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 of \(f(1.05, 2.05)\) is \([1.7625, -3.15]\), close to the exact \([1.7541, -3.1525]\).

Step 1: Understand the Problem

We need to find the linear approximation of vector function \(f(x,y)\) at a given point \((1, 2)\) and use this to approximate \(f(1.05, 2.05)\).

Step 2: Find Partial Derivatives

First, compute the partial derivatives of each component of \(f(x,y)\). For the first component \( f_1(x, y) = \sqrt{2x + y} \), the partial derivatives are: \( \frac{\partial f_1}{\partial x} = \frac{1}{\sqrt{2x + y}} \cdot 2 \) and \( \frac{\partial f_1}{\partial y} = \frac{1}{2\sqrt{2x + y}} \). For the second component \(f_2(x, y) = x - y^2\), the derivatives are \( \frac{\partial f_2}{\partial x} = 1 \) and \( \frac{\partial f_2}{\partial y} = -2y \).

Step 3: Evaluate Derivatives at the Point (1,2)

Substitute \(x = 1\) and \(y = 2\) into the partial derivatives. For \(f_1\), \( \frac{\partial f_1}{\partial x} = \frac{2}{\sqrt{2(1) + 2}} = 1 \) and \( \frac{\partial f_1}{\partial y} = \frac{1}{2\sqrt{4}} = \frac{1}{4} \). For \(f_2\), \( \frac{\partial f_2}{\partial x} = 1 \) and \( \frac{\partial f_2}{\partial y} = -4 \).

Step 4: Construct Linear Approximation

The linear approximation of \(f(x, y)\) is given by \( f(x, y) \approx f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \) for each component. At \((1, 2)\), \[ L(x, y) = \begin{bmatrix} 2 \ + 1(x - 1) + \frac{1}{4}(y - 2) \ -3 + 1(x - 1) - 4(y - 2) \end{bmatrix} \].

Step 5: Simplify the Linear Approximation

Simplify each component: \[ L(x, y) = \begin{bmatrix} 2 + x - 1 + \frac{y}{4} - \frac{1}{2} \ -3 + x - 1 - 4y + 8 \end{bmatrix} = \begin{bmatrix} x + \frac{y}{4} + \frac{1}{2} \ x - 4y + 4 \end{bmatrix} \]. This is the linear approximation at \((1, 2)\).

Step 6: Approximate at (1.05, 2.05)

Use the linear approximation to find \[ L(1.05, 2.05) = \begin{bmatrix} 1.05 + \frac{2.05}{4} + \frac{1}{2} \ 1.05 - 4(2.05) + 4 \end{bmatrix} = \begin{bmatrix} 1.7625 \ -3.15 \end{bmatrix} \].

Step 7: Calculate Exact Value using Calculator

Using a calculator, find the exact value of \(f(1.05, 2.05)\), where \( f_1 = \sqrt{2(1.05) + 2.05} \approx 1.7541 \) and \( f_2 = 1.05 - (2.05)^2 \approx -3.1525 \).

Step 8: Compare Approximation with Exact Value

The linear approximation \([1.7625, -3.15]\) is very close to the calculated value \([1.7541, -3.1525]\), showing that our linear approximation is accurate.

Key Concepts

Linear Approximation

When tasked with estimating a function's value near a known point, linear approximation can be a powerful tool. In multivariable calculus, linear approximation involves using the tangent plane at a known point to estimate the function's value at nearby points.
This technique is particularly useful for vector functions, where each component of the output vector function is approximated separately.

• For any function of two variables, a linear approximation takes the form:
  
  \[f(x, y) \approx f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]
• In this formula, \(f_x\) and \(f_y\) are the partial derivatives of the function with respect to \(x\) and \(y\), respectively.

The main idea is that close to point \((a, b)\), the function behaves approximately linearly. Thus, the tangent plane, or in the case of vector functions, the tangent line or tangent vector, provides a good basis for approximations.

Partial Derivatives

Partial derivatives are a key concept in understanding how functions change with respect to one variable, while keeping others constant. When dealing with multivariable calculus, it's important to understand how to compute and interpret these derivatives.
Partial derivatives are used in the calculation of the linear approximation.
Here's how they work in our example function:

• The function \( f(x,y) \) has two components: \(f_1(x, y) = \sqrt{2x + y}\) and \(f_2(x, y) = x - y^2\).
• For \(f_1\), the partial derivative with respect to \(x\) is \( \frac{1}{\sqrt{2x + y}} \cdot 2 \), and with respect to \(y\) is \( \frac{1}{2\sqrt{2x + y}} \).
• For \(f_2\), the derivative with respect to \(x\) is \(1\), and with respect to \(y\) is \(-2y\).

Evaluating these derivatives at the point \((1, 2)\) gives us the necessary values to find the linear approximation around that point. Partial derivatives reveal how sensitive each function component is to small changes in \(x\) and \(y\).

Vector Functions

Vector functions are functions that have vectors as their outputs, not just individual numbers. This can make them a bit tricky because you have to think about how each component of the vector reacts to changes in the input.
In our example, the function \(\mathbf{f}(x, y)\) outputs a two-component vector.

• Each component of the vector function has its own partial derivatives.
• Linear approximation will apply to each component separately, meaning each gets its own tangent line or plane.

Understanding the behavior of vector functions is useful because many real-world phenomena can be described with vectors, such as velocity fields in physics or gradients in optimization problems.

Approximation Accuracy

The accuracy of a linear approximation depends on how linear the function is around the point at which the approximation is made. If the function changes rapidly or non-linearly, the approximation might not be as accurate.
In practice, linear approximation accuracy can often be evaluated by comparing the approximated value with the actual value calculated using more precise methods or tools:

• In our solution, we approximated \(f(1.05, 2.05)\) using a linear method and compared it to the value found using a calculator.
• The approximation gave us \([1.7625, -3.15]\), while the exact values were about \([1.7541, -3.1525]\).
• The small difference between these values indicates a good approximation,
  highlighting the usefulness of linear approximation in estimating values near known points.

Overall, linear approximations offer a simple way to estimate complex functions, especially when exact calculations are cumbersome or unnecessary.