Problem 22 - Applied Calculus

Question

Suppose that \(f(t)\) is a function with \(f(25)=3.6\) and \(f^{\prime}(25)=-0.2 .\) Estimate \(f(26)\) and \(f(30)\)

Step-by-Step Solution

Answer

\(f(26) \approx 3.4\) and \(f(30) \approx 2.6\).

Step 1: Understanding the Problem

We have a function \(f(t)\) with known values \(f(25) = 3.6\) and its derivative \(f'(25) = -0.2\). We need to estimate \(f(26)\) and \(f(30)\).

Step 2: Using Linear Approximation

To estimate \(f(t)\) near \(t = 25\), we use a linear approximation: \(f(t) \approx f(25) + f'(25) \cdot (t - 25)\). This formula helps us estimate values of the function for \(t\) close to 25.

Step 3: Calculating \(f(26)\)

Substitute \(t = 26\) into the linear approximation formula: \[f(26) \approx f(25) + f'(25) \cdot (26 - 25) = 3.6 + (-0.2) \cdot 1 = 3.6 - 0.2 = 3.4\]. Thus, the estimated value of \(f(26)\) is 3.4.

Step 4: Calculating \(f(30)\)

Substitute \(t = 30\) into the linear approximation formula: \[f(30) \approx f(25) + f'(25) \cdot (30 - 25) = 3.6 + (-0.2) \cdot 5 = 3.6 - 1 = 2.6\]. Thus, the estimated value of \(f(30)\) is 2.6.

Key Concepts

Derivative

The derivative is a core concept in calculus, representing how a function changes as its input changes. In essence, it's the function's instantaneous rate of change. Imagine you're driving a car; the derivative is like the speedometer, showing how fast you're going at any given moment. It's an especially useful tool for:

• Analyzing how a small change in one variable affects another
• Understanding the slope of the tangent line to a curve at a point
• Solving real-world problems where change is involved, like physics and economics

In the context of our problem, given that the derivative at a point is \(f'(25) = -0.2\), this tells us that for every small increase in \(t\), the function \(f(t)\) is expected to decrease by 0.2. Derivatives lay the groundwork for predicting and understanding changes, making them indispensable in mathematics and applied sciences.

Function Estimation

Function estimation involves finding approximate values of the function at certain points, especially when exact computation is difficult or impossible. Linear approximation, also known as linearization, is a powerful tool for this purpose. It uses:

• The value of the function at a known point
• The slope of the tangent (derivative) at that point

The formula for linear approximation is:\[f(t) \approx f(a) + f'(a) \cdot (t - a)\]This simple yet effective method works best for points close to \(a\), where \(a\) is a point with known values. In our example, knowing \(f(25) = 3.6\) and \(f'(25) = -0.2\) allows us to estimate values like \(f(26)\) and \(f(30)\). This real-world problem-solving strategy is particularly useful when dealing with functions that are tough to compute directly.

Calculus Problem Solving

Calculus problem solving often involves using a variety of tools and techniques to tackle real-world questions. In this exercise, our task was to estimate function values at certain points using given data. Here's how we approached it:

• Recognized the known values of the function, \(f(25) = 3.6\)
• Used the derivative, \(f'(25) = -0.2\), to understand the rate of change
• Applied the linear approximation method to estimate \(f(26)\) and \(f(30)\)

Calculus problem solving emphasizes step-by-step logical reasoning. Understanding each step's purpose and how it contributes to the overall solution is vital. Whether estimating functions, analyzing graphs, or solving differential equations, the skills acquired through calculus are widely applicable to scientific research, engineering, and beyond. This problem highlights how a structured approach helps demystify complex mathematical questions.