Problem 26 - 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. $$f(x, y)=12-4 x^{2}-8 y^{2} ;(-1,4) ; \text { estimate } f(-1.05,3.95)$$

Step-by-Step Solution

Answer

In this exercise, we have found the linear approximation (tangent plane) of the function $$f(x, y) = 12 - 4x^2 - 8y^2$$ at point \((-1, 4)\), using the partial derivatives and evaluating them at the given point. The tangent plane is $$L(x, y) = -116 + 8(x + 1) - 64(y - 4)$$. We then used this tangent plane to estimate the function value at the nearby point \((-1.05, 3.95)\), resulting in an estimated value of $$f(-1.05, 3.95) \approx -119.6$$.

Step 1: Calculate the partial derivatives of the function

First, we need to find the partial derivatives of the function with respect to x and y. These will be used to create the tangent plane at the given point. $$f_x(x, y) = \frac{\partial{f}}{\partial{x}} = -8x$$ $$f_y(x, y) = \frac{\partial{f}}{\partial{y}} = -16y$$

Step 2: Evaluate the partial derivatives at the given point

Evaluate the partial derivatives at the given point \((-1, 4)\): $$f_x(-1, 4) = -8(-1) = 8$$ $$f_y(-1, 4) = -16(4) = -64$$

Step 3: Calculate the linear approximation (tangent plane)

The linear approximation (tangent plane) of the function at \((-1, 4)\) is given by the equation: $$L(x, y) = f(-1, 4) + f_x(-1, 4)(x + 1) + f_y(-1, 4)(y - 4)$$ Evaluate the function at \((-1, 4)\): $$f(-1, 4) = 12 - 4(-1)^2 - 8(4)^2 = -116$$ Now, plug in the values of the function and its partial derivatives into the linear approximation equation: $$L(x, y) = -116 + 8(x + 1) - 64(y - 4)$$

Step 4: Estimate the function value at the nearby point

Use the linear approximation to estimate the function value at \((-1.05, 3.95)\): $$L(-1.05, 3.95) = -116 + 8(-1.05 + 1) - 64(3.95 - 4)$$ Calculate the value: $$L(-1.05, 3.95) = -116 - 0.4 - 3.2$$ The estimated function value is: $$f(-1.05, 3.95) \approx -119.6$$

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial for comprehending how a function behaves in multiple dimensions. A partial derivative of a function of several variables is the derivative with respect to one variable, keeping the others constant.
So, for a function like \( f(x, y) = 12 - 4x^2 - 8y^2 \), you have to focus on one variable at a time.

• To find the partial derivative with respect to \( x \), treat \( y \) as constant. For \( f_x(x, y) = \frac{\partial f}{\partial x} = -8x \).
• To find the partial derivative with respect to \( y \), treat \( x \) as constant. For \( f_y(x, y) = \frac{\partial f}{\partial y} = -16y \).

The process allows us to see how sensitive the function \( f \) is to changes only in \( x \) or \( y \). Evaluating these derivatives at a particular point, like \((-1, 4)\), helps us understand the slope of the function in both the \( x \) and \( y \) directions, which is essential for finding tangent planes.

Tangent Plane

The concept of a tangent plane is an extension of the tangent line to functions of two variables. For a given surface, this plane “touches” the surface only at the given point and mirrors the surface locally.To find the tangent plane of \( f(x, y) \) at a specific point, use its partial derivatives:1. Calculate \( f_x \) and \( f_y \) at the point \((-1, 4)\). - \( f_x(-1, 4) = 8 \) - \( f_y(-1, 4) = -64 \)2. Use the point-area formula for the tangent plane:
L(x, y) = \( f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \)For this problem, it turns into:\[ L(x, y) = -116 + 8(x + 1) - 64(y - 4) \]This equation describes the plane that approximates the function \( f \) near \((-1, 4)\), providing a local linearization of the surface.

Function Estimation

Function estimation helps us predict the value of a complex function at a nearby point, using simpler calculations. Linear approximation provides a way to estimate these values near a given point by using the tangent plane.Here’s the approach:

• First, solve for the linear approximation of \( f(x, y) \) using the tangent plane formula obtained earlier.
• Next, substitute the nearby point \((-1.05, 3.95)\) into \( L(x, y) \).

For this exercise, the calculation is:- Substitute into the equation: \( L(-1.05, 3.95) = -116 + 8(-1.05 + 1) - 64(3.95 - 4) \)- Simplify: \( L(-1.05, 3.95) = -116 - 0.4 - 3.2 \)The result, \( -119.6 \), closely estimates the original function \( f \) at \((-1.05, 3.95)\). This method is highly effective for making quick predictions without calculating the exact function value directly.