Problem 11 - Advanced Engineering Mathematics

Question

(a) \(\psi(x, y)=\operatorname{Im}(\sin z)=\cos x \sinh y\) and \(\psi(x, y)=0\) when \(x=\pm \pi / 2\) or when \(y=0\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)}=\overline{\cos z}=(\cos x \cosh y, \sin x \sinh y)\) (c) \(\cos x \sinh y=c \Rightarrow y=\sinh ^{-1}(c \sec x)\) and the streamlines are shown in the figure.

Step-by-Step Solution

Answer

(a) Occurs at \( x=\pm \pi/2 \) or \( y=0 \); (b) \( \mathbf{V} = (\cos x \cosh y, \sin x \sinh y) \); (c) \( y = \sinh^{-1}(c \sec x) \).

Step 1: Understanding the Function

The function \(\psi(x, y) = \operatorname{Im}(\sin z) = \cos x \sinh y\)represents the imaginary part of \(\sin z\), which is a complex function where \(z = x + iy\). Using the identity \(\sin z = \sin(x + iy) = \sin x \cosh y + i \cos x \sinh y\), the imaginary part is \(\cos x \sinh y\), which means the function \(\psi(x, y) = 0 \) when \(\cos x \sinh y = 0\). This occurs when \(x = \pm \pi / 2\) or \(y = 0\).

Step 2: Analyzing Function \\( \mathbf{V} \\)

The vector \( \mathbf{V} = \overline{G^{\prime}(z)} = \overline{\cos z}\) can be broken down into real and imaginary parts. Using the identity \( \cos z = \cos(x + iy) = \cos x \cosh y - i \sin x \sinh y \), the conjugate gives \( \overline{\cos z} = \cos x \cosh y + i \sin x \sinh y \). Therefore, \( \mathbf{V} \) is represented as \( (\cos x \cosh y, \sin x \sinh y) \).

Step 3: Solving for y

We solve \( \cos x \sinh y = c \) by manipulating the equation to find \( y. \) Thus, if \( \cos x \sinh y = c \), then \( \sinh y = \frac{c}{\cos x} \). Hence, \( y = \sinh^{-1}(\frac{c}{\cos x}) \), which can be expressed as \( y = \sinh^{-1}(c \sec x) \). This equation shows the relationship between \( y \) and \( x \) along the streamlines.

Key Concepts

Complex Functions

Complex functions play a crucial role in complex analysis, dealing with functions that involve complex numbers. In our problem, the function \( \psi(x, y) = \operatorname{Im}(\sin z) = \cos x \sinh y \) extracts the imaginary part of the sine of a complex number \( z = x + iy \). Every complex number consists of a real part and an imaginary part. When you apply the sine function to a complex number, the result is also complex.Here’s how it works in this context:

• The complex function \( \sin z \) is expressed as \( \sin(x + iy) = \sin x \cosh y + i \cos x \sinh y \).
• The imaginary part, \( i \cos x \sinh y \), is what we focus on here. So, \( \psi(x, y) = \cos x \sinh y \).
• This function value becomes zero under certain conditions – specifically when \( \cos x \sinh y = 0 \). This happens when either \( x = \pm \pi / 2 \) or \( y = 0 \), simplifying the understanding of these functions.

By understanding these conditions, students can better visualize how different mathematical properties relate to each other in the complex plane. It comes down to knowing where these functions intersect zeros, useful for many advanced applications in engineering and physics.

Vector Analysis

Vector analysis gives us the tools to work with functions involving vectors, which are essential in both physics and engineering for modeling and solving spatial problems. In case (b) of the problem, \( \mathbf{V} = \overline{G^{\prime}(z)} = \overline{\cos z} = (\cos x \cosh y, \sin x \sinh y) \),we are working with vector-valued functions derived from complex functions.The computation involves:

• Starting with the complex function \( \cos z = \cos(x + iy) \).
• Splitting it into real and imaginary parts: \( \cos z = \cos x \cosh y - i \sin x \sinh y \).
• The conjugate is simply obtained by changing the sign of the imaginary part: \( \overline{\cos z} = \cos x \cosh y + i \sin x \sinh y \).
• This results in a vector \( \mathbf{V} = (\cos x \cosh y, \sin x \sinh y) \) that characterizes the direction and magnitude of this transformed space.

Understanding how these vectors are constructed helps in visualizing fields and can assist with optimizing processes like electrical fields, fluid dynamics, and more, where direction and magnitude are key components.

Hyperbolic Functions

Hyperbolic functions, much like trigonometric functions, are fundamental in calculus and have applications in various fields including physics and engineering. In our exercise, we use hyperbolic functions to solve for \( y \) in the equation \( \cos x \sinh y = c \).Hyperbolic functions are defined as follows:

• \( \sinh y = \frac{e^y - e^{-y}}{2} \)
• \( \cosh y = \frac{e^y + e^{-y}}{2} \)
• These functions are similar to \( \sin x \) and \( \cos x \) but are based on exponential functions, showcasing different properties.

To solve for \( y \):

• Rearrange the given equation: \( \cos x \sinh y = c \) into \( \sinh y = \frac{c}{\cos x} \).
• Now, express \( y \) explicitly using the inverse hyperbolic sine function: \( y = \sinh^{-1}(\frac{c}{\cos x}) \) or \( y = \sinh^{-1}(c \sec x) \).
• This describes the streamlines in the space, showcasing how the values of \( x \) influence the values of \( y \), and vice versa.

These insights are useful when dealing with hyperbolic equations in real-world problems, particularly where growth rates and time constants play a role, such as in hyperbolic trajectories and scientific wave functions.