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Chapter 5: Problem 12

Compute the are length exactly. $$y=e^{x / 2}+e^{-x / 2},-1 \leq x \leq 1$$

Short Answer

Expert verified
The exact length of the arc is \(2\sqrt{\frac{3}{2}}\)

Step by step solution

01

Calculate the derivative

Firstly, we calculate the derivative of the function y with respect to x, which is \(f'(x)= \frac{1}{2}e^{x/2}-\frac{1}{2}e^{-x/2}\).
02

Square the derivative

Next, we square the derivative and simplify the result \([f'(x)]^2 = \left(\frac{1}{2}e^{x/2} - \frac{1}{2}e^{-x/2}\right)^2 = \frac{1}{4} + \frac{2}{4}- \frac{1}{4}= \frac{1}{2}\).
03

Use the formula to compute the arc length

We now substitute the results from step 1 and 2 into the formula for the length of an arc: \[L=\int_{-1}^{1}\sqrt{1 + (1/2)} dx = \int_{-1}^{1} \sqrt{3/2} dx\]. This simplifies to \( L = \sqrt{ \frac{3}{2} } \left[ x \right]_{-1}^{1} = 2\sqrt{\frac{3}{2}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
In calculus, a derivative represents how a function changes as its input changes. For any given function, the derivative can be thought of as the slope of the tangent line to the curve at any point. In the context of arc length, derivatives are essential since they measure the rate of change of the curve, which helps in determining the length.

For the exponential function in the exercise, where \( y = e^{x/2} + e^{-x/2} \), the derivative was calculated as: \( f'(x) = \frac{1}{2}e^{x/2} - \frac{1}{2}e^{-x/2} \). This shows how each component's rate of change contributes differently. Derivatives allow us to understand these variations and how they influence the overall geometry of the curve.
Exploring the Arc Length Formula
The arc length formula is a powerful tool in calculus, used to determine the distance along a curve from one point to another. It's rather intuitive: you take the curve, break it down into infinitesimally small line segments, and sum them up. The length \( L \) of a curve from \( x = a \) to \( x = b \) is computed as:
  • \( L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \)

In our specific exercise, we used the derivative to find \([f'(x)]^2\) which simplifies the equation inside the square root of the formula. Once we had \(1 + \frac{1}{2} = \frac{3}{2}\), the integral calculus was straightforward, leading us to the total arc length of the curve within the specified bounds.
Diving into Exponential Functions
Exponential functions have the form \(y = a^{x}\), where \(a\) is a constant and \(x\) is the variable. These functions are known for their distinctive property of continuous growth or decay. In our exercise, the function \( y = e^{x/2} + e^{-x/2} \) combines these exponential terms to showcase unique behavior over the interval \([-1, 1]\).

Exponential functions like \( e^{x} \) grow rapidly because they repeatedly multiply a constant factor. Conversely, \( e^{-x} \) models exponential decay, as it successively divides by this factor. Understanding this combination is crucial when calculating derivatives and further simplifying to find the arc length, as each term reacts differently to changes in \( x \).

This symmetry also directly influences the slope, or derivative, as seen in the calculation, impacting the resultant arc length significantly.

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Most popular questions from this chapter

Compute the are length \(L_{1}\) of the curve and the length \(L_{2}\) of the secant line connecting the endpoints of the curve. Compute the ratio \(L_{2} / L_{1}\); the closer this number is to 1 , the straighter the curve is. $$y=\sin x,-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}$$

The base of a solid \(V\) is the region bounded by \(y=e^{-2 x}\) \(y=0, x=0\) and \(x=\ln 5 .\) Find the volume if \(V\) has (a) square cross sections and (b) semicircular cross sections perpendicular to the \(x\) -axis.

A football punt follows the path \(y=\frac{1}{15} x(60-x)\) yards. Sketch a graph. How far did the punt go horizontally? How high did it go? Compute the arc length. If the ball was in the air for 4 seconds, what was the ball's average velocity?

Briefly explain what it means when your CAS returns \(\int f(x) d x\) when asked to evaluate the indefinite integral \(\int f(x) d x\)

Use the given table of values to estimate the volume of the solid formed by revolving \(y=f(x), 0 \leq x \leq 3,\) about the \(x\) -axis. $$\begin{array}{|l|l|l|l|l|l|l|l|} \hline x & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ \hline f(x) & 2.0 & 1.2 & 0.9 & 0.4 & 1.0 & 1.4 & 1.6 \\ \hline \end{array}$$
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