Problem 57 - Brief Applied Calculus

Question

Learning Theory In a psychology experiment, a person could memorize \(x\) words in \(f(x)=2 x^{2}-x\) seconds (for \(0 \leq x \leq 10)\). a. Find \(f^{\prime}(x)\) by using the definition of the derivative. b. Find \(f^{\prime}(5)\) and interpret it as an instantaneous rate of change in the proper units.

Step-by-Step Solution

Answer

\( f'(x) = 4x - 1 \); \( f'(5) = 19 \) seconds per word.

Step 1: Understanding the function

The function given is \( f(x) = 2x^2 - x \). This function models the time it takes for a person to memorize \( x \) words.

Step 2: Definition of the Derivative

By the definition, the derivative of \( f(x) \) at any point \( x \) is given by \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \). This represents the rate of change of the function.

Step 3: Calculate \( f(x+h) \)

First, calculate \( f(x+h) \). Substitute \( x+h \) into the original function: \( f(x+h) = 2(x+h)^2 - (x+h) \). Expand this expression to simplify it. \( f(x+h) = 2(x^2 + 2xh + h^2) - x - h = 2x^2 + 4xh + 2h^2 - x - h \).

Step 4: Calculate \( f(x+h) - f(x) \)

Subtract \( f(x) \) from \( f(x + h) \): \( f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - x - h) - (2x^2 - x) = 4xh + 2h^2 - h \). Simplify this to: \( f(x+h) - f(x) = h(4x + 2h - 1) \).

Step 5: Limit of the Difference Quotient

Now compute the derivative as: \( f'(x) = \lim_{{h \to 0}} \frac{h (4x + 2h - 1)}{h} = \lim_{{h \to 0}} (4x + 2h - 1) \). As \( h \to 0 \), this simplifies to: \( f'(x) = 4x - 1 \).

Step 6: Calculate \( f'(5) \)

Substitute \( x = 5 \) into the derivative \( f'(x) = 4x - 1 \): \( f'(5) = 4(5) - 1 = 20 - 1 = 19 \).

Step 7: Interpretation of \( f'(5) \)

The value of \( f'(5) = 19 \) represents the instantaneous rate of change of the time with respect to the number of words memorized. In proper units, it means that at \( x = 5 \) words, the time required to memorize is increasing at a rate of 19 seconds per additional word.

Key Concepts

Instantaneous Rate of Change

The idea of instantaneous rate of change is central to understanding derivatives in calculus. It tells us how fast a function is changing at a specific point.
For example, in the context of our psychology experiment, it shows how quickly the time to memorize words changes as more words are added.

Think of it as the speedometer on a car, showing how fast you are going at an exact moment.

• When you calculate the derivative at a particular point, you find the instantaneous rate of change.
• In the original exercise, we calculated this rate at the point where 5 words are memorized.
• This gives the rate of change in seconds per additional word, indicating how the process includes more words to memorize.

It's essential in numerous applications, helping us understand systems that change over time.

Limit of a Function

The limit is a foundational concept in calculus that allows us to define a derivative.
Limits help in understanding how a function behaves as it approaches a certain value or point.
In our derivative definition, the limit was used to determine the slope of the tangent line to the curve at any given point.

Let's explore how limits work:

• We use limits to understand what happens when we make very tiny changes (like approaching zero).
• In the solution, the expression \( \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \) represents the limit of the difference quotient.
• Calculating this limit gives us the derivative \( f'(x) \), revealing the rate of change.

The limit helps smooth out errors and fixes undefined expressions, offering precision in calculus problems.

Differentiation Techniques

Differentiation techniques are crucial in solving calculus problems involving rate of change.
They provide us with methods to find derivatives of different types of functions.
This process involves using the limit of a function as well as specific rules to calculate derivatives efficiently.

Some basic differentiation techniques include:

• Power Rule: Useful for polynomial functions, where you multiply the power by the coefficient and reduce the power by one.
• Chain Rule: Applies when you have a function inside another function, helping to differentiate complex compositions.
• Product and Quotient Rules: Help differentiate products or quotients of functions.

In our problem, the derivative was calculated directly using the definition, but these techniques often simplify derivative calculations in practice.