Question

Obtain Eqn. (6.3.35) formally by differentiating: (a) \(\arg (x, y)=\cos ^{-1} \frac{x}{\sqrt{x^{2}+y^{2}}}\) (b) \(\arg (x, y)=\sin ^{-1} \frac{y}{\sqrt{x^{2}+y^{2}}}\) (c) \(\arg (x, y)=\tan ^{-1} \frac{y}{x}\) Where do these formulas come from? What is the disadvantage of using any one of them to define \(\arg (x, y) ?\)

Short answer

In summary, we have formally differentiated the three given expressions of the argument and showed that the combination of their partial derivatives with respect to x and y gives Eqn. (6.3.35). The given expressions originate from the trigonometric functions and the definition of polar coordinates. However, using any single formula to define the argument is disadvantageous, as each has a specific range of values and may not provide the correct angle for all points in the plane.

Step-by-step solution

Differentiate the expression with respect to x

We know that the given expression is \(\arg (x, y) = \cos ^{-1} \frac{x}{\sqrt{x^{2}+y^{2}}}\), so let's differentiate with respect to x: \(\frac{\partial}{\partial x} \cos ^{-1} \frac{x}{\sqrt{x^{2} + y^{2}}} = -\frac{y}{(x^{2} + y^{2}) \sqrt{1 - \frac{x^2}{x^2 + y^2}}}\)

Differentiate the expression with respect to y

Now, let's differentiate with respect to y: \(\frac{\partial}{\partial y} \cos ^{-1} \frac{x}{\sqrt{x^{2} + y^{2}}} = \frac{x}{(x^{2} + y^{2}) \sqrt{1 - \frac{x^2}{x^2 + y^2}}}\) ##1b. Differentiating Part (b)##

Differentiate the expression with respect to x

The given expression is \(\arg (x, y) = \sin ^{-1} \frac{y}{\sqrt{x^{2} + y^{2}}}\), so let's differentiate with respect to x: \(\frac{\partial}{\partial x} \sin ^{-1} \frac{y}{\sqrt{x^{2} + y^{2}}} = -\frac{y}{(x^{2} + y^{2}) \sqrt{1 - \frac{y^2}{x^2 + y^2}}}\)

Differentiate the expression with respect to y

Now, let's differentiate with respect to y: \(\frac{\partial}{\partial y} \sin ^{-1} \frac{y}{\sqrt{x^{2} + y^{2}}} = \frac{x}{(x^{2} + y^{2}) \sqrt{1 - \frac{y^2}{x^2 + y^2}}}\) ##1c. Differentiating Part (c)##

Differentiate the expression with respect to x

The given expression is \(\arg (x, y) = \tan ^{-1} \frac{y}{x}\), so let's differentiate with respect to x: \(\frac{\partial}{\partial x} \tan ^{-1} \frac{y}{x} = -\frac{y}{x^{2} + y^{2}}\)

Differentiate the expression with respect to y

Now, let's differentiate with respect to y: \(\frac{\partial}{\partial y} \tan ^{-1} \frac{y}{x} = \frac{x}{x^{2} + y^{2}}\) For all the cases, we can observe that combining the partial derivatives with respect to x and y gives the equation (6.3.35), which is \(\frac{\partial \arg (x, y)}{\partial x}+i \frac{\partial \arg (x, y)}{\partial y}=\frac{-y+i x}{x^{2} + y^{2}}\).

Formula Origins and Disadvantages

The given formulas are derived from the trigonometric functions and the definition of polar coordinates. In polar coordinates, a point is represented by its distance from the origin (r) and the angle formed with the positive x-axis (arg(x,y)). The disadvantage of using any one of these formulas to define the argument is that each of them has a specific range of values and may not give the correct angle for all points in the plane. For example, the arctan function is only defined in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\), so it will not give the correct value for points in the second and third quadrants.

Key concepts

Trigonometric Functions

Trigonometric functions like cosine, sine, and tangent are essential in mathematics and are used to describe relationships in right-angled triangles. In this exercise, these functions are used to define the argument of a point \(x, y\) in polar coordinates. The formulas provided in the exercise, such as \(\cos^{-1}\), \(\sin^{-1}\), and \(\tan^{-1}\), are inverse trigonometric functions. These inverse functions help us find the angles when given a ratio of the sides of a triangle. The reverse nature of these functions makes them suitable for finding the angle related to the coordinates of a point.
In more detail, the function \(\cos^{-1}\) grants us the knowledge of an angle when we know the adjacent side and hypotenuse relation, while \(\sin^{-1}\) deals with the relation between opposite side and hypotenuse. The \(\tan^{-1}\) will calculate the angle knowing opposite and adjacent sides. These calculations are valuable in polar coordinate systems where angles need precise computation to ensure they correctly depict the location of a point in Cartesian coordinates.

Polar Coordinates

Polar coordinates provide a unique way to represent points on a plane through distance and angle, rather than using traditional Cartesian coordinates \(x, y\). In polar coordinates, any point can be represented as \(r, \theta\), where \(r\) is the distance from the origin, and \(\theta\) is the angle formed with the positive x-axis. This depiction is particularly useful in fields like physics and engineering where circular and rotational motion is common.
A crucial advantage of polar coordinates is their ability to represent periodic and rotational systems effectively. However, they can be less intuitive and sometimes more complex to work with, especially when translating to Cartesian coordinates. The transformation involves using trigonometric functions to switch between the systems: \(x = r \cdot \cos(\theta)\) and \(y = r \cdot \sin(\theta)\). This relation helps in calculating and understanding the position and angle of points in different coordinate systems, which is useful in the proper example contexts this exercise is derived from.

Arc Functions

Arc functions, also known as inverse trigonometric functions, like \(\cos^{-1}\), \(\sin^{-1}\), and \(\tan^{-1}\), play a pivotal role in deriving angles from a known trigonometric value. In the context of this exercise, arc functions define the angle, or argument, that a point \(x, y\) makes with the positive x-axis. This angle, especially in polar coordinates, helps in locating the precise position of the point in a 2D plane.
The primary characteristic of arc functions is their output range. For instance, \(\sin^{-1}(y)\) results in an angle ranging from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), while \(\tan^{-1}(\frac{y}{x})\) results in the same. This specificity is both a feature and a limitation; the angle values obtained through these functions are restricted. It poses a disadvantage, especially in contexts where continuous angle measurement without breaks is essential. Interpreting these restricted outputs and knowing which function to use requires understanding the context and purpose as seen in the exercise, ensuring accurate representation of all angles around a coordinate point.