Problem 46 - Calculus for Biology and Medicine

Question

Find a linear approximation to $$ \mathbf{f}(x, y)=\left[\begin{array}{c} \sqrt{2 x+y} \\ x-y^{2} \end{array}\right] $$ at \((1,2) .\) Use your result to find an approximation for \(f(1.05,2.05)\), and compare the approximation with the value of \(f(1.05,2.05)\) that you get when you use a calculator.

Step-by-Step Solution

Answer

The linear approximation for \( \mathbf{f}(1.05, 2.05) \) is \( \begin{bmatrix} 2.075 \\ -3.15 \end{bmatrix} \), while the exact values are \( \begin{bmatrix} 2.037 \\ -3.1525 \end{bmatrix} \).

Step 1: Find the Partial Derivatives

First, we need to find the partial derivatives of each component of the vector function \( \mathbf{f}(x, y) \). 1. For the first component \( f_1(x, y) = \sqrt{2x + y} \): - Partial derivative with respect to \( x \), \( \frac{\partial f_1}{\partial x} = \frac{1}{\sqrt{2x + y}} \cdot 2 \) - Partial derivative with respect to \( y \), \( \frac{\partial f_1}{\partial y} = \frac{1}{\sqrt{2x + y}} \)2. For the second component \( f_2(x, y) = x - y^2 \): - Partial derivative with respect to \( x \), \( \frac{\partial f_2}{\partial x} = 1 \) - Partial derivative with respect to \( y \), \( \frac{\partial f_2}{\partial y} = -2y \)These derivatives will be used in the linear approximation formula.

Step 2: Evaluate Derivatives at the Point

Now substitute \( x = 1 \) and \( y = 2 \) into the partial derivatives to find their values at this point:\( \frac{\partial f_1}{\partial x}(1, 2) = \frac{2}{\sqrt{4}} = 1 \)\( \frac{\partial f_1}{\partial y}(1, 2) = \frac{1}{\sqrt{4}} = 0.5 \)\( \frac{\partial f_2}{\partial x}(1, 2) = 1 \)\( \frac{\partial f_2}{\partial y}(1, 2) = -4 \)

Step 3: Write the Linear Approximation

The linear approximation of \( \mathbf{f}(x, y) \) at the point \( (1, 2) \) is given by the formula: \[ \mathbf{f}(x, y) \approx \mathbf{f}(1, 2) + D\mathbf{f}|_{(1,2)} \cdot \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix} \] where \( D\mathbf{f}|_{(1,2)} \) is the Jacobian matrix at \( (1,2) \).Substitute the evaluated partial derivatives and coordinate values: \[ \mathbf{f}(1, 2) = \begin{bmatrix} \sqrt{4} \ 1 - 4 \end{bmatrix} = \begin{bmatrix} 2 \ -3 \end{bmatrix}\] \[ D\mathbf{f}|_{(1,2)} = \begin{bmatrix} 1 & 0.5 \ 1 & -4 \end{bmatrix} \] Thus, the linear approximation is: \[ \mathbf{f}(x, y) \approx \begin{bmatrix} 2 \ -3 \end{bmatrix} + \begin{bmatrix} 1 & 0.5 \ 1 & -4 \end{bmatrix} \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix} \]

Step 4: Calculate the Approximation for \((1.05, 2.05)\)

Substitute \( x = 1.05 \) and \( y = 2.05 \) into the linear approximation formula derived in Step 3.The change in variables is \( \Delta x = 0.05 \) and \( \Delta y = 0.05 \):\[ \mathbf{f}(1.05, 2.05) \approx \begin{bmatrix} 2 \ -3 \end{bmatrix} + \begin{bmatrix} 1 & 0.5 \ 1 & -4 \end{bmatrix} \begin{bmatrix} 0.05 \ 0.05 \end{bmatrix} \] Calculate the matrix multiplication and addition:\[ \begin{bmatrix} 2 \ -3 \end{bmatrix} + \begin{bmatrix} 0.05 + 0.025 \ 0.05 - 0.2 \end{bmatrix} = \begin{bmatrix} 2.075 \ -3.15 \end{bmatrix} \] Thus, our linear approximation for \( \mathbf{f}(1.05, 2.05) \) is \( \begin{bmatrix} 2.075 \ -3.15 \end{bmatrix} \).

Step 5: Comparison with Calculator Result

Calculate \( \mathbf{f}(1.05, 2.05) \) directly using a calculator:\- First component: \( \sqrt{2(1.05) + 2.05} = \sqrt{4.15} \approx 2.037 \)\- Second component: \( 1.05 - (2.05)^2 = 1.05 - 4.2025 = -3.1525 \)\- The exact values are \( \begin{bmatrix} 2.037 \ -3.1525 \end{bmatrix} \).Compare with the linear approximation \( \begin{bmatrix} 2.075 \ -3.15 \end{bmatrix} \).The approximation is close but not exact, indicating the difference between the approximation and the actual computation.

Key Concepts

Partial Derivatives

Partial derivatives are at the heart of understanding how a multi-variable function changes when we vary one of its inputs while keeping the others constant. This is key in functions with more than one variable, such as the vector function in our problem.

For example, consider the function \( f_1(x, y) = \sqrt{2x + y} \). The partial derivative with respect to \( x \) assesses how \( f_1 \) changes as \( x \) changes while \( y \) remains held constant. It tells us the 'slope' or rate of change of \( f_1 \) with a unit change in \( x \).

Similar calculations apply to the partial derivative with respect to \( y \), indicating how \( f_1 \) varies with changes in \( y \).

• The partial derivative of \( f_1 \) with respect to \( x \) was found to be \( \frac{2}{\sqrt{2x + y}} \).
• With respect to \( y \), it was \( \frac{1}{\sqrt{2x + y}} \).

These derivatives are crucial as they become part of the linear approximation framework, informing how we predict changes in the vector function \( \mathbf{f} \) with small variations in \( x \) and \( y \).

Vector Function

Vector functions map input values (often taking on multiple variables) to vectors. In essence, they describe how each input influences multiple outputs simultaneously.

In our example, the vector function \( \mathbf{f}(x, y) = \left[ \begin{array}{c} \sqrt{2 x+y} \ x-y^2 \end{array} \right] \), consists of two components: \( f_1(x, y) \) and \( f_2(x, y) \).

• The first component, \( f_1(x, y) \), provides a scalar value resulting from the square root expression.
• The second component, \( f_2(x, y) \), outputs the difference between \( x \) and the square of \( y \).

Each component function returns a scalar, but together as a vector, they offer a comprehensive picture of how \( \mathbf{f} \) behaves. This allows us to analyze a system's behavior at a specific point, such as \( (1, 2) \), and determine how changes affect the entire vector output.

Jacobian Matrix

The Jacobian matrix is a cornerstone in multi-variable calculus. It is a matrix of all first-order partial derivatives of a vector-valued function and plays a critical role in linear approximations, as it encodes how the function changes in response to small variations in the inputs.

In our specific example, the Jacobian matrix \( D\mathbf{f}|_{(1,2)} \) is given as:\[\begin{bmatrix} 1 & 0.5 \ 1 & -4 \end{bmatrix}\]
This matrix comprises:

• The first row’s elements arising from derivatives of \( f_1 \) with respect to \( x \) and \( y \).
• The second row’s from \( f_2 \).

The presence of the Jacobian is significant when forming the linear approximation equation:\[\mathbf{f}(x, y) \approx \mathbf{f}(1, 2) + D\mathbf{f}|_{(1,2)} \cdot \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix}\]
By multiplying the Jacobian with the change in inputs \([x-1, y-2]^T\), we obtain the linear approximation. This provides insight into the vector function's local behavior near the point \((1, 2)\).