Problem 43 - Calculus for Biology and Medicine

Question

Find a linear approximation to $$\mathbf{f}(x, y)=\left[\begin{array}{l} x^{2}-x y \\ 3 y^{2}-1 \end{array}\right]$$ at \((1,2)\). Use your result to find an approximation for \(f(1.1,1.9)\), and compare the approximation with the value of \(f(1.1,1.9)\) that you get when you use a calculator.

Step-by-Step Solution

Answer

The linear approximation at \((1.1, 1.9)\) is \((-0.9, 9.8)\), closely approximating the true value \((-0.88, 10.83)\).

Step 1: Find the partial derivatives

First, find the partial derivatives of the function \( \mathbf{f}(x, y) \). The function \( \mathbf{f}(x, y) \) consists of two components: the first component \( f_1(x, y) = x^2 - xy \), and the second component \( f_2(x, y) = 3y^2 - 1 \). For the first component \( f_1 \): - The partial derivative with respect to \( x \) is \( \frac{\partial f_1}{\partial x} = 2x - y \). - The partial derivative with respect to \( y \) is \( \frac{\partial f_1}{\partial y} = -x \). For the second component \( f_2 \): - The partial derivative with respect to \( y \) is \( \frac{\partial f_2}{\partial y} = 6y \). - The partial derivative with respect to \( x \) is irrelevant since \( f_2 \) doesn't depend on \( x \).

Step 2: Evaluate partial derivatives at point (1, 2)

Now, evaluate these partial derivatives at the point \( (1, 2) \). For \( f_1 \): - \( \frac{\partial f_1}{\partial x}(1,2) = 2(1) - 2 = 0 \) - \( \frac{\partial f_1}{\partial y}(1,2) = -(1) = -1 \) For \( f_2 \): - \( \frac{\partial f_2}{\partial y}(1,2) = 6(2) = 12 \)

Step 3: Write the Jacobian matrix and find the linear approximation

The Jacobian matrix \( J \) at point \( (1, 2) \) is \[J = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ 0 & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & -1 \ 0 & 12 \end{bmatrix} \]The linear approximation about \( (1, 2) \) is then given by \[L(x, y) = \mathbf{f}(1, 2) + J \cdot \begin{bmatrix}x - 1 \ y - 2 \end{bmatrix}\]Computing \( \mathbf{f}(1, 2) \): - \( f_1(1, 2) = 1^2 - 1(2) = -1 \) - \( f_2(1, 2) = 3(2)^2 - 1 = 11 \) Thus, \[L(x, y) = \begin{bmatrix} -1 \ 11 \end{bmatrix} + \begin{bmatrix} 0 & -1 \ 0 & 12 \end{bmatrix} \cdot \begin{bmatrix}x - 1 \ y - 2 \end{bmatrix}= \begin{bmatrix} -1 - (y - 2) \ 11 + 12(y - 2) \end{bmatrix}\]which simplifies to \[\begin{bmatrix} 1 - y \ 12y - 13 \end{bmatrix}\]

Step 4: Approximate value at (1.1, 1.9)

Using the linear approximation, calculate the value at \( (1.1, 1.9) \): - Insert \(x = 1.1\) and \(y = 1.9\) into \( L(x, y) = \begin{bmatrix} 1 - y \ 12y - 13 \end{bmatrix} \): \[L(1.1, 1.9) = \begin{bmatrix} 1 - 1.9 \ 12(1.9) - 13 \end{bmatrix} = \begin{bmatrix} -0.9 \ 9.8 \end{bmatrix}\]

Step 5: Calculate actual value at (1.1, 1.9) and compare

Now calculate the exact value of \( \mathbf{f}(1.1, 1.9) \) using a calculator: - \( f_1(1.1, 1.9) = (1.1)^2 - 1.1(1.9) = 1.21 - 2.09 = -0.88 \) - \( f_2(1.1, 1.9) = 3(1.9)^2 - 1 = 10.83 \) Thus, \( \mathbf{f}(1.1, 1.9) = \begin{bmatrix} -0.88 \ 10.83 \end{bmatrix} \). The linear approximation gives \( L(1.1, 1.9) = \begin{bmatrix} -0.9 \ 9.8 \end{bmatrix} \), which is a close approximation.

Key Concepts

partial derivatives

Partial derivatives play a crucial role in understanding how a function behaves as we vary one of its input variables while keeping others fixed. For the multivariable function \( \mathbf{f}(x, y) \), partial derivatives measure the rate at which the function changes with respect to each of its variables.When we have a function with components, like \( f_1(x,y) = x^2 - xy \) and \( f_2(x,y) = 3y^2 - 1 \), each component requires its own set of partial derivatives.

• For \( f_1 \), the partial derivative with respect to \( x \) is found by treating \( y \) as a constant, resulting in \( \frac{\partial f_1}{\partial x} = 2x - y \).
• The partial derivative with respect to \( y \) is calculated by holding \( x \) constant, giving \( \frac{\partial f_1}{\partial y} = -x \).
• For \( f_2 \), since it depends only on \( y \), the partial derivative with respect to \( y \) is \( 6y \), while the derivative with respect to \( x \) is zero.

Understanding these derivatives helps in forming the Jacobian matrix, essential for linear approximations.

Jacobian matrix

The Jacobian matrix is a powerful tool when dealing with functions of multiple variables, particularly for linear approximations. It is formed by arranging all partial derivatives of a multi-component function into a matrix.

Forming the Jacobian Matrix

For a function \( \mathbf{f}(x, y) = [f_1(x,y), f_2(x,y)] \), the Jacobian matrix \( J \) is:\[J = \begin{bmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{bmatrix}\]

Evaluating the Jacobian at Specific Points

To find the Jacobian at a specific point, say \((1, 2)\), substitute these coordinates into the partial derivatives.

• For \( f_1 \) at \((1, 2)\): The partial derivatives are \( \frac{\partial f_1}{\partial x} = 0 \) and \( \frac{\partial f_1}{\partial y} = -1 \).
• For \( f_2 \) at \((1, 2)\): Since \( f_2 \) doesn't depend on \( x \), the derivative is zero, and \( \frac{\partial f_2}{\partial y} = 12 \).

Hence, the Jacobian matrix becomes:\[J = \begin{bmatrix} 0 & -1 \ 0 & 12 \end{bmatrix}\]This matrix is pivotal for expressing how the function changes linearly around the given point.

function evaluation

Function evaluation is the process of determining the output of a function given specific input values. For linear approximation, we begin by evaluating the original function, \( \mathbf{f}(x, y) \), at the point of interest and then use the Jacobian matrix to find the approximation.

Evaluating the Function at a Point

Given \( \mathbf{f}(x, y) = [x^2 - xy, 3y^2 - 1] \), at \((1, 2)\), we compute:

• For \( f_1(1, 2) = 1^2 - 1 \cdot 2 = -1 \)
• For \( f_2(1, 2) = 3 \cdot 2^2 - 1 = 11 \)

Thus, \( \mathbf{f}(1, 2) = [-1, 11] \).

Using Linear Approximation

The linear approximation \( L(x, y) \) is formed by:\[ L(x, y) = \mathbf{f}(1, 2) + J \cdot \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix} \]Substituting the Jacobian and the offsets \((x-1, y-2)\), you get the linear approximation:\[L(x, y) = \begin{bmatrix} 1 - y \ 12y - 13 \end{bmatrix}\]For inputs \((1.1, 1.9)\), substituting yields:

• \( f_1(1.1, 1.9) \approx -0.9 \)
• \( f_2(1.1, 1.9) \approx 9.8 \)

The approximation gives a good insight into the function's behavior near the initial point.