Problem 41 - Calculus for Biology and Medicine

Question

Find a linear approximation to each func\mathrm{tion } \(f(x, y)\) at the indicated point. \(\mathbf{f}(x, y)=\left[\begin{array}{l}\frac{x}{y} \\\ \frac{y}{x}\end{array}\right]\) at \((1,2)\)

Step-by-Step Solution

Answer

The linear approximation is: \[ \left( \begin{array}{c} \frac{1}{2}(x - 1) - \frac{1}{4}(y - 2) + \frac{1}{2} \\ -2(x - 1) + 1(y - 2) + 2 \end{array} \right) \]

Step 1: Identify the Func ion

We have the function \[\mathbf{f}(x, y)=\left[\begin{array}{l}\frac{x}{y} \ \frac{y}{x}\end{array}\right] \] a vector function. Our task is to find a linear approximation around the point \((1, 2)\).

Step 2: Compute Partial Derivatives

Find the partial derivatives of \(f_1(x, y) = \frac{x}{y}\) and \(f_2(x, y) = \frac{y}{x}\). For \(f_1\), \[\frac{\partial f_1}{\partial x} = \frac{1}{y} \ \frac{\partial f_1}{\partial y} = -\frac{x}{y^2} \] and for \(f_2\),\[\frac{\partial f_2}{\partial x} = -\frac{y}{x^2} \ \frac{\partial f_2}{\partial y} = \frac{1}{x}\]

Step 3: Evaluate Partial Derivatives at the Point

Substitute \((1, 2)\) into the partial derivatives:\[\frac{\partial f_1}{\partial x}\Big|_{(1,2)} = \frac{1}{2} \ \frac{\partial f_1}{\partial y}\Big|_{(1,2)} = -\frac{1}{4} \] and \[\frac{\partial f_2}{\partial x}\Big|_{(1,2)} = -\frac{2}{1^2} = -2, \ \frac{\partial f_2}{\partial y}\Big|_{(1,2)} = \frac{1}{1} = 1\]

Step 4: Gradient Vector Evaluation

The gradient vector at the point \((1, 2)\) is: \[abla f(1, 2) = \left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{4} \ -2 & 1 \end{array}\right]\]

Step 5: Compute Linear Approximation

The linear approximation of \(\mathbf{f}(x, y)\) at the point \((1,2)\) is given by: \[L(x, y) = \mathbf{f}(1, 2) + abla f(1, 2) \cdot \left( \begin{array}{c} x - 1 \ y - 2 \end{array} \right)\] Compute \(\mathbf{f}(1,2)\): \[\mathbf{f}(1,2) = \left[ \begin{array}{c} \frac{1}{2} \ 2 \end{array} \right]\] Thus, \[L(x, y) = \left[ \begin{array}{c} \frac{1}{2} \ 2 \end{array} \right] + \left[ \begin{array}{cc} \frac{1}{2} & -\frac{1}{4} \ -2 & 1 \end{array} \right] \left( \begin{array}{c} x - 1 \ y - 2 \end{array} \right)\]This simplifies to: \( L(x, y) \approx \left[ \begin{array}{c} \frac{1}{2}(x - 1) - \frac{1}{4}(y - 2) + \frac{1}{2} \ -2(x - 1) + 1(y - 2) + 2 \end{array} \right] \)

Key Concepts

Partial Derivatives

When studying multivariable functions like our given vector function, partial derivatives are essential tools. They represent how the function changes as one specific variable changes, while all other variables remain constant. For example, for the function component \( f_1(x, y) = \frac{x}{y} \), the partial derivative \( \frac{\partial f_1}{\partial x} \) tells us how \( f_1 \) changes with \( x \) when \( y \) is held constant. Similarly, \( \frac{\partial f_1}{\partial y} \) shows the change in \( f_1 \) with \( y \) when \( x \) remains the same.

• Calculate each derivative separately: For a composite function, calculate the derivative relative to one variable at a time.
• Substitute the specific point: Once you have the derivatives, substitute the point of interest to determine the value of the derivative at that point.

Partial derivatives form the building blocks for other concepts like the gradient vector, which we will discuss next.

Gradient Vector

The gradient vector is a collection of partial derivatives which provides direction and rate of the fastest increase for a function at a given point. For a function \( f(x, y) \), the gradient is denoted as \( abla f(x, y) \) and is a vector that contains all of its partial derivatives. In our exercise, the gradient vector for the vector function components \( f_1(x, y) = \frac{x}{y} \) and \( f_2(x, y) = \frac{y}{x} \) was computed and evaluated at the point \((1, 2)\).

• Gradient composition: The gradient vector \( abla f \) is represented as \( \left[ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right] \).
• Directional benefits: It gives both the direction of the steepest ascent and the rate of increase of the function.

Using the gradient vector is essential for constructing linear approximations and optimization problems.

Vector Function

A vector function is a mathematical function that takes in one or more variables and returns a vector. In our exercise, the function \( \mathbf{f}(x, y) = \left[ \begin{array}{c} \frac{x}{y} \ \frac{y}{x} \end{array} \right] \) takes two inputs and outputs a vector with two components. Vector functions are useful in representing complex relationships and changes that occur across multiple dimensions.

• Input and output: A vector function maps each set of input values (here \((x, y)\)) to a vector, which can vary according to the function rules.
• Applications: They are used across various fields such as physics, engineering, and computer graphics to model real-world phenomena.

The concept of linear approximation in vector functions helps in finding estimations of functions at specific points, simplifying the complexity by focusing on local linear behavior around a given point.