Problem 41 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(h(t)=\left(e^{-t}+e^{t}\right)^{5} \quad\) for \(-1 \leq t \leqq 1\)

Step-by-Step Solution

Answer

The function \( h(t) \) evaluated at \( t = -1 \) and \( t = 1 \) is approximately 286.56.

Step 1 - Identify the Function

The given function is \( h(t) = (e^{-t} + e^{t})^{5} \). It is defined for the interval \( -1 \leq t \leq 1 \).

Step 2 - Simplify the Exponential Terms

Recognize that \( e^{-t} \) can be written as \( \frac{1}{e^{t}} \). However, for the sake of differentiation later, we will keep them in the exponential form.

Step 3 - Differentiate the Function

Use the chain rule to differentiate the function. Let \( u = e^{-t} + e^{t} \), then \( h(t) = u^{5} \). Differentiate with respect to \( t \): \[ \frac{d}{dt} h(t) = \frac{d}{dt} (u^{5}) = 5u^{4} \frac{du}{dt} \] where \( \frac{du}{dt} = -e^{-t} + e^{t} \). Thus, \[ \frac{d}{dt} h(t) = 5(e^{-t} + e^{t})^{4}(-e^{-t} + e^{t}) \].

Step 4 - Evaluate at Specific Points

Evaluate the function and its derivative at the endpoints \( t = -1 \) and \( t = 1 \). For \( t = -1 \): \[ h(-1) = (e^{1} + e^{-1})^{5} \]. Calculate \( e^{1} \approx 2.718 \) and \( e^{-1} \approx 0.368 \), thus \( h(-1) \approx (2.718 + 0.368)^{5} \approx (3.086)^{5} \approx 286.56 \). For \( t = 1 \): \[ h(1) = (e^{-1} + e^{1})^{5} = h(-1) \] due to the symmetry, so \( h(1) \approx 286.56 \).

Key Concepts

Exponential Functions

Exponential functions play a key role in many areas of calculus, particularly when dealing with growth and decay problems. In our exercise, we have two exponential functions: \(e^{-t}\) and \(e^{t}\). The variable \(t\) is present in the exponent, which makes these functions unique. When handling exponential functions, remember that:

• \(e^{t}\) grows rapidly as \(t\) increases.
• \(e^{-t}\) decays as \(t\) increases.
• \(e^{a+b} = e^{a} \times e^{b}\)
• \(e^{0} = 1\)

To manage our task better, we keep the exponentials in their original form for differentiation purposes.

Calculus

Calculus provides us with the tools to explore changes and motion. When we talk about differentiation, we are concerned with finding the rate at which one quantity changes with respect to another. In our exercise, we use differentiation to find how the function \(h(t) = (e^{-t} + e^{t})^{5}\) changes as \(t\) varies within \([-1, 1]\). Calculus allows us to:

• Function manipulation.
• Evaluate limits.
• Differentiate exponential, logarithmic, and trigonometric functions.

It’s a powerful way to understand the behavior of functions across different points.

Derivative Evaluation

In calculus, evaluating the derivative means finding the rate of change of a function. For the given function \(h(t)\), we determine its derivative with respect to \(t\). By applying the chain rule, we find the derivative of the inner function first and then the outer function. For our specific function, we let
\(u = e^{-t} + e^{t}\)
Then calculate \(\frac{du}{dt}\) which results in
\(\frac{du}{dt} = -e^{-t} + e^{t}\).
The complete derivative of our function \(h(t) = u^{5}\) is then derived as:
\( \frac{d}{dt} (u^{5}) = 5u^{4} \cdot \frac{du}{dt}\). Replacing \(u\) back, the full expression becomes:
\( \frac{d}{dt} h(t) = 5(e^{-t} + e^{t})^{4}(-e^{-t} + e^{t})\).
This shows how we handle it step by step for clarity.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When one function is composed inside another, we use the chain rule to find the derivative of the full expression. Here's how we applied the chain rule in our function:

• Let \(u = e^{-t} + e^{t}\).
• Differentiate the outer function \(u^{5}\) giving \(5u^{4}\).
• Multiply by the derivative of the inner function \(u\) which equals \(-e^{-t} + e^{t}\).

By systematically applying the chain rule, we efficiently found the derivative of the complete function.

Function Evaluation

To understand the behavior of our function \(h(t)\), we need to evaluate it at specific points. This typically involves plugging values into the function and observing the output. For our function, we evaluated it at the endpoints \(t = -1\) and \(t = 1\):
At \(t = -1\):
\(h(-1) = (e^{1} + e^{-1})^{5}\)
Approximating with \(e^{1} \approx 2.718\) and \(e^{-1} \approx 0.368\),
\(h(-1) \approx (2.718 + 0.368)^{5} \approx 286.56\).
Similarly, for \(t = 1\):
\(h(1) = (e^{-1} + e^{1})^{5} = h(-1)\) due to symmetry, we get the same output as
\(h(-1) \approx 286.56\). This symmetrical evaluation helps confirm our understanding.