Problem 66 - Calculus Early Transcendentals

Question

Show that the arc length of the catenary \(y=\cosh x\) over the interval \([0, a]\) is \(L=\sinh\) \(a.\)

Step-by-Step Solution

Answer

Question: Calculate the arc length of the curve \(y = \cosh x\) over the interval \([0, a]\). Answer: The arc length of the catenary \(y=\cosh x\) over the interval \([0,a]\) is \(L=\sinh a\).

Step 1: Find the derivative of \(y\) with respect to \(x\).

Let \(y = \cosh x\). To find the derivative of \(y\) with respect to \(x\), we differentiate it: $$ \frac{dy}{dx} = \frac{d(\cosh x)}{dx} = \sinh x $$

Step 2: Substitute the derivative into the arc length formula.

Now that we have the derivative, we can substitute it into the arc length formula: $$L = \int_{0}^{a}\sqrt{1+(\frac{dy}{dx})^2} dx = \int_{0}^{a}\sqrt{1+(\sinh x)^2} dx $$

Step 3: Simplify the integrand.

To simplify the integrand, we use the identity: $$ \cosh^2 x - \sinh^2 x = 1 $$ Therefore, we can rewrite the integrand as: $$\sqrt{1 + (\sinh x)^2} = \sqrt{\cosh^2 x} = \cosh x$$ So, the arc length integral becomes: $$L = \int_{0}^{a} \cosh x\, dx$$

Step 4: Evaluate the integral.

Now we have to evaluate the integral: $$ L = \int_{0}^{a} \cosh x\, dx = \sinh x \Big|_{0}^a = \sinh a - \sinh 0$$ Since \(\sinh 0 = 0\), we have: $$L = \sinh a$$ So, the arc length of the catenary \(y=\cosh x\) over the interval \([0,a]\) is \(L=\sinh a\).

Key Concepts

Catenary

A catenary is the curve that a hanging flexible chain or cable assumes when supported only at its ends and acted upon by a uniform gravitational force. You can imagine it like the shape a piece of string makes when you hold it up from both ends. It appears naturally in the real world in structures like suspension bridges or power lines. The catenary has a distinctive U-shape, and mathematically, it is described by the hyperbolic cosine function,

• The equation for a standard catenary is given by: \( y = rac{a}{2}(e^{x/a} + e^{-x/a}) \),
• where \( e \) is the base of the natural logarithm, and \( a \) is a constant.

In problems like the one at hand, dealing with arc length, we're often interested in finding the length of such a curve over a certain interval. The simplicity of the catenary's form allows for elegant solutions that involve essential calculus concepts such as derivatives and integrals.

Hyperbolic Functions

Hyperbolic functions, much like the trigonometric functions, are a set of functions that appear frequently in mathematical descriptions of geometric and physical systems. They are defined using the exponential function and provide a set of counterparts to the familiar sine, cosine, and tangent functions:

• The hyperbolic cosine function, \( \ cosh x \), is defined as \( (e^x + e^{-x})/2 \).
• Similarly, the hyperbolic sine function \( \ sinh x \) is defined as \( (e^x - e^{-x})/2 \).

These functions have important identities, such as \( \ cosh^2 x - sinh^2 x = 1 \), which are fundamental in solving calculus problems effectively.
In the context of arc length calculations, when given a curve described by a hyperbolic function like \( y = \ cosh x \), these identities simplify the integration process and allow for results that are useful in practical applications, such as determining the span of bridges or the sag of power lines.

Integration

Integration is a core concept in calculus representing the accumulation of quantities. It is essentially the reverse process of differentiation and is used to find areas under curves or the total accumulated value across a range. In the problem at hand, integration is used to find the arc length of the curve defined by \( y = \ cosh x \):

• The arc length formula \ involves \( \int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^2}dx \).
• The interval of integration is \([0, a]\), reflecting the segment of the catenary we are interested in.

By initially finding the derivative and substituting it into the arc length formula, we're able to transform the problem into a solvable integral. The use of identities such as \( \cosh^2 x - \sinh^2 x = 1 \) simplifies the integrand, reducing the computational complexity and allowing us to find a neat expression for the arc length as \( L = \sinh a \).
Thus, integration connects our understanding of the behavior of curves with real-world measurements.

Derivative

The derivative is a fundamental operation in calculus, representing the rate of change of a function. It’s how we understand steepness, slopes, and rates of increase or decrease. To find the arc length of a curve, like our catenary \( y = \ cosh x \), we first need to determine its derivative with respect to \( x \).

• For \( y = \ cosh x \), the derivative is \( \ sinh x \).

This result comes straight from the definition of hyperbolic functions, where the derivative of the hyperbolic cosine is the hyperbolic sine. Thus, \( \frac{d}{dx}(\cosh x) = \sinh x \).
Once we have the derivative, it plays a crucial role in formulating the arc length integral. The derivative captures how rapidly the curve ascends or descends, fundamentally influencing how we compute the total path length the curve traces. Therefore, derivatives are key to transitioning from local properties of a function to a global understanding, such as total length.