Problem 8 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=\left(t^{2}+3\right) e^{t} $$

Step-by-Step Solution

Answer

Derivative is \( \frac{dy}{dt} = e^t(2t + t^2 + 3) \).

Step 1: Identify the Rule to Use

Given the function \( y = (t^2 + 3)e^t \), we notice it's a product of two functions: \( u(t) = t^2 + 3 \) and \( v(t) = e^t \). To differentiate these, we apply the Product Rule, which states \( \frac{dy}{dt} = u'v + uv' \).

Step 2: Differentiate Each Function Sub-component

Differentiate \( u(t) = t^2 + 3 \) to get \( u'(t) = 2t \). For \( v(t) = e^t \), since the derivative of \( e^t \) is itself, \( v'(t) = e^t \).

Step 3: Apply the Product Rule

Substitute the derivatives into the Product Rule formula: \( \frac{dy}{dt} = (2t)e^t + (t^2 + 3)e^t \).

Step 4: Simplify the Expression

Combine the terms: \( \frac{dy}{dt} = 2te^t + t^2e^t + 3e^t \). Group them by the common factor \( e^t \) to get \( \frac{dy}{dt} = e^t(2t + t^2 + 3) \).

Key Concepts

Product Rule

When finding derivatives, especially with functions that are products of two simpler functions, like in our exercise, we use the Product Rule. The Product Rule is a vital differentiation technique used in calculus. It helps us find the derivative of two multiplied functions. If you have two functions, say \( u(t) \) and \( v(t) \), their derivative is given by:

• \( \frac{d}{dt}(uv) = u'v + uv' \)

Here, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively. So, we first take the derivative of the first function and multiply it by the second, and then take the derivative of the second function and multiply it by the first. This rule is essential when your function isn’t a simple polynomial, but instead a product like \( (t^2 + 3)e^t \), that combines a polynomial and an exponential function. Learning to apply the Product Rule correctly enables you to tackle a wide range of problems in differentiation.

Differentiation

Differentiation is a fundamental process in calculus used to understand how a function changes at any given point. It involves finding the derivative, which gives the slope or rate of change of a function. In our example, \( y = (t^2 + 3)e^t \), differentiation helps us determine how this function behaves.

• Identify individual components of the function that need differentiating.
• Apply different rules for different forms, like the Product Rule for products and chain rule for compositions.

In simple terms, when you differentiate, you look at each piece of your function, determine how it changes, and then put those changes together in a meaningful way. For \( u(t) = t^2 + 3 \), its derivative is \( 2t \). Meanwhile, \( v(t) = e^t \), being an exponential function, differentiates to \( e^t \). Differentiation thus provides us with powerful tools to analyze dynamic systems and changes.

Exponential Function

Exponential functions are a unique and fascinating family of functions. They involve constants raised to a variable power, and they have special properties that simplify differentiation and other analyses. Specifically, \( e^t \), or \( e \), Euler's number, is the base of the natural logarithm and is approximately equal to 2.71828.

• The derivative of \( e^t \) is \( e^t \) itself, making it exceptionally straightforward to work with.
• Exponential functions grow at an increasing rate, which is why they appear in population models, finance, and other fields.

In our problem, we see \( e^t \) used as part of the function \( y = (t^2 + 3)e^t \). This component, due to its nature, keeps its form when differentiated. Such behavior significantly simplifies calculus, especially in complex functions, making exponential functions a crucial aspect of mathematical modeling and problem-solving. Recognizing exponential forms and being comfortable differentiating them is key to mastering calculus.