Problem 13 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ f(t)=\frac{5}{t}+\frac{6}{t^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(t) = -\frac{5}{t^2} - \frac{12}{t^3} \).

Step 1: Identify the Function Form

The function given is \( f(t) = \frac{5}{t} + \frac{6}{t^2} \). Note that both terms involve a variable in the denominator, which can alternatively be written with negative exponents: \( 5t^{-1} + 6t^{-2} \).

Step 2: Apply the Power Rule

For each term, use the power rule for derivatives: if \( g(t) = kt^n \), then \( g'(t) = nkt^{n-1} \). Start with the first term \( 5t^{-1} \): the derivative is \( -1 \cdot 5 \cdot t^{-2} = -5t^{-2} \).

Step 3: Differentiate the Second Term

Now differentiate \( 6t^{-2} \) using the power rule: \( -2 \cdot 6 \cdot t^{-3} = -12t^{-3} \).

Step 4: Write the Final Derivative Expression

Combine the derivatives from Step 2 and Step 3: the derivative of the entire function is \( f'(t) = -5t^{-2} + (-12t^{-3}) = -5t^{-2} - 12t^{-3} \).

Step 5: Simplify the Derivative Expression

The expression \( -5t^{-2} - 12t^{-3} \) can be rewritten in fraction form: \( -\frac{5}{t^2} - \frac{12}{t^3} \). This is the simplified derivative.

Key Concepts

Derivative

In calculus, the derivative is a fundamental concept that represents the rate at which a function changes. It gives us a way to describe the behavior of a function's graph. When you differentiate a function, you're essentially figuring out how the function's output changes for small changes in input. This process helps in understanding whether the function is increasing or decreasing at a specific point.

Consider the function given in the exercise:

• The function is: \( f(t) = \frac{5}{t} + \frac{6}{t^2} \)
• The goal is to find \( f'(t) \), which is the derivative of this function.

Understanding derivatives is crucial for solving problems in physics, engineering, and economics, wherever a change in one quantity depends on another. This derivative tells us how the function's values change as the variable \( t \) varies.

Power Rule

The power rule is one of the simplest and most important tools for finding derivatives. It states that if you have a function of the form \( g(t) = at^n \), then its derivative is \( g'(t) = nat^{n-1} \). This rule makes it easy to calculate the derivative of polynomial functions, where each term is a power of the variable.

In the exercise, each term of the function \( f(t) \) can be handled using the power rule:

• For the term \( 5t^{-1} \), apply the power rule: - The derivative is \( -1 \times 5 \times t^{-2} = -5t^{-2} \)
• Similarly for \( 6t^{-2} \): - The derivative is \( -2 \times 6 \times t^{-3} = -12t^{-3} \)

The power rule is especially handy because you can directly apply it to each term in the function separately, simplifying the process significantly.

Negative Exponent

Negative exponents often appear in functions involving division by a variable, as seen in the exercise. Recall that a negative exponent indicates a reciprocal. So, \( t^{-1} \) is essentially \( \frac{1}{t} \).

• This is why \( \frac{5}{t} \) can be written as \( 5t^{-1} \).
• Similarly, \( \frac{6}{t^2} \) is \( 6t^{-2} \).

Changing the terms of the function into forms with negative exponents is an important step in making it easier to differentiate using the power rule. By recognizing when and how to rewrite terms with negative exponents, you turn potentially tricky expressions into manageable ones.

Differentiation

Differentiation is the process of finding the derivative of a function. It's a key operation in calculus and is used extensively in many scientific and engineering applications. Differentiation involves applying rules, such as the power rule, to determine how a function changes based on its input variables.

• In the context of the exercise, differentiation allows us to transform the function \( f(t) \) into \( f'(t) \), which describes its rate of change.
• The process involved using negative exponents and applying the power rule, finally yielding \( -5t^{-2} - 12t^{-3} \) as the derivative.

This derivative tells us important information about the behavior of the original function over its domain. Differentiation is a fundamental concept that serves as a building block for more advanced topics like integration and differential equations.