Problem 25 - 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

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

Step 1: Understand the Problem

We are given the function value at a specific point, \(f(25) = 3.6\), and the derivative of the function at that point, \(f'(25) = -0.2\). We need to estimate the function values \(f(26)\) and \(f(30)\).

Step 2: Use the Derivative for Linear Approximation

The derivative \(f'(t)\) represents the rate of change of the function \(f(t)\). Therefore, we can use linear approximation (or linearization) to estimate changes in the function value over small intervals: \(f(t + riangle t) \approx f(t) + f'(t) \times riangle t\).

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

First, let's estimate \(f(26)\). Here, \(t = 25\) and \(\triangle t = 1\). Using the linear approximation formula, we have: \[f(26) \approx f(25) + f'(25) \times 1 = 3.6 + (-0.2) \times 1 = 3.6 - 0.2 = 3.4\].

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

Now, estimate \(f(30)\). Here, \(t = 25\) and \(\triangle t = 5\). Again using the linear approximation formula, we have: \[f(30) \approx f(25) + f'(25) \times 5 = 3.6 + (-0.2) \times 5 = 3.6 - 1.0 = 2.6\].

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. Think of it as the function's speedometer, showing how quickly the function's value is rising or falling. A positive derivative means the function is increasing, while a negative derivative points to a decrease. The formal definition involves limits, where we take the limit of the ratio of the change in function values to the change in input values as the change in input approaches zero. Mathematically, if \( f(t) \) is a function, then its derivative \( f'(t) \) is given by:
\[ f'(t) = \lim_{{h \to 0}} \frac{{f(t+h) - f(t)}}{h} \]
In the exercise, we're given that \( f'(25) = -0.2 \). This tells us that at \( t = 25 \), the function is decreasing at a rate of 0.2 units per unit of time. This information is crucial for estimating the function's future values.

Function Estimation

In many situations, we don't need the exact value of a function but an estimate that is close enough, especially when working with difficult or unknown functions. Linear approximation is a simple but powerful method for estimating function values.
Using the linear approximation or linearization method, we can estimate the value of a function near a known point. The idea is rooted in the formula:

• \( f(t + \Delta t) \approx f(t) + f'(t) \times \Delta t \)

This formula takes advantage of the derivative, using it to "predict" what the function might look like slightly ahead of the known point, assuming it maintains a straight-line path. It works well for small changes in \( t \) because many functions are nearly linear over tiny intervals.
In the provided exercise example, we used this concept to estimate \( f(26) \) and \( f(30) \). For \( f(26) \), since our change was relatively small (\