91Ó°ÊÓ

The derivative is a core concept in calculus, used to measure how a function changes as its input changes. Think of it as a tool to find the rate of change or slope of a curve at any given point. In the context of our exercise, we find the derivative of the exponential functions given:

• \(f(t) = e^{-t^2}\) to get \(f'(t) = -2te^{-t^2}\)
• \(f(t) = e^{-t^3}\) to get \(f'(t) = -3t^2e^{-t^3}\)

This step is crucial because the derivative is needed for calculating the relative rate of change. It tells us how fast the function values are changing at any value of \( t \). A positive derivative means the function is increasing, whereas a negative derivative indicates the function is decreasing.

The chain rule is a fundamental technique in calculus used for finding the derivative of a composite function. A composite function is when one function is plugged into another. If you have a function \( f(g(x)) \), where \( f \) and \( g \) are functions, the chain rule tells you how to take its derivative. It states that the derivative is: \[ f'(g(x)) \cdot g'(x) \]In our exercise, the chain rule is applied to both exponential functions, because they are composed of an exponential function \( e^u \) and a polynomial inside, such as \( -t^2 \) or \( -t^3 \). - For \( f(t) = e^{-t^2} \), setting \( u = -t^2 \) gives \( u'(t) = -2t \).- Similarly, for \( f(t) = e^{-t^3} \), setting \( u = -t^3 \) results in \( u'(t) = -3t^2 \).This chain rule application simplifies tackling complex derivatives by breaking them into manageable parts.

An exponential function is a mathematical function of the form \( f(x) = e^x \). In our case, we deal with \( e^{-t^2} \) and \( e^{-t^3} \), which are examples of exponential decay functions. These functions are characterized by their rapid increase or decrease, depending on the sign of the exponent. Exponential functions have a unique property: their rate of change is proportional to their current value. This is significant when calculating relative rates because, essentially, the exponential still remains after its derivative application. For example:

• \( f'(t) = -2te^{-t^2} \) retains the \( e^{-t^2} \) part of the original function

Understanding this helps to make sense of the formulas derived during the solution to the exercise.

Evaluating a function at a point means substituting a specific value into the function to determine its output at that particular point. This is the final step in finding the particular rate of change for a specific input in our exercise.Once we have the expressions for the relative rate of change,

• \( -2t \) for \( e^{-t^2} \)
• \( -3t^2 \) for \( e^{-t^3} \)

we substitute the given values of \( t \) to find the specific rates of change at those points. - For \( t = 10 \) in the first function, we substitute 10 for \( t \), giving \( -2(10) = -20 \). - Similarly, for \( t = 5 \) in the second function, substitute 5 to find \( -3(5)^2 = -75 \). This step ties all the previous steps together by applying the formula to get a tangible, real-world number for the rate of change.