Problem 36 - Calculus an Applied Approach

Question

Use the limit definition to find the derivative of the function. $$ f(t)=t^{3}+t^{2} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(t) = t^{3} + t^{2}\) is \(f'(t) = 3t^{2} + 2t\)

Step 1: Write the Function with \(h\) term

We write the function with a \(h\) term added to \(x\),\[f(x+h) = (x+h)^{3} + (x+h)^{2}\]

Step 2: Expand and Simplify the Function

Now we expand and simplify this function. We get,\[f(x+h) = x^{3} + 3x^{2}h + 3xh^{2} + h^{3} + x^{2} + 2xh + h^{2}\]

Step 3: Substitute into the Limit Definition

Substitute \(f(x+h)\) and \(f(x)\) into the limit definition. We get,\[f'(x) = \lim_{h \to 0} \frac{(x^{3} + 3x^{2}h + 3xh^{2} + h^{3} + x^{2} + 2xh + h^{2}) - (x^{3} + x^{2})}{h}\]which simplifies to,\[f'(x) = \lim_{h \to 0} \frac{3x^{2}h + 3xh^{2} + h^{3} + 2xh + h^{2}}{h}\]

Step 4: Simplify the Limit Expression

Now we divide the entire numerator by h,\[f'(x) = \lim_{h \to 0} 3x^{2} + 3xh + h^{2} + 2x + h\]

Step 5: Take the Limit

Now take the limit as \(h\) approaches 0, \[f'(x) = 3x^{2} + 0 + 0 + 2x + 0\]which simplifies to,\[f'(x) = 3x^{2} + 2x\]

Key Concepts

Limit Definition

The limit definition of a derivative is a fundamental concept in calculus. It allows us to find the derivative of a function at any given point. Think of it as a mathematical way to find the slope of the tangent line to a curve at a specific point. This is done by considering a small change in the input value of the function, often represented as \(h\).

• The formula for the derivative \(f'(x)\) using the limit definition is:

\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]Here, \(f(x+h)\) represents the function after a small change \(h\) is applied.

• By evaluating this limit, we determine how the function changes momentarily as the input changes slightly.
• When \(h\) approaches zero, the calculation reveals the function's instantaneous rate of change.

In the original exercise, we used this definition to find the derivative of a polynomial function, demonstrating the application of these principles.

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number powers. They can consist of one or more terms combined together. Familiar examples include linear functions, quadratic functions, and cubic functions, like the one in our exercise: \(f(t) = t^3 + t^2\).

• Polynomials are represented by combining terms that are products of constants and variables raised to any non-negative integer power.
• They are continuous and very smooth, making them ideal candidates for differentiation.

The exercise involves a cubic and a quadratic term, showing how variables can combine to form more complex functions.

• The power and simplicity of polynomial functions make them essential in calculus as they are simple to differentiate.
• Understanding the structure of a polynomial helps in applying differentiation techniques effectively.

Differentiation Techniques

Differentiation is the process of finding a derivative, which provides valuable information about a function's behavior. In this exercise, we saw the use of the limit definition to differentiate a polynomial function. Apart from this, several rules and techniques simplify the differentiation process:

• Power Rule: For a term \(x^n\), its derivative is given by \(nx^{n-1}\). This rule applies directly to each term in a polynomial.
• Sum Rule: The derivative of a sum of functions is simply the sum of their derivatives.

In the given problem, these rules help simplify the derivative once it is set up using the limit definition.

• First, expand the function using algebra; this step allows for applying the rule conveniently.
• Then, simplify and evaluate the limit for \(h\) approaching zero to determine the function's slope.

By understanding and applying these techniques, we can uncover important characteristics of polynomial functions efficiently.