Problem 54 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Solution of \(\frac{x d x+y d y}{x d y-y d x}=\frac{\sqrt{1-\left(x^{2}+y^{2}\right)}}{\sqrt{x^{2}+y^{2}}}\) is (A) \(\sin ^{-1} \sqrt{x^{2}+y^{2}}=c\) (B) \(\tan ^{-1} \frac{y}{x}=c\) (C) \(\sin ^{-1} \sqrt{x^{2}+y^{2}}=\tan ^{-1} \frac{y}{x}+c\) (D) None of these

Step-by-Step Solution

Answer

(C) \(\sin^{-1}\sqrt{x^2+y^2}=\tan^{-1}\frac{y}{x}+c\)."

Step 1: Simplify the Equation

The given expression is \( \frac{x \, dx + y \, dy}{x \, dy - y \, dx} = \frac{\sqrt{1-(x^2 + y^2)}}{\sqrt{x^2 + y^2}} \). We aim to equate the left side and the right side components. Notice that the left side resembles a differential form in polar coordinates.

Step 2: Convert to Polar Coordinates

Let \( x = r \cos \theta \) and \( y = r \sin \theta \), then \( dx = \cos \theta \, dr - r \sin \theta \, d\theta \) and \( dy = \sin \theta \, dr + r \cos \theta \, d\theta \). Substitute these into the left side of the equation.

Step 3: Evaluate Left-hand Side

Substituting the derivatives, we have \( x \, dx + y \, dy = r \, dr \) and \( x \, dy - y \, dx = r^2 \, d\theta \). Thus, the left side becomes \( \frac{r \, dr}{r^2 \, d\theta} = \frac{dr}{r \, d\theta} = \frac{1}{r} \frac{dr}{d\theta} \).

Step 4: Express Right-hand Side in Terms of r

The right side simplifies to \( \frac{\sqrt{1-r^2}}{r} \) since \( x^2 + y^2 = r^2 \).

Step 5: Integrate Both Sides

Equating the simplified equations from Steps 3 and 4, we get \( \frac{dr}{r} = \frac{\sqrt{1-r^2}}{r} \, d\theta \). This leads to \( \int \frac{dr}{\sqrt{1-r^2}} = \int d\theta \).

Step 6: Solve the Integral

The integral \( \int \frac{dr}{\sqrt{1-r^2}} \) is known to be \( \sin^{-1}(r) \) and \( \int d\theta = \theta \). Thus, we have \( \sin^{-1}(r) = \theta + C \), where \( C \) is a constant.

Step 7: Identify r with Original Variables

Since \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}(\frac{y}{x}) \), the solution simplifies to \( \sin^{-1}(\sqrt{x^2 + y^2}) = \tan^{-1}(\frac{y}{x}) + C \).

Key Concepts

Polar Coordinates

Polar coordinates are an alternative system to the standard Cartesian coordinates. They are especially useful for problems involving symmetry or circular motion. In this system, a point is determined by a distance from the origin (the radius) and an angle from the positive x-axis.

• Radius ( ): Represents the distance from the origin to a specific point. It is given by the formula \( r = \sqrt{x^2 + y^2} \), where \( x \) and \( y \) are Cartesian coordinates.
• Angle (\theta): Indicates the direction of the point from the origin, calculated as \( \theta = \tan^{-1}(\frac{y}{x}) \).

Understanding polar coordinates simplifies complex transformations, which is crucial for differential equations like the one in our original exercise. In terms of differential equations, converting to polar coordinates can reveal symmetries that simplify integration and other operations.

Integral Calculus

Integral Calculus is the branch of mathematics dealing with integrals, focusing on the accumulation of quantities and the areas under and between curves. In solving differential equations, integrals are essential tools.For our exercise, we encountered the integral \( \int \frac{dr}{\sqrt{1-r^2}} \).

• This integral is known as the arcsine integral, because its solution is \( \sin^{-1}(r) \). This is derived from the fact that the derivative of \( \sin^{-1}(r) \) is \( \frac{1}{\sqrt{1-r^2}} \).
• Equally, the integral \( \int d\theta \) results easily in \( \theta \), since the derivative of \( \theta \) with respect to itself is simply 1.

By calculating integrals, we can find relationships between variables, vital for finding solutions to the original differential equation.

Trigonometric Functions

Trigonometric Functions are mathematical functions of an angle. They relate angles to the lengths of the sides of right triangles and are fundamental in various branches of mathematics.In the context of this exercise:

• Sine (\sin): Described our radius expression in the form of an inverse sine, \( \sin^{-1}(\sqrt{x^2 + y^2}) \), highlighting the connection between circular measurements and polar radii.
• Tangent (\tan): Appeared when identifying angles, \( \tan^{-1}(\frac{y}{x}) \) as part of the polar transformation, showing the angle of the vector from the positive x-axis.

Recognizing these trigonometric identities allows for smoother transitions between Cartesian and polar coordinates. It also aids in understanding how they contribute to reshaping differential equations, as demonstrated in this solution.