Problem 40 - Calculus of a Single Variable

Question

Approximating Function Values In Exercises \(37-40,\) use differentials to approximate the value of the expression. Compare your answer with that of a calculator. $$ (2.99)^{3} $$

Step-by-Step Solution

Answer

The function \((2.99)^3\) can be approximated as 27.27 using differentials. This approximation should be close to the calculator's result.

Step 1: Define Function and Differential

First, define a function \(f(x) = x^3\). Then, compute the differential \(df\) to be \(3x^2 dx\).

Step 2: Choose a Known Point to Approximate From

Choosing \(x=3\) as a known point allows for simpler calculations, in this case. Compute \(f(3) = 27\) and notice that \(dx = 2.99 - 3 = -0.01\).

Step 3: Apply Differential Approximation

With the previously computed values, we can apply the differential approximation, obtaining \(f(2.99) \approx f(3) + df = 27 - 3*(3^2)*(-0.01) = 27 + 0.27 = 27.27\).

Step 4: Compare With Calculator

A comparison with the result obtained by a calculator for \(2.99^3\) should give something close to 27. The comparison helps to understand the accuracy of the approximation used.

Key Concepts

Differentials

Differentials are a fundamental concept in differential calculus. They help us approximate small changes in a function's value. Imagine you have a function, say, \(f(x) = x^3\). When you slightly alter \(x\) by a tiny amount \(dx\), the differential \(df\) gives us an approximation of how much \(f(x)\) will change. This is calculated using \(df = f'(x) \cdot dx\), where \(f'(x)\) is the derivative of \(f(x)\).
In the context of our exercise, knowing \(df = 3x^2 \cdot dx\) allows you to predict how \(x^3\) behaves when \(x\) is slightly closer or further from 3. Differentials are particularly useful in scenarios where exact values are complex to calculate or measurements are subject to variability.

Function Approximation

In mathematical terms, function approximation involves estimating a function's output with another simpler function. Often, this means using linearization, a technique grounded in calculus. Linearization simplifies the computations by approximating the function's behavior around a given point using its tangent.
Coming back to our exercise, when we approximate \((2.99)^3\), we use the known point \(x = 3\). Since exact calculations can be cumbersome, we use the differential to find \(f(2.99) \approx f(3) + df\). This gives a quick estimate, allowing us to avoid tedious calculations while still maintaining a reasonable level of accuracy.

Calculator Comparison

Comparing calculated approximations to exact values derived from calculators is invaluable in learning and understanding mathematics. Calculators can quickly crunch numbers to provide precise outcomes. However, when using calculus, the purpose is to gain insight into the method and its applicability.
In this exercise, computing \((2.99)^3\) gives a realistic view of how differentials approximate the function versus the exact calculator outcome. This not only demonstrates the practical use of differential calculus but also provides insight into the approximation's effectiveness and accuracy. Such comparisons are essential in validating mathematical techniques.

Derivative Applications

Derivatives have many applications in various fields ranging from physics to economics. In calculus, they measure the rate at which a quantity changes. In simpler terms, a derivative tells you how one variable changes relative to another.
For problem-solving, derivatives help us understand and predict dynamics in systems by approximating changes through differentials. In our example, the derivative \(f'(x) = 3x^2\) was crucial in estimating how \((2.99)^3\) diverges from \(3^3\). This principle extends beyond this exercise, providing tools for optimizing functions, calculating slopes, and even analyzing motion.