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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.3.71a

A power line is attached at the same height to two utility poles that are separated by a distance of 100 ft; the power line follows the curve ƒ(x) = a cosh x/a. Use the following steps to find the value of a that produces a sag of 10 ft midway between the poles. Use a coordinate system that places the poles at x = ±50.


a. Show that a satisfies the equation cosh 50/a − 1 = 10/a.

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Understand the problem setup: The power line follows the curve \(f(x) = a \cosh \frac{x}{a}\), and the poles are located at \(x = \pm 50\). The sag is the vertical distance between the height of the poles and the lowest point of the cable, which occurs at \(x=0\) because \(\cosh\) has its minimum there.
Express the height of the poles: At \(x = \pm 50\), the height of the cable is \(f(50) = a \cosh \frac{50}{a}\). Since the poles are at the same height, both ends have this height.
Express the sag: The lowest point of the cable is at \(x=0\), where \(f(0) = a \cosh 0 = a \cdot 1 = a\). The sag is the difference between the height at the poles and the lowest point, so sag \(= f(50) - f(0) = a \cosh \frac{50}{a} - a\).
Set the sag equal to 10 ft: According to the problem, the sag is 10 ft, so write the equation \(a \cosh \frac{50}{a} - a = 10\).
Rearrange the equation to isolate terms: Divide both sides by \(a\) to get \(\cosh \frac{50}{a} - 1 = \frac{10}{a}\), which is the equation that \(a\) satisfies.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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3m
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Catenary Curve and Hyperbolic Cosine Function

The power line forms a catenary curve described by ƒ(x) = a cosh(x/a), where cosh is the hyperbolic cosine function. This curve models the shape of a flexible chain or cable hanging under its own weight, and understanding its properties is essential to relate the sag and horizontal distance.
추천 영상:
5:53
Graph of Sine and Cosine Function

Coordinate System and Boundary Conditions

Placing the poles at x = ±50 sets the horizontal span of the cable, and the sag is measured at the midpoint x = 0. Using these boundary conditions allows us to express the sag in terms of the parameter a and set up an equation involving cosh(50/a) to solve for a.
추천 영상:
05:32
Intro to Polar Coordinates

Deriving and Solving the Sag Equation

The sag is the vertical drop from the pole height to the lowest point of the cable. By substituting x = 0 and x = ±50 into the catenary equation and using the sag value, we derive the equation cosh(50/a) − 1 = 10/a. Solving this transcendental equation yields the parameter a.
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Solving Logarithmic Equations
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