Problem 41 - Calculus Single Variable

Question

Calculate the arc length of the graph of the given function over the given interval. (In these exercises, the functions have been contrived to permit a simplification of the radical in the arc length formula.) $$ f(x)=e^{x} / 2+e^{-x} / 2+7 \quad I=[0,1] $$

Step-by-Step Solution

Answer

The arc length is \( \frac{1}{2} \left(e - \frac{1}{e}\right) \).

Step 1: Recall the Arc Length Formula

The formula for the arc length of the curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]For the function \( f(x) = \frac{e^x}{2} + \frac{e^{-x}}{2} + 7 \) over the interval \( [0,1] \), we will first need to find the derivative \( f'(x) \).

Step 2: Differentiate the Function

Differentiate the function \( f(x) \) to find \( f'(x) \):\[f'(x) = \frac{d}{dx}\left(\frac{e^x}{2} + \frac{e^{-x}}{2} + 7\right)\]Using basic differentiation rules:\[f'(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}\]

Step 3: Compute \((f'(x))^2\)

We need to calculate \((f'(x))^2\):\[(f'(x))^2 = \left(\frac{1}{2}e^x - \frac{1}{2}e^{-x}\right)^2\]Expanding this expression:\[= \left(\frac{1}{2}e^x\right)^2 - 2 \cdot \frac{1}{2}e^x \cdot \frac{1}{2}e^{-x} + \left(\frac{1}{2}e^{-x}\right)^2\]\[= \frac{1}{4}e^{2x} - \frac{1}{4} + \frac{1}{4}e^{-2x}\]

Step 4: Simplify the Expression Inside the Integral

Simplify \(1 + (f'(x))^2\):\[1 + (f'(x))^2 = 1 + \frac{1}{4}e^{2x} - \frac{1}{4} + \frac{1}{4}e^{-2x}\]Combine the terms:\[= \frac{3}{4} + \frac{1}{4}e^{2x} + \frac{1}{4}e^{-2x}\]Recognize this as a perfect square:\[= \left(\frac{1}{2}e^x + \frac{1}{2}e^{-x}\right)^2\]

Step 5: Insert into Arc Length Integral and Evaluate

Substitute into the arc length formula:\[L = \int_{0}^{1} \sqrt{\left(\frac{1}{2}e^x + \frac{1}{2}e^{-x}\right)^2} \, dx\]\( \sqrt{a^2} = |a| \), and since \( e^x \) is positive,\[L = \int_{0}^{1} \left(\frac{1}{2}e^x + \frac{1}{2}e^{-x}\right) \, dx\]Evaluate the integral:\[L = \left[ \frac{1}{2} \left(e^x - e^{-x}\right) \right]_{0}^{1}\]\[L = \frac{1}{2} \left((e^1 - e^{-1}) - (1 - 1)\right)\]\[L = \frac{1}{2} (e - \frac{1}{e})\]

Step 6: Final Answer

The arc length \( L \) of \( f(x) \) from \( x = 0 \) to \( x = 1 \) is:\[L = \frac{1}{2} \left(e - \frac{1}{e}\right)\]

Key Concepts

Differentiation

Differentiation is a mathematical process that determines how a function changes as its input changes. For any curve represented by a function \( y = f(x) \), differentiation helps us find the slope or rate of change of the function at any given point. This is crucial for calculating the arc length of a curve.

When we differentiate our specific function \( f(x) = \frac{e^x}{2} + \frac{e^{-x}}{2} + 7 \), we essentially apply rules of differentiation to each component. Here's why each part is differentiated as it is:

• The derivative of \( \frac{e^x}{2} \) is \( \frac{1}{2} e^x \) because the derivative of \( e^x \) is \( e^x \), and the constant multiplier comes along for the ride.
• Similarly, for \( \frac{e^{-x}}{2} \), the derivative is \( -\frac{1}{2} e^{-x} \). The additional negative sign arises because of the chain rule, considering the exponent is \(-x\).
• The constant 7 just becomes 0 when differentiated, since constants don't change.

With these rules, we find that \( f'(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x} \). Differentiation sets the stage for understanding how the shape of the curve contributes to its length.

Arc Length Formula

The arc length formula is used to calculate how long a path or curve is from one point to another. For a curve defined by a function \( y = f(x) \), the arc length \( L \) from \( x = a \) to \( x = b \) is given by the integral:\[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \; dx\]This formula emerges from the Pythagorean theorem, considering infinitely small segments of the curve. Each segment can be thought of as the hypotenuse of a right triangle where the other sides are small changes in x and y.

For our function \( f(x) \), after differentiating, we calculate \( \left(\frac{dy}{dx}\right)^2 \) to make the formula complete. Specifically, \( \left(\frac{1}{2}e^x - \frac{1}{2}e^{-x}\right)^2 \) simplifies and helps us express \( \left(\frac{dy}{dx}\right)^2 \) in a form that makes integration feasible. This manipulation ensures that the expression inside the square root can be simplified and integrated accurately. Recognizing perfect squares in expressions, as done here, significantly simplifies calculation.

Integration

Integration is the mathematical process that essentially "adds up" tiny pieces to find a whole. In the context of calculating arc length, integration takes the role of adding up infinitesimal segments of the curve over the interval \([0, 1]\). Each little piece, defined by our rearranged formula \[L = \int_{0}^{1} \left(\frac{1}{2}e^x + \frac{1}{2}e^{-x}\right) \, dx\]represents the linear path over the tiny change in x.

To solve the integral, we apply integral rules to each component. The integral of \( e^x \) is simply \( e^x \), and for \( e^{-x} \), the integral is \(-e^{-x}\), which accounts for the negative exponent. These rules allow us to evaluate the expression and find how far the curve stretches from \( x = 0 \) to \( x = 1 \).

Here, performing integration over the defined interval results in finding the total arc length, incorporating each element's contribution, and thereby providing us with the final answer: \[L = \frac{1}{2} \left(e - \frac{1}{e}\right)\]This result gives a precise measure of how much the curve "stretches" over the specified interval, combining both differentiation and integration seamlessly.