• \( \frac{\partial f}{\partial x} = \frac{-2y}{(x-y)^2} \)
• \( \frac{\partial f}{\partial y} = \frac{-2x}{(x-y)^2} \)

By understanding these rates of change, we can better approximate function values near the point of interest.

Quotient Rule

The quotient rule is essential when differentiating a function that is the division of two other functions. When you have a function that looks like \( g(x) / h(x) \), the rule helps you find its derivative.Here's the formula for the quotient rule:\[\left( \frac{g}{h} \right)' = \frac{g' \cdot h - g \cdot h'}{h^2}\]In this problem, our function \( f(x, y) = \frac{x+y}{x-y} \) requires the quotient rule.
By substituting \( g = x + y \) and \( h = x - y \) into the formula, we differentiate with respect to \( x \) or \( y \) as needed.

• When finding \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant during differentiation.
• When finding \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.

The quotient rule can be complex but following step by step helps manage this complexity.

Tangent Plane

The tangent plane is a linear approximation of a function at a given point. It provides a simple way to estimate function values close to this point with just a flat plane.For a function \( f(x, y) \), the equation for the tangent plane at a particular point \((a, b)\) is:\[L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(a, b)(x-a) + \frac{\partial f}{\partial y}(a, b)(y-b)\]This formula uses the function value as well as its partial derivatives at the chosen point.

• In our example, with the point (3, 2), and \( f(3, 2) = 5 \), the tangent plane simplification helps us estimate nearby values.
• Thus, at (2.95, 2.05), the estimated function value using the tangent plane is very close to the actual value, approximately \( 4.9 \).

The tangent plane is especially useful for functions that are too complex to calculate exactly at every point, making approximation easier.