Problem 23 - Calculus: Early Transcendentals

Question

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( f(t) = 2.5t^2 + 6t \)

Step-by-Step Solution

Answer

Derivative is \( 5t+6 \); both function and derivative have domain \( \mathbb{R} \).

Step 1: Understand the Problem

We need to find the derivative of the function \( f(t) = 2.5t^2 + 6t \) using the definition of the derivative. We will then determine the domain of the function and its derivative.

Step 2: Recall the Definition of Derivative

The derivative of a function \( f(t) \) at a point \( t \) can be found using the limit \[f'(t) = \lim_{{h \to 0}} \frac{f(t+h) - f(t)}{h}.\]

Step 3: Apply the Definition

Substitute \( f(t) \) into the definition: 1. Calculate \( f(t+h) = 2.5(t+h)^2 + 6(t+h) \).2. Simplify \( f(t+h) \) to get \( 2.5(t^2 + 2th + h^2) + 6t + 6h \).3. This simplifies to \( 2.5t^2 + 5th + 2.5h^2 + 6t + 6h \).4. Subtract \( f(t) = 2.5t^2 + 6t \) from this to get \( 5th + 2.5h^2 + 6h \).

Step 4: Simplify the Difference Quotient

Dividing by \( h \), the difference quotient becomes \[\frac{5th + 2.5h^2 + 6h}{h} = 5t + 2.5h + 6.\]

Step 5: Take the Limit as h Approaches Zero

As \( h \to 0 \), the expression simplifies to \[f'(t) = 5t + 6.\] Thus, the derivative of the function \( f(t) \) is \( f'(t) = 5t + 6 \).

Step 6: Determine the Domain of the Function and Its Derivative

The original function \( f(t) = 2.5t^2 + 6t \) is a polynomial, which has a domain of all real numbers, \( \mathbb{R} \). The derivative \( f'(t) = 5t + 6 \), being a linear polynomial, also has a domain of all real numbers, \( \mathbb{R} \).

Key Concepts

Definition of Derivative

The derivative of a function helps us understand how that function changes at a particular point. Imagine you are driving a car and you want to know how fast you are going at an exact moment — that's similar to finding a derivative.
For the function \( f(t) \), the derivative at a point \( t \) is found using a special formula:
\[ f'(t) = \lim_{{h \to 0}} \frac{f(t+h) - f(t)}{h}. \]
This formula calculates the rate of change of the function, or the slope of the tangent line at that point.
Here's a simple way to think about the steps involved in using this definition:

• Compute \( f(t+h) \), which is like evaluating the function at a nearby point.
• Subtract \( f(t) \) from \( f(t+h) \) to find the difference in function values.
• Divide this difference by \( h \) to find the average rate of change over that tiny interval.
• Take the limit as \( h \to 0 \). This captures the instantaneous rate of change, yielding the derivative \( f'(t) \).

When applying this to our function \( f(t) = 2.5t^2 + 6t \), we followed these steps precisely and found that the derivative is \( f'(t) = 5t + 6 \).

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables and coefficients. They can have terms like \( t^2 \), \( t \), and constants.
In our example, the function \( f(t) = 2.5t^2 + 6t \) is a polynomial because it consists of powers of the variable \( t \), with the highest power being 2.
Polynomials are generally expressed in the form:
\[ a_n t^n + a_{n-1} t^{n-1} + \ldots + a_1 t + a_0 \]
with each \( a_i \) being a real number coefficient.

• The degree of a polynomial is determined by its highest power. In this case, \( f(t) \) is a quadratic polynomial (degree 2) because of the \( t^2 \) term.
• Finding the derivative of a polynomial is straightforward because each term can be differentiated individually.
• The domain of a polynomial is all real numbers \( \mathbb{R} \), meaning any real number \( t \) can be plugged into the function.

These properties make polynomials common in calculus, helping to easily understand their behavior and graphs.

Calculus Concepts

Calculus is the mathematical study of change, and two main concepts are the derivative and the integral. In this exercise, we focus on derivatives, which give us the slope of a function at any specific point.
Let's summarize some essential calculus concepts involved here:

• Limits: Limits help us find the value that a function approaches as the input approaches a certain point. In the definition of the derivative, \( \lim_{{h \to 0}} \) gives the exact rate of change at a point.
• Continuity: A function is continuous if it doesn't have breaks or jumps. Polynomial functions like \( f(t) = 2.5t^2 + 6t \) are continuous everywhere because you can draw them without lifting your pen.
• Linear functions: The derivative we found, \( f'(t) = 5t + 6 \), is a linear function, which means it represents constant change.

Understanding these core calculus concepts helps in analyzing how functions change and behave, paving the way for more advanced studies in optimization, motion, and other dynamic scenarios.