Problem 71 - Calculus for Biology and Medicine

Question

Show that if $$ f(x)=\frac{e^{x}+e^{-x}}{2} $$ then the length of the curve \(f(x)\) between \(x=0\) and \(x=a\) for any \(a>0\) is given by \(f^{\prime}(a)\).

Step-by-Step Solution

Answer

The curve length from \( x=0 \) to \( x=a \) equals \( f'(a) \).

Step 1: Understand the Problem

We want to show that the length of the curve of the function \( f(x) = \frac{e^x + e^{-x}}{2} \) from \( x = 0 \) to \( x = a \) is equal to the derivative of function \( f(x) \) evaluated at \( x = a \).

Step 2: Recall the Arc Length Formula

The arc length of a curve \( y = f(x) \) between \( x = a \) and \( x = b \) is given by \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \] In our case, we integrate from \( x = 0 \) to \( x = a \).

Step 3: Find the Derivative of \( f(x) \)

First, compute the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} \]

Step 4: Calculate \((f'(x))^2\)

Compute \[ (f'(x))^2 = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^x - e^{-x})^2}{4} \]Simplifying further: \[ = \frac{e^{2x} - 2 + e^{-2x}}{4} \]

Step 5: Simplify the Expression Inside the Square Root

Now include 1 and simplify inside the square root for the arc length formula:\[ 1 + (f'(x))^2 = 1 + \frac{e^{2x} - 2 + e^{-2x}}{4} \]Simplifying gives: \[ = \frac{4 + e^{2x} - 2 + e^{-2x}}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4} \]

Step 6: Recognize as a Perfect Square

Notice that \[ \frac{e^{2x} + 2 + e^{-2x}}{4} = \left( \frac{e^x + e^{-x}}{2} \right)^2 \]This tells us that\[ \sqrt{1 + (f'(x))^2} = \frac{e^x + e^{-x}}{2} = f(x) \]

Step 7: Set Up the Integral for Arc Length

Using the simplified expression, the arc length becomes:\[ L = \int_{0}^{a} f(x) \, dx = \int_{0}^{a} \frac{e^x + e^{-x}}{2} \, dx \]

Step 8: Evaluate the Integral

Evaluate the integral for the function \[ \int \frac{e^x + e^{-x}}{2} \, dx = \frac{1}{2} \left( \int e^x \, dx + \int e^{-x} \, dx \right) \]The integrals become:\[ \frac{1}{2}(e^x - e^{-x}) + C \]So we calculate:\[ \left[ \frac{1}{2}(e^x - e^{-x}) \right]_0^a \]Calculating from 0 to \( a \):\[ = \left( \frac{e^a - e^{-a}}{2} \right) - \left( \frac{e^0 - e^{0}}{2} \right) = \frac{e^a - e^{-a}}{2} \]

Step 9: Relate Back to \( f'(a) \)

Notice that :\[ \frac{e^a - e^{-a}}{2} = f'(a) \] Thus, the arc length from \( 0 \) to \( a \) is indeed \( f'(a) \).

Step 10: Conclusion: Verify the Final Result

We have verified that the length of the curve \( f(x) \) from \( x=0 \) to \( x=a \) is given by \( f'(a) \). This matches the problem's requirement.

Key Concepts

Derivative

A derivative measures how a function changes as its input changes. Think of it as the function's "slope" at any given point. Finding the derivative allows us to understand how the shape of a curve, like our function \(f(x) = \frac{e^x + e^{-x}}{2}\), behaves between different intervals.

For our specific function, we calculate the derivative in a straightforward manner by applying rules of differentiation you may already know:

• The derivative of \(e^x\) is \(e^x\).
• The derivative of \(e^{-x}\) is \(-e^{-x}\).

Thus, the derivative of \(f(x)\) becomes:\[f'(x) = \frac{e^x - e^{-x}}{2}\]This formula tells us how quickly the value of \(f(x)\) is changing for given values of \(x\). By finding \(f'(x)\), we also lay the groundwork for comparing it to the arc length of the function.

Integral Calculus

Integral calculus allows us to find the total effect, or the cumulative sum, of changes over a certain interval. It is the mathematical principle that helps us to calculate things like the area under a curve, or in this case, the arc length of a curve from one point to another.

When we want to determine the length of a curve described by \(y = f(x)\) over an interval \([a, b]\), we use the arc length formula:\[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\]In our problem, this becomes evaluating the integral from \(x = 0\) to \(x = a\).

Our derivative \(f'(x)\), was previously calculated as \(\frac{e^x - e^{-x}}{2}\), so the square of this expression helps us determine what to integrate. After some simplification, the expression inside the integral turns out to be a perfect square that matches our original function, \(f(x)\). Because of this neat trick, we can easily integrate over the interval to find the length of the curve.

Exponential Functions

Exponential functions have the form \(e^{x}\) or \(e^{-x}\), where \(e\) is the base of natural logarithms. These functions exhibit rapid growth or decay patterns, depending on whether the exponent is positive or negative.

The function given in the exercise, \(f(x) = \frac{e^x + e^{-x}}{2}\), involves both exponential growth and decay. It is a form of a hyperbolic cosine function \(\cosh(x)\).

• \(e^x\) contributes to exponential growth.
• \(e^{-x}\) introduces exponential decay.

Together, they create a function that is symmetric around the vertical axis, which has useful properties for curve and arc calculations. By integrating or differentiating \(f(x)\), we engage with these exponential parts to analyze its curve effectively. Exponential functions like this one often crop up in problems related to growth models and waveforms, making them a fundamental part of calculus and a tool you will use often.