Problem 27 - Calculus for Biology and Medicine

Question

Find the linear approximation of $$f(x, y)=\ln \left(x^{2}-3 y\right)$$ at \((1,0)\), and use it to approximate \(f(1.1,0.1)\). Using a calculator, compare the approximation with the exact value of \(f(1.1,0.1)\).

Step-by-Step Solution

Answer

The linear approximation of \( f(1.1, 0.1) \) is \(-0.1\), close to the exact value \(-0.0943\).

Step 1: Compute Partial Derivatives

To find the linear approximation, we first need the partial derivatives of the function \( f(x, y) = \ln(x^2 - 3y) \). The partial derivative with respect to \( x \) is \( f_x(x, y) = \frac{2x}{x^2 - 3y} \) and with respect to \( y \) is \( f_y(x, y) = \frac{-3}{x^2 - 3y} \).

Step 2: Evaluate Partial Derivatives at the Point

Evaluate the partial derivatives at the point \((1, 0)\):1. \( f_x(1, 0) = \frac{2 \times 1}{1^2 - 3 \times 0} = 2 \).2. \( f_y(1, 0) = \frac{-3}{1^2 - 3 \times 0} = -3 \).

Step 3: Construct the Linear Approximation

The linear approximation \( L(x, y) \) of the function at \((1, 0)\) is given by:\[ L(x, y) = f(1, 0) + f_x(1, 0)(x - 1) + f_y(1, 0)(y - 0) \]First, calculate \( f(1, 0) = \ln(1^2 - 3 \times 0) = \ln(1) = 0 \). Substituting all values gives:\[ L(x, y) = 0 + 2(x - 1) - 3y = 2x - 2 - 3y \].

Step 4: Approximate \( f(1.1, 0.1) \) Using the Linear Approximation

Using the linear approximation formula, approximate \( f(1.1, 0.1) \):\[ L(1.1, 0.1) = 2(1.1) - 2 - 3(0.1) = 2.2 - 2 - 0.3 = -0.1 \].

Step 5: Calculate the Exact Value of \( f(1.1, 0.1) \)

Calculate the exact value of \( f(1.1, 0.1) \) using a calculator:\[ f(1.1, 0.1) = \ln((1.1)^2 - 3 \times 0.1) = \ln(1.21 - 0.3) = \ln(0.91) \approx -0.0943 \].

Step 6: Compare the Approximation with the Exact Value

The linear approximation gives \(-0.1\), and the exact value is approximately \(-0.0943\). Thus, the approximation is very close to the exact value, with a small difference between them.

Key Concepts

Partial Derivatives

In calculus, understanding how a multivariable function changes in different directions is crucial, and that's where **partial derivatives** come into play. For the function given, \( f(x, y) = \ln(x^2 - 3y) \), we need to find how it changes with respect to each variable independently while keeping the other constant. This process helps in constructing the linear approximation.

• The partial derivative with respect to \( x \), denoted \( f_x(x, y) \), reflects the rate of change of \( f \) in the direction of the \( x \)-axis. For our function, it's computed as \( \frac{2x}{x^2 - 3y} \).
• Similarly, the partial derivative with respect to \( y \), denoted \( f_y(x, y) \), shows the rate of change along the \( y \)-axis. It's obtained as \( \frac{-3}{x^2 - 3y} \).

Both these derivatives are crucial as they are needed to approximate the function using linearization. By evaluating these partial derivatives at a specific point, such as \((1, 0)\), we get a snapshot of how the function behaves at that spot.

Exact Value Comparison

After acquiring the linear approximation, comparing the estimated value to the **exact value** provides insight into the accuracy of our approximation.

In the given problem, the linear approximation at the point \((1.1, 0.1)\) yields a result of \(-0.1\). To check the accuracy, we compute the exact value using a calculator, obtaining approximately \(-0.0943\).

• Exact value computations involve using the initial function \( f(x, y) = \ln(x^2 - 3y) \) for the exact point.
• The difference between the exact and approximated values is small, showing that linear approximations are useful for estimating nearby points.

Exact value comparison helps confirm that our linear model is a good representation of the function in the neighborhood of the point where we're approximating.

Function Evaluation

When working with functions, **function evaluation** simply means determining the output for a given set of input values. With linear approximations, this step is fundamental in comparing estimated values to actual ones.

For the exercise, evaluating the function at the point \((1, 0)\) gives us the starting value of the linear approximation. Calculating \( f(1, 0) = \ln(1^2 - 3 \times 0) = \ln(1) = 0 \), we find it sets the base level for our approximation.

• Function evaluation is necessary to anchor the linear approximation calculation.
• It utilizes the derived formulas from partial derivatives to assess the function at different points.

Overall, function evaluation alongside linear approximations provides an efficient way to understand and estimate the behavior of complex functions near specific points.