Problem 3 - Applied Calculus

Question

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(y=5 t^{2}+4 e^{t}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = 10t + 4e^t \).

Step 1: Understand the Problem

We are given a function \(y = 5t^2 + 4e^t\) and asked to differentiate it with respect to \(t\). To do this, we will use the rules for differentiating power functions and exponential functions.

Step 2: Differentiate the Power Term

The first term is \(5t^2\). To differentiate \(5t^2\) with respect to \(t\), use the power rule which states \(\frac{d}{dt}(t^n) = nt^{n-1}\). Applying this, we get \( \frac{d}{dt}(5t^2) = 5 \cdot 2t^{2-1} = 10t \).

Step 3: Differentiate the Exponential Term

The second term is \(4e^t\). The derivative of \(e^t\) with respect to \(t\) is simply \(e^t\). Therefore, differentiating \(4e^t\) gives us \( \frac{d}{dt}(4e^t) = 4 \cdot e^t = 4e^t \).

Step 4: Combine the Results

Now that we have differentiated both terms, we combine them to find the derivative of the entire function. Thus, \( \frac{dy}{dt} = 10t + 4e^t \).

Key Concepts

Power Rule

The Power Rule is a fundamental tool in differentiation that is particularly useful for handling polynomial expressions. When you have a term like \( t^n \), the Power Rule allows you to find its derivative easily. The rule states that the derivative of \( t^n \) is \( nt^{n-1} \). This means you multiply the exponent by the coefficient and decrease the exponent by one.

For example, differentiating \( 5t^2 \) results in:

• Multiply the exponent, 2, by the coefficient, 5, to get 10.
• Decrease the exponent by one to get \( t^{2-1} = t^1 = t \).
• Thus, the derivative \( \frac{d}{dt}(5t^2) = 10t \).

Using the Power Rule simplifies finding derivatives of terms with whole number exponents, helping solve problems efficiently and accurately.

Exponential Function

Exponential functions have the form \( a e^{bt} \), where \( a \) and \( b \) are constants and \( e \) is Euler's number, approximately 2.718. These functions grow increasingly quickly as the variable increases, making them important in various scientific and financial models. When it comes to differentiation, exponential functions have a unique property: the derivative of \( e^t \) with respect to \( t \) is \( e^t \) itself.

For an exponential function like \( 4e^t \), the differentiation is straightforward:

• The coefficient 4 remains unaffected.
• The derivative of \( e^t \) is \( e^t \), so the result is \( 4e^t \).

This property simplifies working with exponential functions, as their rate of change remains consistent with the function's original form, making calculus problems involving these functions much easier to handle.

Calculus

Calculus is the branch of mathematics that focuses on change. It's divided mainly into two categories: differentiation and integration. Differentiation deals with finding how a function changes at any point, while integration focuses on the accumulation of quantities. Differentiation, which we explore today, is a process that allows us to find the derivative of a function.

In problems like differentiating \( y = 5t^2 + 4e^t \), calculus helps us discover how \( y \) changes as \( t \) changes. This is done by applying differentiation rules such as the Power Rule and handling each term according to its form.

The basics of differentiation involve:

• Identifying the type of function or term, like polynomial or exponential.
• Applying the relevant rules (e.g., power or exponential rules).
• Summing up the derivatives of individual terms to find the derivative of the entire expression.

Through calculus, particularly differentiation, we gain insights into dynamic behaviors and can model real-world situations effectively.