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 \([-0.9, 9.8]\). The actual value is \([-0.88, 10.83]\), showing the approximation is quite close.

Step 1: Find Partial Derivatives

Calculate the partial derivatives of the functions in \( \mathbf{f}(x, y) \). For the first component \( f_1(x, y) = x^2 - xy \):- \( \frac{\partial f_1}{\partial x} = 2x - y \)- \( \frac{\partial f_1}{\partial y} = -x \)For the second component \( f_2(x, y) = 3y^2 - 1 \):- \( \frac{\partial f_2}{\partial x} = 0 \)- \( \frac{\partial f_2}{\partial y} = 6y \).

Step 2: Evaluate Partial Derivatives at (1,2)

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

Step 3: Construct Linear Approximation

Use the evaluated partial derivatives to construct the linear approximation formula:\[ \mathbf{L}(x, y) = \mathbf{f}(1, 2) + \begin{bmatrix} \frac{\partial f_1}{\partial x}(1, 2) & \frac{\partial f_1}{\partial y}(1, 2) \ \frac{\partial f_2}{\partial x}(1, 2) & \frac{\partial f_2}{\partial y}(1, 2) \end{bmatrix} \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix} \].Calculating \( \mathbf{f}(1, 2) \):- \( f_1(1, 2) = 1^2 - 1\times 2 = -1 \)- \( f_2(1, 2) = 3\times 2^2 - 1 = 11 \)So \( \mathbf{f}(1, 2) = \begin{bmatrix} -1 \ 11 \end{bmatrix} \).Plug in the values:\[ \mathbf{L}(x, y) = \begin{bmatrix} -1 \ 11 \end{bmatrix} + \begin{bmatrix} 0 & -1 \ 0 & 12 \end{bmatrix} \begin{bmatrix} x - 1 \ y - 2 \end{bmatrix}. \]

Step 4: Simplify the Linear Approximation

Simplify the matrix multiplication to find the linear approximation:\[ \mathbf{L}(x, y) = \begin{bmatrix} -1 - (y - 2) \ 11 + 12(y - 2) \end{bmatrix} \].This simplifies to:\[ \mathbf{L}(x, y) = \begin{bmatrix} 1 - y \ 11 + 12y - 24 \end{bmatrix} = \begin{bmatrix} 1 - y \ 12y - 13 \end{bmatrix}. \]

Step 5: Approximate f(1.1, 1.9) Using Linear Approximation

Plug \( x = 1.1 \) and \( y = 1.9 \) into the linear approximation:\[ \mathbf{L}(1.1, 1.9) = \begin{bmatrix} 1 - 1.9 \ 12\times 1.9 - 13 \end{bmatrix} = \begin{bmatrix} -0.9 \ 9.8 \end{bmatrix}. \]

Step 6: Calculate Actual Value of f(1.1, 1.9)

Use a calculator to find the exact value of \( \mathbf{f}(1.1, 1.9) \):- \( f_1(1.1, 1.9) = 1.1^2 - 1.1\times 1.9 = 1.21 - 2.09 = -0.88 \)- \( f_2(1.1, 1.9) = 3\times 1.9^2 - 1 = 3\times 3.61 - 1 = 10.83 \)So, \( \mathbf{f}(1.1, 1.9) = \begin{bmatrix} -0.88 \ 10.83 \end{bmatrix} \).

Step 7: Compare Approximation with Actual Value

Compare the linear approximation to the actual calculated values:- The linear approximation gives \( \mathbf{L}(1.1, 1.9) = \begin{bmatrix} -0.9 \ 9.8 \end{bmatrix} \).- The actual value is \( \mathbf{f}(1.1, 1.9) = \begin{bmatrix} -0.88 \ 10.83 \end{bmatrix} \).Notice that the linear approximation is close to the actual value.

Key Concepts

Partial Derivatives

In the world of calculus, partial derivatives play a crucial role in understanding multivariable functions. When a function depends on multiple variables, each of these variables can change independently. A partial derivative represents how the function changes as one specific variable changes, while all other variables remain constant.
For instance, consider a function like \( f(x, y) \). To find the partial derivative with respect to \( x \), we treat \( y \) as a constant and differentiate with respect to \( x \) alone. Similarly, to find the partial derivative with respect to \( y \), we treat \( x \) as a constant.

• Symbolic Representation: Partial derivatives are represented as \( \frac{\partial f}{\partial x} \) for \( x \) and \( \frac{\partial f}{\partial y} \) for \( y \).
• Importance: These calculations help us determine the rate of change of the function along the respective axes.

Understanding and calculating partial derivatives are the first steps in analyzing how a function behaves, especially when aiming for applications like linear approximation.

Multivariable Calculus

Multivariable calculus expands upon the principles of single-variable calculus by focusing on functions involving multiple inputs. Functions like \( \mathbf{f}(x, y) \), with more than one independent variable, are a fundamental aspect of this branch of calculus.
Multivariable calculus allows us to:

• Mimic concepts of derivatives and integrals but apply them to multiple dimensions.
• Analyze how changes in one variable affect an entire function while factoring in the complex interplay of all variables involved.
• Create graphical representations to better understand these relationships in a multi-dimensional space.

The field finds significant applications in areas like physics, engineering, and economics, where systems often depend on several interacting factors.

Mathematical Function Evaluation

Evaluating a mathematical function means plugging specific values into the function's formula and performing the calculations to get an exact result. This process helps determine the precise output of a function when the inputs are known.
Let's apply this to the function \( \mathbf{f}(x, y) \) specified in the exercise:

• Plugging Values: To find \( f_1(1.1, 1.9) \), you'd calculate \( 1.1^2 - 1.1 \times 1.9 \).
• Performing Calculations: This would simplify to \( 1.21 - 2.09 \), resulting in -0.88 as the output of the first component.
• Interpreting Results: Similarly, evaluate \( f_2(1.1, 1.9) \) and interpret the results to understand the output of the second component.

Evaluations like these are essential for verifying the accuracy of approximations and predictions in practical scenarios.