Problem 8 - Calculus with Concepts in Calculus

Question

Approximate the value of \(f\) at the given point. $$ f(x, y, z)=x y z^{2} ;(-2.1,1.01,0.989) $$

Step-by-Step Solution

Answer

The approximate value of \(f\) at the given point is -2.0737.

Step 1: Determine the Function to Approximate

We are given the function \(f(x, y, z) = x y z^2\) and the point \((-2.1, 1.01, 0.989)\). To approximate \(f\) at this point, we need to substitute these values into the function.

Step 2: Substitute the Variables

Substitute \(x = -2.1\), \(y = 1.01\), and \(z = 0.989\) into the function: \[ f(-2.1, 1.01, 0.989) = (-2.1) \cdot (1.01) \cdot (0.989)^2 \]

Step 3: Calculate \(z^2\)

First, find the square of \(z\): \[0.989^2 = 0.978121\].

Step 4: Simplify the Multiplication

Multiply the values obtained:\[ f(-2.1, 1.01, 0.989) = (-2.1) \times (1.01) \times 0.978121 \].

Step 5: Compute the Final Result

Perform the multiplication step-by-step: 1. Calculate \((-2.1) \times (1.01) = -2.121\).2. Then calculate \(-2.121 \times 0.978121 = -2.0737\).

Step 6: Provide the Approximation

The approximate value of \(f(x, y, z)\) at the point \((-2.1, 1.01, 0.989)\) is -2.0737.

Key Concepts

Function Approximation

Function approximation is a fundamental technique in multivariable calculus and analysis. It aims to find an estimate of a function's value when direct computation might be complex or impractical.
For the function given by \(f(x, y, z) = x y z^2\), our goal was to approximate \(f\) at a specific point \((-2.1, 1.01, 0.989)\). This involves substituting the provided variables into the function and simplifying it to get an approximate value.
The concept of approximation is important in various fields such as engineering and physics, where exact values are sometimes impossible to calculate directly.

• This is especially true in multivariable functions where three or more variables are involved, and precision is vital for modeling real-world situations.

Substitution Method

The substitution method is a straightforward approach used to simplify calculations or solve functions.
For the exercise \(f(x, y, z) = x y z^2\), we substitute the given values of \(x = -2.1\), \(y = 1.01\), and \(z = 0.989\) directly into the equation.
This method is simple but powerful, allowing us to break down more complex functions into easier computations.

• During substitution, it’s crucial to carefully replace each variable with the given numbers to maintain accuracy in later steps.
• This can be particularly helpful in classroom settings for teaching students about managing multiple variables at once.

Multiplication of Decimals

Multiplication of decimals is a skill that is often used when dealing with functions like \(f(x, y, z) = x y z^2\).
After substituting, calculations involved multiplying \(-2.1\), \(1.01\), and \(0.978121\) (derived from the square of 0.989).
When multiplying decimals, it is important to:

• Count the total number of decimal places in both numbers. This helps to correctly place the decimal in the answer.
• Break down the multiplication into simpler steps, such as multiplying two numbers first before including the third.

Practicing the multiplication of decimals strengthens numerical skills and ensures the accuracy of function approximations in real-world applications.

Approximation Techniques

Approximation techniques are foundational in computations where exact values are challenging to find. In multivariable calculus, these techniques are essential, but can vary based on the complexity of the function and the closeness needed.

• Approximating \(f\) at \((-2.1, 1.01, 0.989)\) involves using straightforward numeric calculations.
• Linear approximations or polynomial approximations are sometimes used in more advanced settings to achieve better precision.

In our exercise, the use of straightforward arithmetic is an excellent example of how simplifying complex calculations with approximation methods makes solving and understanding functions more manageable.