Problem 45 - Calculus for Biology and Medicine

Question

Find a linear approximation to $$ \mathbf{f}(x, y)=\left[\begin{array}{c} (x-y)^{2} \\ 2 x^{2} y \end{array}\right] $$ at \((2,-3)\). Use your result to find an approximation for \(f(1.9,-3.1)\), and compare the approximation with the value of \(f(1.9,-3.1)\) that you get when you use a calculator.

Step-by-Step Solution

Answer

Linear approximation: \( \begin{bmatrix} 26 \\ -22.4 \end{bmatrix} \). Exact: \( \begin{bmatrix} 25 \\ -22.316 \end{bmatrix} \). Approximation is close.

Step 1: Determine Partial Derivatives

First, we need to compute the partial derivatives of each component of \( \mathbf{f}(x, y) \). For the first component \( f_1(x, y) = (x-y)^2 \), compute \( \frac{\partial f_1}{\partial x} = 2(x-y) \) and \( \frac{\partial f_1}{\partial y} = -2(x-y) \). For the second component \( f_2(x, y) = 2x^2y \), compute \( \frac{\partial f_2}{\partial x} = 4xy \) and \( \frac{\partial f_2}{\partial y} = 2x^2 \).

Step 2: Evaluate Derivatives at Point

We now evaluate these partial derivatives at the point \( (2, -3) \). For \( f_1 \), we have \( \frac{\partial f_1}{\partial x} (2, -3) = 10 \) and \( \frac{\partial f_1}{\partial y} (2, -3) = -10 \). For \( f_2 \), we have \( \frac{\partial f_2}{\partial x} (2, -3) = -24 \) and \( \frac{\partial f_2}{\partial y} (2, -3) = 8 \).

Step 3: Construct Linear Approximation

The linear approximation of \( \mathbf{f}(x, y) \) at \( (2, -3) \) can be expressed in vector form: \[ \mathbf{L}(x, y) = \mathbf{f}(2, -3) + abla \mathbf{f}(2, -3) \cdot \begin{bmatrix} x-2 \ y+3 \end{bmatrix} \]. Calculating \( \mathbf{f}(2, -3) \), we get \( \begin{bmatrix} 25 \ -24 \end{bmatrix} \). The gradient matrix is \[\begin{bmatrix} 10 & -10 \ -24 & 8 \end{bmatrix}\].

Step 4: Apply Linear Approximation

Calculate \[ \mathbf{L}(1.9, -3.1) = \begin{bmatrix} 25 \ -24 \end{bmatrix} + \begin{bmatrix} 10 & -10 \ -24 & 8 \end{bmatrix} \begin{bmatrix} -0.1 \ -0.1 \end{bmatrix} \]. This yields \[ \begin{bmatrix} 25 + 1 \begin{align*}\-24 + 1.6 \end{align*} \end{bmatrix} = \begin{bmatrix} 26 \ -22.4 \end{bmatrix} \].

Step 5: Calculate Exact Values Using Calculator

Now calculate \( \mathbf{f}(1.9, -3.1) \) using a calculator or directly evaluating: For the first component: \( (1.9 - (-3.1))^2 \) gives \( 25 \). For the second component: \( 2 \cdot 1.9^2 \cdot (-3.1) \) gives approximately \( -22.316 \). Hence, \( \mathbf{f}(1.9, -3.1) \approx \begin{bmatrix} 25 \ -22.316 \end{bmatrix} \).

Step 6: Compare Results

The linear approximation at \( (1.9, -3.1) \) is \( \begin{bmatrix} 26 \ -22.4 \end{bmatrix} \), while the exact calculation is approximately \( \begin{bmatrix} 25 \ -22.316 \end{bmatrix} \). The approximation is close, especially in the second component, but not exactly the same.

Key Concepts

Partial Derivatives

A partial derivative represents the rate of change of a multivariable function with respect to one variable while keeping others constant. For vector functions like the one given, each component is treated independently.

• To find the partial derivatives of the first component \( f_1(x, y) = (x-y)^2 \), we compute:
  • \( \frac{\partial f_1}{\partial x} = 2(x-y) \)
  • \( \frac{\partial f_1}{\partial y} = -2(x-y) \)
• For the second component \( f_2(x, y) = 2x^2y \), we find:
  • \( \frac{\partial f_2}{\partial x} = 4xy \)
  • \( \frac{\partial f_2}{\partial y} = 2x^2 \)

These derivatives help in understanding how the function behaves as each variable changes, and they are crucial for constructing the linear approximation.

Gradient Matrix

The gradient is a vector that contains all the partial derivatives of a function, providing a direction and rate for the fastest increase of the function.

For vector functions, we compile these gradients into a matrix form known as the gradient matrix. For our given function, evaluated at the point \((2, -3)\):

• First component evaluated gives:
  • \( \frac{\partial f_1}{\partial x}(2, -3) = 10 \)
  • \( \frac{\partial f_1}{\partial y}(2, -3) = -10 \)
• Second component evaluated gives:
  • \( \frac{\partial f_2}{\partial x}(2, -3) = -24 \)
  • \( \frac{\partial f_2}{\partial y}(2, -3) = 8 \)

This results in the gradient matrix being:\[\begin{bmatrix}10 & -10 \-24 & 8\end{bmatrix}\]This matrix represents how the function's output changes concerning small variations in \(x\) and \(y\).

Vector Functions

Vector functions map input vectors to output vectors, making them essential in multivariable calculus. They are particularly useful in representing multiple outputs from the same set of inputs.

In the problem given:

• The vector function \( \mathbf{f}(x, y) \) has components that yield different results depending on \(x\) and \(y\).
• These functions produce vector outputs, which can be thought of as mapping points in one space to another.

Linear approximation helps by providing a very close linear model of these functions near a specific point \((2, -3)\).

This is achieved by evaluating the vector function at a chosen point and applying the gradient matrix, helping us predict the function's behavior near this point, demonstrated here as \( \mathbf{L}(1.9, -3.1) \). It's an estimated value based on the actual behavior of the function around the approximation point.