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Recall that \(\cosh b t=\left(e^{b t}+e^{-b t}\right) / 2\) and \(\sinh b t=\left(e^{b t}-e^{-b t}\right) / 2 .\) In each of Problems 7 through 10 find the Laplace transform of the given function; \(a\) and \(b\) are real constants. $$ \cosh b t $$

Short Answer

Expert verified
Answer: The Laplace transform of \(\cosh bt\) is given by \(\frac{1}{2} \left(\frac{1}{s-b} + \frac{1}{s+b}\right)\).

Step by step solution

01

Write the function using exponential terms

Recall the definition of \(\cosh bt\). To proceed with the Laplace transform, rewrite the given function in terms of exponential functions: $$ \cosh bt = \frac{e^{bt} + e^{-bt}}{2} $$
02

Find the Laplace transform of each term

Since the Laplace transform is a linear operation, we can find the Laplace transform of \(e^{bt}\) and \(e^{-bt}\) separately and then divide the sum by 2. Recall the Laplace transform formula: $$ \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) dt $$ Using this formula for \(e^{bt}\), we have: $$ \mathcal{L}\{e^{bt}\} = \int_0^\infty e^{-st} e^{bt} dt = \int_0^\infty e^{(b-s)t} dt $$ Now, to find the Laplace transform for \(e^{-bt}\): $$ \mathcal{L}\{e^{-bt}\} = \int_0^\infty e^{-st} e^{-bt} dt = \int_0^\infty e^{-(b+s)t} dt $$
03

Evaluate the Laplace transforms

Now, we need to evaluate the above integrals. For \(e^{(b-s)t}\): $$ \mathcal{L}\{e^{bt}\} = \int_0^\infty e^{(b-s)t} dt = \frac{1}{(b-s)}\left[e^{(b-s)t}\right]_0^\infty $$ As long as \(b<s\), we can evaluate the limit at infinity to get: $$ \frac{1}{(b-s)}\left[-1\right] = \frac{1}{(s-b)} $$ For \(e^{-(b+s)t}\): $$ \mathcal{L}\{e^{-bt}\} = \int_0^\infty e^{-(b+s)t} dt = \frac{1}{(b+s)}\left[e^{-(b+s)t}\right]_0^\infty $$ This gives us: $$ \frac{1}{(b+s)}\left[-1\right] = \frac{1}{(s+b)} $$
04

Combine the Laplace transforms and divide by 2

Now that we have found the Laplace transforms of both exponential terms, we need to combine them and divide by 2 to obtain the Laplace transform of \(\cosh bt\). Therefore: $$ \mathcal{L}\{\cosh bt\} = \frac{1}{2} \left(\frac{1}{s-b} + \frac{1}{s+b}\right) $$ So the Laplace transform of \(\cosh bt\) is given by: $$ \frac{1}{2} \left(\frac{1}{s-b} + \frac{1}{s+b}\right) $$

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions we use commonly in mathematics. While trigonometric functions involve circular calculations, hyperbolic functions deal with hyperbolas. The most common hyperbolic functions are
  • \(\cosh t\), or hyperbolic cosine, defined as \(\cosh t = (e^t + e^{-t})/2\)
  • \(\sinh t\), or hyperbolic sine, defined as \(\sinh t = (e^t - e^{-t})/2\)
These functions are used to describe the shape of hanging cables, are involved in exponential growth models, and often pop up in engineering and physics calculations. They are similar to their circular function counterparts but have properties suited for models involving hyperbolas and growth or decay processes.
The hyperbolic cosine, \(\cosh bt\), is significant because it can easily be represented in terms of exponential functions, making it simpler to apply transformations like the Laplace transform, often required to solve differential equations.
Exponential Functions
Exponential functions are mathematical expressions in which variables appear in the exponent. They have the general form \(e^{at}\), where \(a\) is a constant and \(e\) is Euler's number, approximately 2.718.
Exponential functions are crucial in modeling growth and decay processes, such as population growth and radioactive decay. They play a vital role in hyperbolic functions as they form the basis of the definitions of hyperbolic sine and hyperbolic cosine.
Given the definition of \(\cosh bt = (e^{bt} + e^{-bt})/2\), we can see how exponential functions simplify the computation of hyperbolic functions. Their neat property of differentiation \(\frac{d}{dt} e^{at} = ae^{at} \) makes them particularly handy in calculus, especially in integral transforms such as the Laplace transform.
Integrals
Integrals are fundamental in calculus, depicting the area under a curve. They come into play when calculating continuous accumulations like area, volume, and sums of infinite series.
The Laplace transform integrates functions in relation to exponential decay, represented as \(\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) dt\). This integral is a powerful tool for transforming differential equations into algebraic equations, making them much easier to solve.
In our exercise, we employ integrals to determine the Laplace transform of exponential terms \(e^{bt}\) and \(e^{-bt}\), capitalizing on the beauty of the exponential property in these transforms. By evaluating these specific integrals, we transform complex time-based functions, like hyperbolic sines and cosines, into simpler algebraic forms in the frequency domain.

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Most popular questions from this chapter

Find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related. \(y^{\prime \prime}+3 y^{\prime}+2 y=f(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0 ; \quad f(t)=\left\\{\begin{array}{ll}{1,} & {0 \leq t<10} \\ {0,} & {t \geq 10}\end{array}\right.\)

Use the Laplace transform to solve the given initial value problem. $$ y^{\prime \prime}-4 y^{\prime}+4 y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Use the Laplace transform to solve the given initial value problem. $$ y^{\mathrm{w}}-y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=0 $$

Find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{array}{ll}{0,} & {t<1} \\ {t^{2}-2 t+2,} & {t \geq 1}\end{array}\right. $$

The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The geometrical configuration is shown in Figure \(6.6 .2 .\) The starting point \(P(a, b)\) is joined to the terminal point \((0,0)\) by the arc \(C .\) Arc length \(s\) is measured from the origin, and \(f(y)\) denotes the rate of change of \(s\) with respect to \(y:\) $$ f(y)=\frac{d s}{d y}=\left[1+\left(\frac{d x}{d y}\right)^{2}\right]^{1 / 2} $$ Then it follows from the principle of conservation of energy that the time \(T(b)\) required for a particle to slide from \(P\) to the origin is $$ T(b)=\frac{1}{\sqrt{2 g}} \int_{0}^{b} \frac{f(y)}{\sqrt{b-y}} d y $$

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