Chapter 6: Problem 66
Show that the arc length of the catenary \(y=\cosh x\) over the interval \([0, a]\) is \(L=\sinh a\).
Short Answer
Expert verified
Answer: The arc length of the catenary \(y = \cosh x\) over the interval \([0, a]\) is \(L = \sinh a\).
Step by step solution
01
Find the derivative of y with respect to x
Given the function \(y = \cosh x\), we can find its derivative \(\frac{dy}{dx}\) using the following rule:
\(\frac{d}{dx} (\cosh x) = \sinh x\)
So, we have:
\(\frac{dy}{dx} = \sinh x\)
02
Substitute the derivative into the formula for arc length
Now we can substitute the derivative of \(y\) with respect to \(x\) into the arc length formula:
\(L = \int_0^a \sqrt{1 + (\frac{dy}{dx})^2} dx = \int_0^a \sqrt{1 + (\sinh x)^2} dx\)
03
Simplify the expression under the square root
Note that using the identity \(\cosh^2 x - \sinh^2 x = 1\), we can rewrite the expression under the square root as follows:
\(\sqrt{1 + (\sinh x)^2} = \sqrt{\cosh^2 x} = \cosh x\)
Thus, the arc length formula becomes:
\(L = \int_0^a \cosh x dx\)
04
Integrate and find the arc length
Now, we can integrate \(\cosh x\) over the interval \([0, a]\):
\(L = \int_0^a \cosh x dx = [\sinh x]_0^a = \sinh a - \sinh 0 = \sinh a - 0 = \sinh a\)
Therefore, the arc length of the catenary \(y = \cosh x\) over the interval \([0, a]\) is \(L = \sinh a\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Functions
Hyperbolic functions are analogues of the trigonometric functions but for the hyperbola, much like trigonometric functions are for the circle. Two primary hyperbolic functions are hyperbolic cosine (\( \cosh x \) ) and hyperbolic sine (\( \sinh x \) ). These functions are defined using exponential functions:
- The hyperbolic cosine, \( \cosh x = \frac{e^x + e^{-x}}{2} \), represents the curve of a hanging cable or chain, known as a catenary.
- The hyperbolic sine, \( \sinh x = \frac{e^x - e^{-x}}{2} \), captures the shape variance compared to straight lines.
Integration
Integration is a crucial concept in calculus used to find quantities that add up over a range, such as area under curves, accumulated values, and in this case, arc length. To integrate a function mathematically means to determine the indefinite integral, or antiderivative, of a function.
In our context, we integrate the function \( \cosh x \) over a given interval using the fundamental theorem of calculus. The formula for integration that we use here is:
The definite integral is evaluated by calculating \( \sinh x \) at \( a \) and then subtracting the value at \( 0 \). This solution seamlessly connects integration with arc length computation.
In our context, we integrate the function \( \cosh x \) over a given interval using the fundamental theorem of calculus. The formula for integration that we use here is:
- For a given function \( f(x) \), the indefinite integral is \( \int f(x) \, dx \).
- To find the definite integral from \( a \) to \( b \), you evaluate the antiderivative at the two boundaries and subtract: \( \left[F(b) - F(a)\right] \).
The definite integral is evaluated by calculating \( \sinh x \) at \( a \) and then subtracting the value at \( 0 \). This solution seamlessly connects integration with arc length computation.
Derivative
Understanding derivatives involves knowing how functions change as their input changes. Derivatives help us understand the rate of change of a function.
In this exercise, when we first determine the derivative of \( y = \cosh x \), we get \( \frac{dy}{dx} = \sinh x \). This reflects how \( y \) changes with respect to \( x \). To find this derivative, we use a basic rule of differentiation for hyperbolic functions, which follows a similar process as differentiating trigonometric functions:
By using derivatives, we set the groundwork for further steps in calculus, such as determining geometry-related measures like arc lengths, which tell us about the curve's actual path when assessing the hyperbolic shape.
In this exercise, when we first determine the derivative of \( y = \cosh x \), we get \( \frac{dy}{dx} = \sinh x \). This reflects how \( y \) changes with respect to \( x \). To find this derivative, we use a basic rule of differentiation for hyperbolic functions, which follows a similar process as differentiating trigonometric functions:
- Using the rule, \( \frac{d}{dx} (\cosh x) = \sinh x \).
By using derivatives, we set the groundwork for further steps in calculus, such as determining geometry-related measures like arc lengths, which tell us about the curve's actual path when assessing the hyperbolic shape.
Arc Length Formula
The arc length formula gives us a straightforward way to calculate the distance along a curve over a specific interval.
For the function \( y = \cosh x \), we use the arc length formula given by:
This step is crucial as it turns what could be a complex integral into a simpler one by showing that:
For the function \( y = \cosh x \), we use the arc length formula given by:
- \( L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx \)
This step is crucial as it turns what could be a complex integral into a simpler one by showing that:
- \( \sqrt{1 + (\sinh x)^2} = \cosh x \)