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The hyperbolic functions cosh \(z\) and \(\sinh z\) are defined as follows: $$\cosh z=\frac{e^{z}+e^{-z}}{2} \quad \text { and } \quad \sinh z=\frac{e^{z}-e^{-z}}{2}$$ for any \(z,\) real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of \(z\). (b) Show that \(\cosh z=\cos (i z)\). What is the corresponding relation for \(\sinh z ?\) (c) What are the derivatives of cosh \(z\) and \(\sinh z ?\) What about their integrals? ( \(\mathbf{d}\) ) Show that \(\cosh ^{2} z-\sinh ^{2} z=1\) (e) Show that \(\int d x / \sqrt{1+x^{2}}=\operatorname{arcsinh} x\). [Hint: One way to do this is to make the substitution \(x=\sinh z .]\)

Short Answer

Expert verified
(b) \( \sinh z \) corresponds to \( i \sin z \); (c) Derivatives: \( \cosh' = \sinh \), \( \sinh' = \cosh \); Integrals: \( \int \cosh = \sinh + C \), \( \int \sinh = \cosh + C \); (d) \( \cosh^2 - \sinh^2 = 1 \); (e) \( \int dx/\sqrt{1+x^2} = \mathrm{arcsinh } x \).

Step by step solution

01

Sketch the Functions

To sketch the functions \( \cosh z \) and \( \sinh z \), plot them over a real range, for example, \( z \) from \(-3\) to \(3\):- \( \cosh z \) is an even function, symmetric around the y-axis, resembling a "U" shape starting from 1 when \( z = 0 \). Endpoints rise exponentially as \( z \) increases or decreases. \( \sinh z \) is an odd function, passing through the origin, it resembles a reflected exponential graph that increases positively in positive z, and negatively in negative z.
02

Show Identity for \( \cosh z \)

Given the identity \( \cos(i z) = \cosh z \), substitute \( i z \) in \( \cos x = \frac{e^{ix} + e^{-ix}}{2} \),\( \cos(i z) = \frac{e^{-z} + e^{z}}{2} \). This matches \( \cosh z = \frac{e^{z} + e^{-z}}{2} \). Both expressions are equal, proving the identity.
03

Identity for \( \sinh z \)

To find \( \sinh(i z) \), substitute \( x = i z \) in \( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \), yielding \( \sin(i z) = \frac{e^{-z} - e^{z}}{2i} = -i \cdot \sinh z \). Thus, \( \sinh(i z) = i \sin z \).
04

Derivatives and Integrals

The derivative of \( \, \cosh z \) is \( \sinh z \), because differentiating \( \frac{e^{z} + e^{-z}}{2} \) results in \( \frac{e^{z} - e^{-z}}{2} = \sinh z \). The derivative of \( \sinh z \) is \( \cosh z \) since differentiating \( \frac{e^{z} - e^{-z}}{2} \) results in \( \frac{e^{z} + e^{-z}}{2} = \cosh z \). For integrals, \( \int \cosh z \, dz = \sinh z + C \) and \( \int \sinh z \, dz = \cosh z + C \).
05

Prove Hyperbolic Identity

Use the expressions: \( \cosh^2 z - \sinh^2 z = \left(\frac{e^{z} + e^{-z}}{2}\right)^2 - \left(\frac{e^{z} - e^{-z}}{2}\right)^2 \). Simplify to \( \frac{(e^{2z} + 2 + e^{-2z}) - (e^{2z} - 2 + e^{-2z})}{4} = \frac{4}{4} = 1 \). This proves the identity.
06

Evaluate Integral

Use the substitution \( x = \sinh z \), hence, \( dx = \cosh z \, dz \). The integral becomes \( \int d x / \sqrt{1+x^{2}} = \int d(\sinh z) / \sqrt{\cosh^2 z} \) which simplifies to \( \int dz = z + C \). Since \( \sinh^{-1} x = z \), the integral equals \( \operatorname{arcsinh} x + C \).

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Key Concepts

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

Cosh and Sinh Identities
Hyperbolic functions, such as cosh and sinh, are similar to trigonometric functions like cosine and sine, but they differ in significant ways due to their foundation in hyperbolas rather than circles. Hyperbolic cosine, or \( \cosh z \), is defined as \( \cosh z = \frac{e^z + e^{-z}}{2} \), and hyperbolic sine, or \( \sinh z \), is defined as \( \sinh z = \frac{e^z - e^{-z}}{2} \). These identities are integral in showing how hyperbolic functions can be used to mimic complex trigonometric identities. By understanding how these functions are constructed through exponential expressions, you can better grasp their behavior. For instance, these expressions reveal that \( \cosh z \) is inherently an even function, meaning it behaves the same for \( z \) and \( -z \). Conversely, \( \sinh z \) is an odd function, illustrating symmetry about the origin.

Moreover, an interesting identity can be drawn where \( \cosh z \) relates to trigonometric functions: notably, \( \cosh z \) can be shown to equal to \( \cos(iz) \). This comes from substituting \( iz \) into the general trigonometric cosine function. Likewise, the function \( \sinh z \) has a complex relationship with the trigonometric sine function. Through these identities, hyperbolic functions demonstrate how closely linked they are with circular trigonometric functions, yet offer unique benefits, particularly in solving differential equations and modeling real-world problems such as catenary curves in physics.
Differentiation and Integration of Hyperbolic Functions
Both differentiation and integration in calculus are essential tools when working with hyperbolic functions. The derivatives and integrals of hyperbolic functions mimic those of their trigonometric counterparts, but they have their own elegant rules and patterns.

For derivatives:
  • The derivative of \( \cosh z \) is \( \sinh z \). Differentiating the definition of \( \cosh z = \frac{e^z + e^{-z}}{2} \) gives us \( \frac{e^z - e^{-z}}{2} \), which is \( \sinh z \).
  • Similarly, the derivative of \( \sinh z \) is \( \cosh z \). By differentiating \( \sinh z = \frac{e^z - e^{-z}}{2} \), we obtain \( \frac{e^z + e^{-z}}{2} \), the definition of \( \cosh z \).
For integrals:
  • The integral of \( \cosh z \) with respect to \( z \) is \( \sinh z + C \).
  • The integral of \( \sinh z \) with respect to \( z \) is \( \cosh z + C \).
These differentiations and integrations demonstrate their simplicity and beauty. They also highlight a surprising symmetry: whereas integration and differentiation often involve more complex functions, hyperbolic functions maintain an almost effortless balance. Learning these derivatives and integrals can aid you significantly, especially when encountering more complex integrals in calculus and higher-level applied math.
Hyperbolic Function Relationships
Hyperbolic functions have relationships that are crucial to both their theoretical development and application in solving real-world problems. One of the foundation stones of hyperbolic function relationships is their parallels to classic trigonometric identities, but with key differences.

One of the most vital identities is \( \cosh^2 z - \sinh^2 z = 1 \). This equation resembles the Pythagorean identity from trigonometry, \( \cos^2 x + \sin^2 x = 1 \). This identity can be derived from the definitions of \( \cosh z \) and \( \sinh z \) by using algebraic manipulation:
  • Expand each square: \( (\cosh z)^2 = \left(\frac{e^z + e^{-z}}{2}\right)^2 \) and \( (\sinh z)^2 = \left(\frac{e^z - e^{-z}}{2}\right)^2 \).
  • By subtracting these squared terms, the resulting simplification shows that the exponential terms cancel out, ultimately equating this identity to one.
This identity is not only mathematically beautiful but also practically beneficial, often used in fields such as physics and engineering, especially when modeling hyperbolic structures or processes.

Moreover, these relationships enable substitutions that simplify integration, such as the integral \( \int \frac{dx}{\sqrt{1 + x^2}} = \operatorname{arcsinh} x \). By substituting \( x = \sinh z \), the calculation benefits from the identity above, leading to a straightforward evaluation. Understanding these relationships enhances comprehension and solves complex integrals and differential equations effectively.

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Most popular questions from this chapter

A charged particle of mass \(m\) and positive charge \(q\) moves in uniform electric and magnetic fields, \(\mathbf{E}\) and \(\mathbf{B}\), both pointing in the \(z\) direction. The net force on the particle is \(\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})\). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion.

For any complex number \(z=x+i y,\) the real and imaginary parts are defined as the real numbers \(\operatorname{Re}(z)=x\) and \(\operatorname{Im}(z)=y .\) The modulus or absolute value is \(|z|=\sqrt{x^{2}+y^{2}}\) and the phase or angle is the value of \(\theta\) when \(z\) is expressed as \(z=r e^{i \theta} .\) The complex conjugate is \(z^{*}=x-i y\) (This last is the notation used by most physicists; most mathematicians use \(\bar{z}\).) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, and sketch \(z\) and \(z^{*}\) in the complex plane: (a) \(z=1+i \quad\) (b) \(z=1-i \sqrt{3}\) (d) \(z=5 e^{i \omega t}\) (c) \(z=\sqrt{2} e^{-i \pi / 4}\) In part (d), \(\omega\) is a constant and \(t\) is the time.

There are certain simple one-dimensional problems where the equation of motion (Newton's second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity \(v\), that is, \(F=F(v)\). Write down Newton's second law and separate the variables by rewriting it as \(m d v / F(v)=d t .\) Now integrate both sides of this equation and show that $$t=m \int_{v_{0}}^{v} \frac{d v^{\prime}}{F\left(v^{\prime}\right)}$$ Provided you can do the integral, this gives \(t\) as a function of \(v\). You can then solve to give \(v\) as a function of \(t .\) Use this method to solve the special case that \(F(v)=F_{\mathrm{o}},\) a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and \(2.9 .\)

where the primes denote successive derivatives of \(f(x)\). (Depending on the function this series may converge for any increment \(\delta\) or only for values of \(\delta\) less than some nonzero "radius of convergence.") This theorem is enormously useful, especially for small values of \(\delta\), when the first one or two terms of the series are often an excellent approximation. \(^{11}\) (a) Find the Taylor series for \(\ln (1+\delta)\). (b) Do the same for cos \(\delta\). (c) Likewise sin \(\delta\). (d) And \(e^{\delta}\).

Consider an object that is coasting horizontally (positive \(x\) direction) subject to a drag force \(f=-b v-c v^{2} .\) Write down Newton's second law for this object and solve for \(v\) by separating variables. Sketch the behavior of \(v\) as a function of \(t .\) Explain the time dependence for \(t\) large. (Which force term is dominant when \(t\) is large?)

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