Problem 18 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The linearization of the function at (1, -1) is \(-2x + 2y + 2\).

Step 1: Understand the concept of linearization

Linearization is the process of approximating a function using its tangent plane at a specific point. For a function \(f(x, y)\), the linearization \(L(x, y)\) at point \((x_0, y_0)\) is given by the formula: \[L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0).\]

Step 2: Evaluate the function at the given point

The given function is \(f(x, y) = 2xy\). First, evaluate \(f(x, y)\) at the point \((1, -1)\): \[f(1, -1) = 2 imes 1 imes (-1) = -2.\]

Step 3: Calculate the partial derivatives of the function

Compute the partial derivatives \(f_x\) and \(f_y\).- The partial derivative with respect to \(x\) is \(f_x(x, y) = \frac{\partial}{\partial x}(2xy) = 2y\).- The partial derivative with respect to \(y\) is \(f_y(x, y) = \frac{\partial}{\partial y}(2xy) = 2x\).

Step 4: Evaluate the partial derivatives at the given point

Evaluate the partial derivatives at \((1, -1)\):- \(f_x(1, -1) = 2 imes (-1) = -2\).- \(f_y(1, -1) = 2 imes 1 = 2\).

Step 5: Write the linearization of the function

Using the linearization formula \[L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0),\] substitute the values:\[L(x, y) = -2 + (-2)(x - 1) + (2)(y + 1).\]Simplify the expression to get the linearization:\[L(x, y) = -2 - 2x + 2 + 2y + 2 = -2x + 2y + 2.\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, particularly useful when dealing with functions of multiple variables. Imagine you have a surface described by a function with two variables, such as a landscape on a map. A partial derivative helps us understand how changes in one specific direction (either horizontally or vertically) affect the value of the function at a particular point, assuming the other direction remains constant.

To compute the partial derivative with respect to one of the variables, you differentiate the function by treating all other variables as constants. In the given exercise, the function is given as \(f(x, y) = 2xy\). We need partial derivatives with respect to \(x\) and \(y\):

• For \(x\): \(f_x(x, y) = \frac{\partial}{\partial x}(2xy) = 2y\).
• For \(y\): \(f_y(x, y) = \frac{\partial}{\partial y}(2xy) = 2x\).

These derivatives allow us to understand how the function changes as we move infinitely small distances along the \(x\) or \(y\)-axes.

Tangent Plane

The concept of a tangent plane extends from the idea of a tangent line, but it is applied to surfaces rather than curves. Imagine a smooth surface, like a hill. A tangent plane at a specific point on this surface would be a flat surface that 'just touches' the hill at that point, without cutting into it.

In the context of our exercise, the tangent plane provides a linear approximation of the function near the point \((1, -1)\). The equation of the tangent plane is formed using the function value at the given point and the values of the partial derivatives, as these derivatives represent the slope of the surface in the direction of each axis:

• The equation for the tangent plane is given by: \[L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]
• Using the provided values: \(L(x, y) = -2 - 2(x - 1) + 2(y + 1)\)
• Simplifying further gives: \(L(x, y) = -2x + 2y + 2\)

This equation represents the tangent plane, capturing the behavior of the surface near the point \((1, -1)\). It provides a flat approximation that mimics the function's behavior around that point.

Approximation

Approximation is a powerful tool in calculus and becomes incredibly useful when dealing with complex functions. It allows us to simplify problems by using a closely related, but simpler, model to make calculations more manageable. In the case of the exercise, we approximate a curvy surface using a flat plane.

The linearization, or tangent plane, is effectively an approximation of the function \(f(x, y)\) near the point \((x_0, y_0)\). In this case, we examined the function at \((1, -1)\), which simplifies to a line. This approximation can be used to predict the values of the function for points close to \((1, -1)\), where the curvature of the original surface does not behave significantly different from the tangent plane itself:

• The linear approximations are effective for small changes and are generally used near the point of tangency.
• This is especially useful when precise calculations are difficult, and an approximate value suffices.
• The simplified equation \(L(x, y) = -2x + 2y + 2\) supports quick computations.

This concept ensures that even without the detailed calculation intense process, students can analyze and utilize complex surfaces efficiently near specific points.