Problem 210 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The linear approximation is \( L(x, y) = \frac{16}{3} - \frac{2}{3}x - \frac{7}{3}y \).

Step 1: Identify the Function and the Given Point

The function given is the two-variable function \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \). We need to find the linear approximation of this function at the point \( P(2,1) \).

Step 2: Calculate the Partial Derivatives

To find the linear approximation, we first need the partial derivatives of \( f \) with respect to \( x \) and \( y \).\Using the chain rule, the partial derivative with respect to \( x \) is: \[ f_x = \frac{-x}{\sqrt{20-x^2-7y^2}} \]The partial derivative with respect to \( y \) is:\[ f_y = \frac{-7y}{\sqrt{20-x^2-7y^2}} \]

Step 3: Evaluate the Partial Derivatives at the Given Point

Next, calculate the values of the partial derivatives at the point \( (2, 1) \).\For \( f_x(2, 1) \):\[ f_x(2, 1) = \frac{-2}{\sqrt{20-2^2-7(1)^2}} = \frac{-2}{\sqrt{20-4-7}} = \frac{-2}{\sqrt{9}} = \frac{-2}{3} \]For \( f_y(2, 1) \):\[ f_y(2, 1) = \frac{-7(1)}{\sqrt{20-2^2-7(1)^2}} = \frac{-7}{\sqrt{9}} = \frac{-7}{3} \]

Step 4: Determine the Function Value at the Given Point

Calculate the value of the function \( f \) at the point \( (2, 1) \):\[ f(2, 1) = \sqrt{20 - 2^2 - 7(1)^2} = \sqrt{9} = 3 \]

Step 5: Write the Linear Approximation Formula

The linear approximation of \( f(x, y) \) at the point \( (2, 1) \) is given by:\[ L(x, y) = f(2, 1) + f_x(2, 1)(x - 2) + f_y(2, 1)(y - 1) \]Substitute the calculated values into this formula:\[ L(x, y) = 3 + \left(\frac{-2}{3}\right)(x - 2) + \left(\frac{-7}{3}\right)(y - 1) \]

Step 6: Simplify the Linear Approximation

Simplify the expression for the linear approximation:\[ L(x, y) = 3 - \frac{2}{3}(x - 2) - \frac{7}{3}(y - 1) \]This gives:\[ L(x, y) = 3 - \frac{2}{3}x + \frac{4}{3} - \frac{7}{3}y + \frac{7}{3} \]Combine the constants:\[ L(x, y) = \frac{16}{3} - \frac{2}{3}x - \frac{7}{3}y \]

Key Concepts

Partial Derivatives

In multivariable calculus, a partial derivative is a derivative where we hold one variable constant and differentiate with respect to another variable. This is particularly useful when dealing with functions of multiple variables, like our function here, \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \).

Consider the partial derivative with respect to \( x \), notated as \( f_x \). For our function, applying the chain rule helps differentiate the square root:

• The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \); thus, the derivative of \( \sqrt{20 - x^2 - 7y^2} \) gives the formula \( f_x = \frac{-x}{\sqrt{20 - x^2 - 7y^2}} \).

Similarly, for \( f_y \), we differentiate with respect to \( y \), treating \( x \) as a constant:

• The result is \( f_y = \frac{-7y}{\sqrt{20 - x^2 - 7y^2}} \).

Partial derivatives help us understand how the function changes as \( x \) or \( y \) changes individually.

Chain Rule

The chain rule is one of the fundamental tools in calculus, especially useful for differentiating compositions of functions. When dealing with a function like \( f(x, y) = \sqrt{20 - x^2 - 7y^2} \), the chain rule enables us to find derivatives of expressions nested within other functions.

For example, when deriving \( f_x \), notice that the outer function is a square root. It's applied to the inner function \( 20 - x^2 - 7y^2 \). The key steps are:

• Differentiate the outer function, treating the inner part as a single unit, resulting in \( \frac{1}{2\sqrt{u}} \).
• Then differentiate the inner function, \( -x^2 - 7y^2 \), to give \( -2x \).

The chain rule combines these to form \( f_x = \frac{-x}{\sqrt{20 - x^2 - 7y^2}} \).
This approach provides a method to handle complex differentiation in multiple variable contexts.

Tangent Plane

In the realm of multivariable calculus, the tangent plane offers a way to approximate a surface near a given point. This concept is essential for understanding linear approximation in functions of two variables.

If we consider the function \( f(x, y) \) at point \( P(2, 1) \), the tangent plane can be thought of as the plane that "just touches" the surface at \( P \), without cutting into it. The formula for the tangent plane is derived using:

• The function value at the point of interest, \( f(2, 1) \).
• The partial derivatives \( f_x \) and \( f_y \), evaluated at \( P \).

The linear approximation, or the equation of the tangent plane, is: \[ L(x, y) = f(2, 1) + f_x(2, 1)(x - 2) + f_y(2, 1)(y - 1) \].
This creates a plane that best approximates the surface \( f(x, y) \) near \( (2, 1) \), providing a simpler, linear form of the function in that small region.