Problem 9 - APEX Calculus

Question

In Exercises \(9-12,\) a function \(2=f(x, y)\) is given. Give the indicated approximation using the total differential. $$ \begin{array}{l} f(x, y)=\sqrt{x^{2}+y} . \text { Approximate } f(2.95,7.1) \text { knowing } \\\ f(3,7)=4 . \end{array} $$

Step-by-Step Solution

Answer

The approximate value of \(f(2.95, 7.1)\) is 3.975.

Step 1: Understand the Total Differential

First, recognize that the total differential of a function \(f(x,y)\) is an approximation of how the function value changes with small changes in \(x\) and \(y\). The formula for the total differential \(df\) is: \[ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy. \] Here, \(dx\) and \(dy\) are small changes in \(x\) and \(y\), respectively.

Step 2: Calculate Partial Derivatives

Compute the partial derivatives of the function \(f(x,y) = \sqrt{x^2 + y}\). The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2+y}}\) and with respect to \(y\) is \(\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^2+y}}\).

Step 3: Evaluate Partial Derivatives at the Known Point

Evaluate the partial derivatives at the point \((x, y) = (3, 7)\). We find: \(\frac{\partial f}{\partial x} \bigg|_{(3, 7)} = \frac{3}{4}\) and \(\frac{\partial f}{\partial y} \bigg|_{(3, 7)} = \frac{1}{8}\).

Step 4: Determine Changes in Variables

We know \((x, y) = (3, 7)\) and want to approximate \(f(2.95, 7.1)\). Therefore, \(dx = 2.95 - 3 = -0.05\) and \(dy = 7.1 - 7 = 0.1\).

Step 5: Compute the Total Differential

Using the total differential formula, compute \(df\): \[ df = \left( \frac{3}{4} \right)(-0.05) + \left( \frac{1}{8} \right)(0.1) = -0.0375 + 0.0125 = -0.025. \]

Step 6: Calculate the Approximation

Once \(df\) is calculated, approximate \(f(2.95, 7.1)\) using the formula \(f(x + dx, y + dy) \approx f(x, y) + df\). Hence, \[ f(2.95, 7.1) \approx f(3, 7) + df = 4 - 0.025 = 3.975. \]

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in differential calculus, especially when dealing with functions of more than one variable. They measure how a function changes as only one of the variables is incremented, while the other variables remain constant.
In mathematical terms, if you have a function, say, \( f(x, y) \), the partial derivative of \( f \) with respect to \( x \) is denoted by \( \frac{\partial f}{\partial x} \). This indicates the rate at which the function \( f \) changes as \( x \) changes, but \( y \) is held constant.

• To compute a partial derivative, differentiate the function with respect to the variable of interest, while treating all other variables as constants.
• For example, to find \( \frac{\partial f}{\partial x} \) when \( f(x, y) = \sqrt{x^2 + y} \), focus on \( x \)'s role in the function, yielding \( \frac{x}{\sqrt{x^2+y}} \).

Partial derivatives provide the groundwork needed to understand the overall behavior of a multivariable function, especially crucial for creating approximations like the total differential.

Function Approximation

Function approximation allows us to estimate the value of a function at a certain point based on the values and changes at a nearby point. One common method of approximation is by using the total differential.
The idea is to use a known or easier-to-compute value, like \( f(3, 7) = 4 \), and small changes in \( x \) and \( y \) (\( dx \) and \( dy \)) to estimate \( f(2.95, 7.1) \).

• The total differential \( df \) gives us a linear approximation of how the function value changes, represented as \( df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \).
• Since the function's change is assumed small, \( df \) can be added to the known function value to get our approximation.
• In practice, if you compute \( df \) as -0.025, an approximate value of the function is \( 3.975 \), from the function value 4 at a known point.

This method simplifies complex calculations by providing approximate values that are often close to actual values, especially when changes \( dx \) and \( dy \) are small.

Differential Calculus

Differential calculus serves as a branch of calculus focused on the concept of differentials, derivatives, and rates of change. It provides tools and techniques to analyze how functions change, which we see in the calculation of total differentials through partial derivatives.

One of the key aspects of differential calculus is it allows for the determination of how small changes in input coordinates can lead to changes in the output of a function. This is expressed through:

• Finding partial derivatives which help in constructing differentials for multivariate functions.
• Using the computed differential (like \( df \)) to predict or approximate function behavior over small intervals.

In the given exercise, differential calculus plays a crucial role in approximating \( f(2.95, 7.1) \) using the total differential of the function \( f(x, y) = \sqrt{x^2 + y} \), based on the information obtained at a known point. Understanding these concepts can particularly aid in fields like physics and engineering, where precise predictions of small changes are often necessary.