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Chapter 7: Problem 33

Length of a Cable An electric cable is hung between two towers that are 40 meters apart (see figure). The cable takes the shape of a catenary whose equation is \(y=10\left(e^{x / 20}+e^{-x / 20}\right), \quad-20 \leq x \leq 20\) where \(x\) and \(y\) are measured in meters. Find the arc length of the cable between the two towers.

Short Answer

Expert verified
The arc length of the cable is \(800 \cdot 2 \sinh(1) \approx 1074.89\) meters.

Step by step solution

01

Understand the Arc Length Formula

Recall the general formula for the arc length of a curve defined by \(y=f(x)\) from \(x=a\) to \(x=b\) is \(L=\int_{a}^{b}\sqrt{1+\left(f'(x)\right)^{2}} dx\). Apply this formula to this particular case.
02

Differentiate the function

Differentiate the given function with respect to \(x\). The derivative of the function \(f(x) = 10\left(e^{x / 20}+e^{-x / 20}\right)\) is \(f'(x) = \frac{1}{2}\left(e^{x / 20} - e^{-x / 20}\right)\).
03

Substitute the derivative into the Arc Length Formula

Substitute \(f'(x)\) into the arc length formula. This will give: \(L=\int_{-20}^{20}\sqrt{1+\left(\frac{1}{2}\left(e^{x / 20} - e^{-x / 20}\right)\right)^{2}} dx\).
04

Simplify the Integral

Simplify the integral to turn it into a form that can be easily integrated. This gives us: \(L = 20 \int_{-20}^{20} \sqrt{\cosh^2\left(\frac{x}{20}\right)}dx \). The integral can be simplified further by using the fact that \(\sqrt{\cosh^2(x)} = \cosh(x)\), resulting in: \(L = 20 \int_{-20}^{20} \cosh\left(\frac{x}{20}\right)dx \).
05

Evaluate the Integral

Evaluate this integral from -20 to 20. Since the antiderivative of \(\cosh(x)\) is \(\sinh(x)\), you will find \(L = 20 \cdot 20 (\sinh(20 / 20) - \sinh(-20 / 20)) = 800 (\sinh(1) - \sinh(-1)) = 800 \cdot 2 \sinh(1)\).

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that involves the study of rates of change and quantities under the action of accumulation. It is divided into two related fields: differential calculus and integral calculus. Differential calculus is primarily concerned with the concept of the derivative, which is used to describe the rate at which quantities change. Integral calculus, on the other hand, deals with the accumulation of quantities, such as areas under curves, and is centered around the concept of the integral.

In our problem, we use calculus to compute the length of a cable hanging between two towers. By employing the principles of integral calculus, we can calculate the total arc length of a catenary curve, which represents the shape of the hanging cable.
Integral Calculus
Integral calculus is all about adding up an infinite number of infinitesimally small amounts to determine the total accumulation of quantity. One common application is to find the area under a curve by integrating the function that defines the curve with respect to its variable.

For example, in the arc length problem, we use integral calculus to sum up the lengths of infinitely small segments of the cable over the interval from one tower to the other. This is accomplished by setting up an integral with limits corresponding to the positions of the towers and an integrand that gives the length of a differential (infinitesimally small) piece of the curve.
Arc Length Formula
The arc length formula is a powerful tool from integral calculus used to calculate the length of a smooth curve between two points. The formula for finding the arc length, L, of a curve defined by a function y=f(x) from x=a to x=b is:
\[L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx\]

Here, f'(x) represents the derivative of function f(x) with respect to x, which provides the slope at any given point along the curve. By squaring this derivative and adding 1, we get a value inside the square root that, when integrated across the interval from a to b, gives us the total length of the curve.
Differentiation
Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes at any given point. In essence, differentiation provides us with the slope of the tangent line to the function at any point on its curve.

In our case, to apply the arc length formula, we first need to differentiate the catenary function y=10(e^{x/20}+e^{-x/20}) with respect to x to obtain f'(x). The derivative, f'(x), then informs us about how steep the cable is at any point between the two towers. This information is crucial as it feeds directly into the arc length formula enabling us to determine the length of the curved cable.

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Most popular questions from this chapter

Think About It Match each integral with the solid whose volume it represents and give the dimensions of each solid. (a) Right circular cone (b) Torus (c) Sphere (d) Right circular cylinder (e) Ellipsoid $$\begin{array}{l}{\text { (i) } 2 \pi \int_{0}^{r} h x d x} \\ {\text { (ii) } 2 \pi \int_{0}^{t} h x\left(1-\frac{x}{r}\right) d x}\end{array}$$ $$\begin{array}{l}{\text { (iii) } 2 \pi \int_{0}^{r} 2 x \sqrt{r^{2}-x^{2}} d x} \\ {\text { (iv) } 2 \pi \int_{0}^{b} 2 a x \sqrt{1-\frac{x^{2}}{b^{2}}} d x} \\ {\text { (v) } 2 \pi \int_{-r}^{r}(R-x)\left(2\sqrt{r^{2}-x^{2}}\right) d x}\end{array}$$

A torus is formed by revolving the graph of \((x-1)^{2}+y^{2}=1\) about the \(y\) -axis. Find the surface area of the torus.

Finding Arc Length In Exercises \(21-30\) , (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) write a definite integral that represents the that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the arc length.arc length of the curve over the indicated interval and observe $$y=\frac{1}{x}, \quad[1,3]$$

In Exercises 43-46, the integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$2 \pi \int_{0}^{6}(y+2) \sqrt{6-y} d y$$

Lateral Surface Area of a Cone A right circular cone is generated by revolving the region bounded by \(y=h x / r\) , \(y=h,\) and \(x=0\) about the \(y\) -axis. Verify that the lateral surface area of the cone is \(S=\pi r \sqrt{r^{2}+h^{2}}\)
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