Problem 8 - Calculus Concepts

Question

a. use the limit definition of the derivative (algebraic method) to write an expression for the rate-of-change function of the given function. b. evaluate the rate of change as indicated. \(\quad s(t)=-2.3 t^{2} ; s^{\prime}(1.5)\)

Step-by-Step Solution

Answer

The rate of change at \(t=1.5\) is \(-6.9\).

Step 1: Understanding the Limit Definition of a Derivative

The limit definition of a derivative is given by \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Our function is \( s(t) = -2.3t^2 \), so we need to set \( f(x) \) to \( s(t) \) and find \( s'(t) \).

Step 2: Apply the Limit Definition to the Function

To find \( s'(t) \), substitute \( s(t) = -2.3t^2 \) into the limit definition: \(s'(t) = \lim_{h \to 0} \frac{-2.3(t+h)^2 - (-2.3t^2)}{h} \).

Step 3: Expand \( (t+h)^2 \) and Simplify the Expression

Expand \((t+h)^2\) to get \(t^2 + 2th + h^2\). Substitute this into the expression:\[-2.3(t^2 + 2th + h^2) + 2.3t^2\].Combine like terms to simplify:\[-2.3t^2 - 4.6th - 2.3h^2 + 2.3t^2\].This further simplifies to:\(-4.6th - 2.3h^2\).

Step 4: Factor and Apply the Limit

Factor out \(h\) from the numerator: \(h(-4.6t - 2.3h)\). Simplify the expression: \(\lim_{h \to 0} (-4.6t - 2.3h)\).As \(h\to 0\), the expression simplifies to \(-4.6t\).

Step 5: Evaluate the Derivative at \(t=1.5\)

Having determined that \(s'(t) = -4.6t\), substitute \(t = 1.5\): \[s'(1.5) = -4.6 \times 1.5\].Calculate the result: \(-6.9\).

Key Concepts

Limit Definition of the Derivative

The limit definition of a derivative is a powerful tool in calculus. It helps us understand how a function behaves at an infinitesimally small level. The formula for the derivative using limits is: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \].This formula captures the "rate of change" of a function at a specific point. Essentially, by taking the limit as \(h\) approaches zero, we are measuring how much the function changes as a result of an infinitesimal change in \(x\).
For the exercise, our function is \(s(t) = -2.3t^2\). Plugging this into the formula, we find the derivative of \(s(t)\), denoted \(s'(t)\). This involves setting up the expression: \[ s'(t) = \lim_{h \to 0} \frac{-2.3(t+h)^2 - (-2.3t^2)}{h} \].
By using this approach, we mathematically define the concept of a derivative, providing a precise way to compute a function's rate of change at any point.

Rate of Change

The rate of change is a fundamental concept in calculus and describes how much one quantity changes in relation to another. In the case of our function \(s(t) = -2.3t^2\), the rate of change tells us how rapidly \(s(t)\) changes as \(t\) changes. In real-world terms, it can represent speed in physics, indicating how fast something is moving.
To find the rate of change of \(s(t)\), we calculate its derivative, \(s'(t)\). The rate of change function, \(s'(t)\), gives us a new function that expresses this relationship. For our example, we derived that \(s'(t) = -4.6t\). This means that at any given time \(t\), the rate of change of \(s(t)\) is \(-4.6t\).
A negative rate of change indicates that \(s(t)\) is decreasing. Knowing the rate of change helps us understand how the function behaves over time and can uncover trends not immediately visible from the original function.

Algebraic Method

The algebraic method involves manipulating the terms in an expression to simplify it. This is crucial when using the limit definition of the derivative. Let's delve into how it is used in the given exercise.
The function \(s(t) = -2.3t^2\) was substituted into the limit form: \[ s'(t) = \lim_{h \to 0} \frac{-2.3(t+h)^2 - (-2.3t^2)}{h} \].To proceed, the algebraic method calls for expanding \((t+h)^2\) into \(t^2 + 2th + h^2\). Substituting this back, we get:\[-2.3(t^2 + 2th + h^2) + 2.3t^2 = -2.3t^2 - 4.6th - 2.3h^2 + 2.3t^2\].
Through combining like terms, simplification leads us to \(-4.6th - 2.3h^2\). Eventually, we factor out \(h\) to simplify further, setting up \(h(-4.6t - 2.3h)\) allowing us to take the limit.By methodically applying these steps, we simplify the calculus problem algebraically, clearing the path to compute the derivative.

Differentiation

Differentiation is the process of finding the derivative of a function. It enables us to determine the rate at which a function's output changes concerning its input. In our case, differentiation was applied to \(s(t) = -2.3t^2\) using the limit definition.
Through applying the limit definition and simplifying the terms, we found that the derivative \(s'(t) = -4.6t\). This new function, \(s'(t)\), represents how fast the original function \(s(t)\) changes for any value of \(t\).
The beauty of differentiation lies in its wide applications. It is used extensively in physics, engineering, economics, and beyond to understand trends, motions, and other dynamic behaviors within systems. Finding a derivative, as we did, demystifies the detailed rate of change of the original block of data or motion.

Function Evaluation

Function evaluation involves calculating the value of a function for given inputs. After finding the derivative, it is often essential to determine the rate of change at particular points of interest.
In this exercise, once we obtained the derivative \(s'(t) = -4.6t\), we evaluated it at \(t = 1.5\). This means substituting \(1.5\) into \(s'(t)\) to find \(s'(1.5)\):\[ s'(1.5) = -4.6 \times 1.5 = -6.9 \].This calculation shows that at \(t = 1.5\), the function \(s(t)\) is decreasing at a rate of \(-6.9\). Evaluating a derivative at specific points enables us to pinpoint exact rates of change, offering critical insights which are invaluable in both theoretical and practical situations.