Problem 210 - Calculus Volume 3

Question

Find the linear approximation of each function at the indicated point. \(\quad f(x, y)=\sqrt{20-x^{2}-7 y^{2}}, \quad P(2,1)\)

Step-by-Step Solution

Answer

The linear approximation is \( L(x, y) = \frac{20}{3} - \frac{2}{3}x - \frac{7}{3}y \).

Step 1: Identify the function and point

The function given is \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \) and we are finding the linear approximation at the point \( P(2, 1) \).

Step 2: Recall the formula for linear approximation

The linear approximation of a function \( f \) at a point \( (a, b) \) is given by:\[ L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \] where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.

Step 3: Compute the function value at the point

Calculate \( f(2, 1) = \sqrt{20 - 2^2 - 7 imes 1^2} = \sqrt{9} = 3 \).

Step 4: Find the partial derivative with respect to x

The partial derivative of \( f \) with respect to \( x \) is:\[ f_x(x, y) = \frac{-x}{\sqrt{20 - x^2 - 7y^2}} \]Evaluate it at \( (2, 1) \):\[ f_x(2, 1) = \frac{-2}{\sqrt{9}} = -\frac{2}{3} \]

Step 5: Find the partial derivative with respect to y

The partial derivative of \( f \) with respect to \( y \) is:\[ f_y(x, y) = \frac{-7y}{\sqrt{20 - x^2 - 7y^2}} \]Evaluate it at \( (2, 1) \):\[ f_y(2, 1) = \frac{-7 \times 1}{\sqrt{9}} = -\frac{7}{3} \]

Step 6: Substitute into the linear approximation formula

Using the formula from Step 2, substitute the known values:\[ L(x, y) = 3 - \frac{2}{3}(x - 2) - \frac{7}{3}(y - 1) \]Simplify:\[ L(x, y) = 3 - \frac{2}{3}x + \frac{4}{3} - \frac{7}{3}y + \frac{7}{3} \]\[ L(x, y) = \frac{20}{3} - \frac{2}{3}x - \frac{7}{3}y \]

Key Concepts

Linear Approximation

Linear approximation is a useful method in multivariable calculus to estimate the value of a function at a point close to a known value. It essentially involves creating a tangent plane to the surface defined by the function at that point.

This tangent plane gives us an approximation of the function's behavior near that point, making it easier to analyze in situations where the actual calculation might be too complex.

The core idea is to take the function's value at a known point and adjust it based on the changes suggested by its derivatives, which leads us to the concept of partial derivatives. Later, you'll see how partial derivatives come into play within the linear approximation formula: \[ L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \]This formula helps us systematically apply this approximation to multivariable functions.

Partial Derivatives

Partial derivatives are central to understanding how a function changes with respect to each variable independently, in a context where multiple variables exist.

For instance, in our function \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \), the partial derivative with respect to \( x \), denoted as \( f_x \), expresses the function's rate of change as \( x \) varies while keeping \( y \) constant.

• The partial derivative with respect to \( x \) is \( f_x(x, y) = \frac{-x}{\sqrt{20 - x^2 - 7y^2}} \).
• The partial derivative with respect to \( y \) is \( f_y(x, y) = \frac{-7y}{\sqrt{20 - x^2 - 7y^2}} \).

These derivatives provide vital information on how the function behaves around the point where they are evaluated. Their values directly contribute to calculating the linear approximation through the formula mentioned earlier.

Function Evaluation

Evaluating a function at a specific point helps in determining its current state or value at that point. In our example, we need to know the value of the function at the point \( P(2, 1) \) to start with linear approximation.

Calculating \( f(2, 1) \), we substitute \( x = 2 \) and \( y = 1 \) in the function \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \).

As we compute, we get: \( f(2, 1) = \sqrt{20 - 2^2 - 7 \times 1^2} = \sqrt{9} = 3 \).

This result provides the foundational value from which the linear approximation is carried out. Only by knowing this initial function value can we properly implement the linear approximation formula to estimate new points.

Point of Tangency

The point of tangency is the specific location on the graph of a function where the tangent plane approximates the surface. For linear approximation, this point, \((a, b)\), serves as the reference point for creating our linear model of the function.

In this case, we are considering the point \( P(2, 1) \) on the function \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \).

Here, the tangent plane provides a simplified linear equation that closely approximates the surface near \( P(2, 1) \).

By finding the partial derivatives at the point of tangency and evaluating them along with the function, we develop insights into how the plane mirrors the surface locally.

This sophisticated mathematical approach allows for significant simplifications, particularly for problems requiring the function's behavior around \((a, b)\) without delving into its full complexity.