Problem 23 - Calculus for Biology and Medicine

Question

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=\ln \left(x^{2}+y\right) ;(1,1)\)

Step-by-Step Solution

Answer

The linearization is \(L(x, y) = \ln(2) + x + \frac{y}{2} - \frac{3}{2}\).

Step 1: Understand the linearization formula

The linearization of a function at a point provides a linear approximation of the function near that point. For a function of two variables, \(f(x, y)\), the linearization around a point \((x_0, y_0)\) is given by the formula: \[L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.

Step 2: Compute the value of the function at the point

Evaluate the function \(f(x, y) = \ln(x^2 + y)\) at the point \((x_0, y_0) = (1, 1)\): \[f(1, 1) = \ln(1^2 + 1) = \ln(2)\]

Step 3: Find partial derivatives

Calculate the partial derivatives of \(f\):- The partial derivative with respect to \(x\): \[f_x(x, y) = \frac{d}{dx} \ln(x^2 + y) = \frac{2x}{x^2 + y}\]- The partial derivative with respect to \(y\): \[f_y(x, y) = \frac{d}{dy} \ln(x^2 + y) = \frac{1}{x^2 + y}\]

Step 4: Evaluate partial derivatives at the point

Substitute \((x_0, y_0) = (1, 1)\) into the partial derivatives:- \(f_x(1, 1) = \frac{2 \times 1}{1^2 + 1} = 1\)- \(f_y(1, 1) = \frac{1}{1^2 + 1} = \frac{1}{2}\)

Step 5: Formulate the linearization

Using the values obtained, substitute into the linearization formula:\[L(x, y) = \ln(2) + 1 \cdot (x - 1) + \frac{1}{2} \cdot (y - 1)\]Simplify the expression:\[L(x, y) = \ln(2) + (x - 1) + \frac{1}{2}(y - 1)\]\[L(x, y) = \ln(2) + x - 1 + \frac{y}{2} - \frac{1}{2}\]\[L(x, y) = \ln(2) + x + \frac{y}{2} - \frac{3}{2}\]

Key Concepts

Partial Derivatives

Imagine you have a function that depends on two variables, like \(f(x, y) = \ln(x^2 + y)\). A partial derivative is a way to see how that function changes if you wiggle just one variable at a time, keeping the other constant. Let's break it down!

When we compute the partial derivative of \(f(x, y)\) with respect to \(x\), denoted as \(f_x(x, y)\), we treat \(y\) as a constant. So the change in \(f\) is primarily concerned with the changes in \(x\). For our function, this calculation gives us:

• \(f_x(x, y) = \frac{2x}{x^2 + y}\)

Similarly, when finding the partial derivative with respect to \(y\), denoted as \(f_y(x, y)\), we treat \(x\) as a constant, focusing on changes in \(y\):

• \(f_y(x, y) = \frac{1}{x^2 + y}\)

By substituting specific values into these expressions, as we did at \(x_0 = 1\) and \(y_0 = 1\), we evaluate how each variable independently impacts the function's value at that point. This step is crucial for creating a linear approximation.

Multivariable Calculus

Multivariable calculus is like regular calculus, but even cooler because it works with functions that depend on two or more variables. Instead of just understanding how things change in the real world with one variable, multivariable calculus lets us explore surfaces and spaces where functions depend on multiple inputs.

In our context, the function \(f(x, y) = \ln(x^2 + y)\) is defined on a 2D plane, where each point \( (x, y) \) gives a specific value of \(f\). The challenge here is to understand how changes in both the \(x\) and \(y\) directions affect the function.

To manage this, we use gradients, directional derivatives, and, as seen here, linear approximations using partial derivatives. This allows us to handle complex variations in surfaces denoted by such functions. Essentially, multivariable calculus brings richness and complexity to our understanding of change, going beyond the relatively simpler realm of functions of a single variable.

Linear Approximation

Linear approximation simplifies complex functions by using a linear equation to estimate the function's behavior around a specific point. Think of it as zooming in on a map so much that a curved road seems straight.

For our function \(f(x, y) = \ln(x^2 + y)\), we use the point \((x_0, y_0) = (1, 1)\). The idea is to replace the complex function with a simpler, linear form that's easy to work with near this point.

Here's the linear approximation formula for functions of two variables:

• \(L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\)

In our solution, we calculate this by plugging in values:

• \(L(x, y) = \ln(2) + 1 \cdot (x - 1) + \frac{1}{2} \cdot (y - 1)\)
• Thus, \(L(x, y) = \ln(2) + x + \frac{y}{2} - \frac{3}{2}\)

This approximation is fantastic for quickly estimating the function's behavior at or near the given point, offering a straightforward technique to make complex calculations manageable.