Problem 20 - Calculus Early Transcendentals

Question

Find the linear approximation of the function \(f(x, y)=\ln (x-3 y)\) at \((7,2)\) and use it to approximate \(f(6.9,2.06) .\) Illustrate by graphing \(f\) and the tangent plane.

Step-by-Step Solution

Answer

The linear approximation at (6.9, 2.06) is 1.72.

Step 1: Find the Partial Derivatives

First, we need to find the partial derivatives of the function \(f(x, y) = \ln(x - 3y)\) with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x = \frac{1}{x - 3y}\). The partial derivative with respect to \(y\) is \(f_y = -\frac{3}{x - 3y}\).

Step 2: Evaluate the Partial Derivatives at Point (7, 2)

Evaluate the partial derivatives at the point \((7, 2)\). For \(f_x\), substitute \(x=7\) and \(y=2\): \(f_x(7, 2) = \frac{1}{7 - 3 \times 2} = 1\). For \(f_y\), substitute \(x=7\) and \(y=2\): \(f_y(7, 2) = -\frac{3}{7 - 3 \times 2} = -3\).

Step 3: Find the Function Value at Point (7, 2)

Calculate the function value at \((7, 2)\), \(f(7, 2) = \ln(7 - 3 \times 2) = \ln(1) = 0\).

Step 4: Write the Equation of the Tangent Plane

The equation of the tangent plane is given by \(L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\). Substituting \(a=7\), \(b=2\), \(f(a, b) = 0\), \(f_x(7, 2) = 1\), \(f_y(7, 2) = -3\), we get \(L(x, y) = 0 + 1(x - 7) - 3(y - 2)\). Simplifying, the tangent plane equation becomes \(L(x, y) = x - 3y + 1\).

Step 5: Use the Linear Approximation to Estimate at (6.9, 2.06)

Substitute \(x = 6.9\) and \(y = 2.06\) into the equation of the tangent plane: \(L(6.9, 2.06) = 6.9 - 3 \times 2.06 + 1 = 6.9 - 6.18 + 1 = 1.72\).

Step 6: Graph the Function and Its Tangent Plane

Graphing \(f(x, y) = \ln(x - 3y)\) along with its tangent plane \(L(x, y) = x - 3y + 1\) will illustrate how the plane touches the surface at the point \((7, 2)\), and closely approximates the function near this point.

Key Concepts

Partial Derivatives

Partial derivatives are a foundational aspect of multivariable calculus. They measure how a function changes as one variable changes while keeping others constant.
Given a function like \(f(x, y) = \ln(x - 3y)\), the partial derivatives are about finding the derivative with respect to \(x\) and \(y\) separately.
This tells us how \(f\) changes as \(x\) or \(y\) changes independently. Here's how you calculate them:

• For \(x\), the partial derivative is \(f_x = \frac{1}{x - 3y}\). This derivative shows how \(f\) changes with respect to \(x\).
• For \(y\), the partial derivative is \(f_y = -\frac{3}{x - 3y}\). This describes the change in \(f\) regarding \(y\).

By evaluating these derivatives at a specific point, you can better understand the behavior of the function at that location. In our exercise, evaluating at \((7, 2)\) gives \(f_x(7, 2) = 1\) and \(f_y(7, 2) = -3\). These results play a crucial role in forming the tangent plane equation.

Tangent Plane

The tangent plane is an essential concept to understand in functions of two variables. Imagine it as the plane that gently touches the 3D graph of a function at a specific point.
Mathematically, it's the linear approximation of the function around that point, which is the exact point where the plane lies flat on the surface of the graph.
The general formula for the tangent plane to a function \(f(x,y)\) at a point \((a, b)\) is:\[ L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \]By plugging the calculated values from the exercise:

• The function value at \((7, 2)\) is \(f(7, 2) = 0\).
• Plug in \(f_x = 1\) and \(f_y = -3\).

We derive the tangent plane: \[ L(x, y) = 0 + 1(x - 7) - 3(y - 2) \] Simplifying this gives us: \[ L(x, y) = x - 3y + 1 \] This plane approximates the function near the point \((7, 2)\). It's a simple linear equation representing how the function's surface appears close to that point.

Function Evaluation

Function evaluation is a straightforward but necessary step in understanding how a function behaves at specific points.
It involves substituting particular values into a function to get a real number output, which tells us the result of the function at those input values.
For \(f(x, y) = \ln(x - 3y)\), if we want to evaluate it at \((7, 2)\), we proceed as follows:

• Substitute \(x = 7\) and \(y = 2\).
• Calculate \(f(7, 2) = \ln(7 - 3 \times 2) = \ln(1)\).
• The value is 0, since \(\ln(1) = 0\).

This evaluation makes it possible to set the constant part of the tangent plane correctly. Function evaluations also allow us to verify how changes in variables affect the function value.

Graphing Functions

Graphing functions involving two variables provides a visual perspective that aids comprehension.
It helps to see how a function behaves across different values, especially near points of interest like where a linear approximation takes place.
In our exercise, the function \(f(x, y) = \ln(x - 3y)\) creates a surface in a three-dimensional space. Graphing this alongside its tangent plane \(L(x, y) = x - 3y + 1\) shows:

• The tangent plane gently touching the graph at the point (7, 2).
• The approximation contact, demonstrating how close the linear model stays to the function nearer to this point of tangency.

Visualizing these graphs helps in

• Understanding where the linear approximation is most accurate.
• Sensing how a small change in \(x\) and \(y\) influences \(f\).

Use graphing tools or software to appreciate these dynamics fully, enhancing clarity and improving decision-making in approximation contexts.