Problem 21 - Applied Calculus

Question

For a function \(f(r, s)\), we are given \(f(50,100)=5.67\), and \(f_{r}(50,100)=0.60\), and \(f_{s}(50,100)=-0.15 .\) Estimate \(f(52,108)\).

Step-by-Step Solution

Answer

The estimated value of \( f(52, 108) \) is 5.67.

Step 1: Understand the problem

We need to estimate the value of the function \( f(r, s) \) at the point \( (52, 108) \) given its value, and partial derivatives at the point \( (50, 100) \). This suggests using a linear approximation approach.

Step 2: Recall the formula for linear approximation

The linear approximation formula for a function \( f \) near the point \( (a, b) \) is: \[ f(x, y) \approx f(a, b) + f_r(a, b)(x - a) + f_s(a, b)(y - b) \] where \( f_r \) and \( f_s \) are the partial derivatives of \( f \) with respect to \( r \) and \( s \) respectively.

Step 3: Apply the given values

Substitute \( r = 50 \), \( s = 100 \), \( x = 52 \), \( y = 108 \) into the linear approximation formula. The approximation becomes: \[ f(52, 108) \approx f(50, 100) + f_r(50, 100)(52 - 50) + f_s(50, 100)(108 - 100) \].

Step 4: Insert known values into the formula

We know from the problem statement that \( f(50, 100) = 5.67 \), \( f_r(50, 100) = 0.60 \), and \( f_s(50, 100) = -0.15 \). Substitute these into the linear approximation formula: \[ f(52, 108) \approx 5.67 + 0.60(2) + (-0.15)(8) \].

Step 5: Calculate the linear approximation

Compute the values: \( 0.60 \times 2 = 1.20 \) and \( (-0.15) \times 8 = -1.20 \). Then, calculate the full expression: \[ 5.67 + 1.20 - 1.20 = 5.67 \].

Key Concepts

Partial Derivatives

Partial derivatives are a cornerstone concept in calculus, especially in the study of multivariable functions. When dealing with functions of more than one variable, like our function \(f(r, s)\), it's useful to understand how the function changes with respect to each variable individually. This is where partial derivatives come in.

• The partial derivative with respect to \(r\), denoted as \(f_r\), measures the rate at which the function \(f\) changes as \(r\) changes, while keeping \(s\) constant.
• Similarly, the partial derivative with respect to \(s\), \(f_s\), indicates the rate of change of \(f\) as \(s\) varies, holding \(r\) fixed.

To determine these rates of change, we differentiate the function with respect to one variable, treating other variables as constants, much like how ordinary derivatives handle single-variable functions. In our exercise, knowing \(f_r(50, 100) = 0.60\) tells us that a small increase in \(r\) alone would increase the value of function \(f\) by 0.60 units for each unit increase in \(r\). Conversely, \(f_s(50, 100) = -0.15\) suggests a small increase in \(s\) decreases \(f\) by 0.15 units for every unit increase in \(s\). Partial derivatives are essential for determining how a function behaves locally around specific points.

Multivariable Calculus

Multivariable calculus is an extension of basic calculus into functions with more than one variable. This branch of mathematics allows us to explore functions with inputs that influence each other and themselves. For instance, a function like \( f(r, s) \) shows how variables \( r \) and \( s \) together influence the output.In a multivariable setting, we often encounter scenarios where a small change in our inputs leads to changes in the output. The concept of linear approximation, like the one used in our problem, capitalizes on this. Linear approximation simplifies complex surfaces of multivariable functions into linear planes by providing an estimate that is manageable and closer to the real function's value at nearby points.Linear approximation uses the idea that for small changes in \(x\) and \(y\) near a point \((a, b)\), the function changes in a manner depicted by a first-degree Taylor polynomial akin to a linear equation: \[ f(x, y) \approx f(a, b) + f_r(a, b)(x - a) + f_s(a, b)(y - b) \]This tool is incredibly powerful in multivariable calculus, allowing us to navigate and predict the behavior of multivariable functions efficiently, such as estimating \( f(52, 108) \) using given data at \( (50, 100) \).

Applied Calculus

Applied calculus takes the theories of calculus and implements them in practical, real-world problems. This application often involves approximations like the linear approximation. In real-world scenarios, it is ideal for situations where exact functions are difficult or impossible to work with, due to complexity or lack of data.Using applied calculus and the concept of linear approximation, as seen in our original problem, allows us to perform estimations based on known rates of change. This is particularly useful in fields such as:

• Engineering, where precise calculations are necessary for design but using simpler estimates accelerates early-stage predictions.
• Economics, where functions model behaviors of supply and demand and small changes can meaningfully affect equilibria.
• Weather prediction, where conditions change over multiple variables and require approximations to make timely forecasts.

In the provided example, the linear approximation formula enables an estimate of \(f(52, 108)\) without needing the exact function form, leveraging given information about its behavior at \((50, 100)\). This approach provides a quick and reasonable assessment, showcasing the power and utility of applied calculus in tackling multivariable problems.