Problem 27 - Calculus for Biology and Medicine

Question

Find the linear approximation of $$ f(x, y)=\sqrt{x+y^{2}} $$ at \((1,0)\), and use it to approximate \(f(1.1,0.1) .\) Using a calculator, compare the approximation with the exact value of \(f(1.1,0.1)\).

Step-by-Step Solution

Answer

The linear approximation of \( f(1.1, 0.1) \) is 1.05, close to the exact value \( \approx 1.053 \).

Step 1: Calculate the Partial Derivatives at the Point

First, find the partial derivatives of the function \( f(x, y) = \sqrt{x + y^2} \). For \( f_x(x, y) \) we have: \[ f_x(x, y) = \frac{1}{2\sqrt{x + y^2}} \]Evaluating at the point (1,0): \[ f_x(1, 0) = \frac{1}{2\sqrt{1 + 0^2}} = \frac{1}{2} \] Next, for \( f_y(x, y) \) : \[ f_y(x, y) = \frac{y}{\sqrt{x + y^2}} \]Evaluating at the point (1,0):\[ f_y(1, 0) = \frac{0}{\sqrt{1 + 0^2}} = 0 \]

Step 2: Write the Linear Approximation Formula

The linear approximation of \( f(x, y) \) at a point \((a, b)\) is given by:\[ L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \] For \( f(x, y) = \sqrt{x + y^2} \) at \((1, 0)\):\[ L(x, y) = \sqrt{1 + 0^2} + \frac{1}{2}(x - 1) + 0(y - 0) \] This simplifies to: \[ L(x, y) = 1 + \frac{1}{2}(x - 1) \]

Step 3: Approximate f(1.1, 0.1) Using the Linear Approximation

Substitute \( x = 1.1 \) and \( y = 0.1 \) into the linear approximation \( L(x, y) \):\[ L(1.1, 0.1) = 1 + \frac{1}{2}(1.1 - 1) \] Calculate the expression:\[ L(1.1, 0.1) = 1 + \frac{1}{2}(0.1) = 1 + 0.05 = 1.05 \]

Step 4: Calculate the Exact Value of f(1.1, 0.1) and Compare

Calculate the exact value using the original function:\[ f(1.1, 0.1) = \sqrt{1.1 + 0.1^2} \]Calculate this expression:\[ f(1.1, 0.1) = \sqrt{1.1 + 0.01} = \sqrt{1.11} \approx 1.053 \]Now, compare the linear approximation \( 1.05 \) to the exact value \( \approx 1.053 \). The linear approximation is close to the exact value.

Key Concepts

Partial Derivatives in Linear Approximation

Partial derivatives play a crucial role when dealing with functions of multiple variables. Just like ordinary derivatives help us understand how a function changes at a point, partial derivatives do the same but with respect to each variable independently. For a function like \[ f(x, y) = \sqrt{x + y^2} \]we can calculate the rate of change along the x-direction and the y-direction independently. This gives us the partial derivatives:

• \( f_x(x, y) = \frac{1}{2\sqrt{x + y^2}} \), indicating how the function changes as x changes while keeping y constant.
• \( f_y(x, y) = \frac{y}{\sqrt{x + y^2}} \), showing how the function changes as y changes while keeping x constant.

To find these derivatives at a specific point, like (1,0), substitute the values into the partial derivative formulas. This helps simplify our calculations when performing linear approximations.

Understanding Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of more than one variable. When we have a function like \( f(x, y) = \sqrt{x + y^2} \), it doesn’t just depend on one input but on two: x and y. This complexity requires a deeper understanding of how changes in these inputs affect the overall output.

Typically, changes in these variables can be analyzed using partial derivatives, helping us understand the "slopes" of the function surface. This is similar to how slopes work in single-variable calculus, but instead of a straight line, it's like a tilted plane in two directions.

In the context of linear approximation, the goal is to use this understanding to create a simpler function that roughly "matches" the behavior of our more complex function at a given point. This technique proves useful for estimating values based on known measurements of the function around a specific point.

Linear Approximation Techniques

Linear approximation is a handy technique for estimating the output of multivariable functions near given points. The technique involves generating a simple linear function \( L(x, y) \),that approximates the behavior of \( f(x, y) \).The formula for this linear approximation is:\[L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]where \((a,b)\) is the point of approximation. The partial derivatives \( f_x \) and \( f_y \) account for how changes in x and y affect the overall function.

In simpler terms, linear approximation takes a complex surface and "flattens" it to a tangent plane at a point, making it easier to work with. For our function \( \sqrt{x + y^2} \)at \((1, 0)\), we derived that \[ L(x, y) = 1 + \frac{1}{2}(x - 1) \]. This assists in approximating the values of \( f(x, y) \) near that point, like finding \( f(1.1, 0.1) \),by using the linear function instead of the original, more complex function.