Problem 40 - Calculus for Biology and Medicine

Question

Find a linear approximation to each function \(f(x, y)\) at the indicated point. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} e^{x} \sin y \\ e^{-y} \cos x \end{array}\right] \text { at }(0,0) $$

Step-by-Step Solution

Answer

The linear approximation of \( \textbf{f}(x, y) \) at \((0, 0)\) is \( \begin{bmatrix} y \\ 1-y \end{bmatrix} \).

Step 1: Identify the General Formula for Linear Approximation

The linear approximation of a function \( extbf{f}(x,y) \) at a point \((x_0, y_0)\) is given by:\[ extbf{f}(x, y) \approx extbf{f}(x_0, y_0) + abla extbf{f}(x_0, y_0) \cdot \begin{bmatrix} x-x_0 \ y-y_0 \end{bmatrix} \]where \( abla extbf{f}(x_0, y_0) \) is the gradient of \( extbf{f} \) at \((x_0, y_0)\).

Step 2: Evaluate the Function at the Point

Substitute \((x_0, y_0) = (0,0)\) into \( extbf{f}(x,y) \) to find \( extbf{f}(0, 0) \):\[ extbf{f}(0, 0) = \begin{bmatrix} e^0 \sin 0 \ e^0 \cos 0 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \]

Step 3: Compute the Partial Derivatives

Calculate the partial derivatives of \( extbf{f}(x, y) \):- For the first component \( f_1(x,y) = e^x \sin y \): - \( \frac{\partial f_1}{\partial x} = e^x \sin y \) - \( \frac{\partial f_1}{\partial y} = e^x \cos y \)- For the second component \( f_2(x,y) = e^{-y} \cos x \): - \( \frac{\partial f_2}{\partial x} = -e^{-y} \sin x \) - \( \frac{\partial f_2}{\partial y} = -e^{-y} \cos x \)

Step 4: Evaluate the Derivatives at the Point

Substitute \((x_0, y_0) = (0, 0)\) into the derivatives:- \( abla f_1(0, 0) = \begin{bmatrix} e^0 \sin 0 \ e^0 \cos 0 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \)- \( abla f_2(0, 0) = \begin{bmatrix} -e^0 \sin 0 \ -e^0 \cos 0 \end{bmatrix} = \begin{bmatrix} 0 \ -1 \end{bmatrix} \)

Step 5: Construct the Gradient Matrix

Combine the evaluated derivatives into a gradient matrix:\[ abla \textbf{f}(0, 0) = \begin{bmatrix} 0 & 1 \ 0 & -1 \end{bmatrix} \]

Step 6: Formulate the Linear Approximation

Using the linear approximation formula:\[ \textbf{f}(x, y) \approx \begin{bmatrix} 0 \ 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \ 0 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} y \ 1-y \end{bmatrix} \]

Step 7: Present the Solution

The linear approximation of the function \( \textbf{f}(x, y) \) at the point \((0, 0)\) is:\[ \begin{bmatrix} y \ 1-y \end{bmatrix} \]

Key Concepts

Multivariable Calculus

Multivariable calculus deals with functions that have more than one input variable, such as both "x" and "y" in a two-variable function. This branch of calculus extends the concepts of single-variable calculus like limits, derivatives, and integrals to functions of several variables. It allows us to analyze how changes in multiple inputs affect the output of a function.
In the context of linear approximation, understanding multivariable calculus is essential because it provides the tools needed to approximate complex functions using simple, linear ones around a point in the input space. This method is particularly useful when dealing with functions that are too complicated to solve exactly, but can be analyzed locally around a specific point using tangent planes or lines.
Multivariable calculus introduces concepts such as partial derivatives and gradient vectors that are crucial for analyzing the rates of change in each variable independently and collectively. These ideas are the foundational building blocks for more advanced topics like optimizing functions of multiple variables, and they're particularly useful in fields like economics, engineering, and physical sciences.

Partial Derivatives

In a multivariable context, the derivative of a function isn't as straightforward as in single-variable calculus, because we have multiple directions in which to take the derivative. That's where partial derivatives come in. They measure the rate of change of the function with respect to one variable, keeping all other variables constant.
For example, given a function \(f(x, y)\), the partial derivative with respect to \(x\) is written as \(\frac{\partial f}{\partial x}\), and it represents how \(f\) changes as \(x\) changes, while \(y\) stays the same. Similarly, \(\frac{\partial f}{\partial y}\) shows the rate of change with respect to \(y\). These derivatives play a critical role in constructing the gradient matrix, which is used for linear approximations.
Understanding partial derivatives is key to exploring how each variable individually impacts the behavior of a multivariable function. In practical applications, partial derivatives help you determine the slope of a surface in specific directions, which is enormously useful in optimization problems and dynamics in physics and engineering.

Gradient Matrix

The gradient matrix, also known as the Jacobian matrix, is a matrix containing all the first-order partial derivatives of a vector-valued function. For a function \( \mathbf{f}(x, y)\) with multiple components, the gradient matrix at a point summarizes the way each input affects each output component.
In our linear approximation problem, constructing the gradient matrix involves calculating the partial derivatives of each component function of \( \mathbf{f} \). For instance, if \( \mathbf{f} \) has two outputs, then the gradient matrix will have two rows for these outputs and two columns for the inputs \(x\) and \(y\). Each entry in this matrix corresponds to the derivative of one component of \( \mathbf{f} \) with respect to one variable.
The gradient matrix is useful because it gives you all the information about how the function behaves locally around a point. In linear approximations, it is used to create a first-order Taylor series expansion, simplifying complex nonlinear behaviors to linear, making analysis more manageable. This simplification is invaluable in computer graphics, machine learning, and other areas requiring high-speed calculations over multivariable spaces.