Problem 5 - A Complete Resource Book in Mathematics for JEE Main

Question

The family of curves represented by \(\frac{d y}{d x}=\frac{x^{2}+x+1}{y^{2}+y+1}\) and the family represented by \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) (A) Touch each other (B) Are orthogonal (C) Are one and the same (D) None of these

Step-by-Step Solution

Answer

The families of curves are orthogonal.

Step 1: Identify the given differential equations

We have two differential equations: \( \frac{dy}{dx} = \frac{x^2 + x + 1}{y^2 + y + 1} \) and \( \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 \). We need to analyze their relationship.

Step 2: Simplify the second equation

Start by rearranging the second equation: \( \frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1} \). This gives us the derivative of \( y \) with respect to \( x \) for the second family of curves.

Step 3: Compare the two derivatives

The derivatives from the two equations are: \( \frac{dy}{dx} = \frac{x^2 + x + 1}{y^2 + y + 1} \) and \( \frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1} \). Notice how one is the negative reciprocal of the other.

Step 4: Determine orthogonality condition

Two families of curves are orthogonal if the product of their slopes at any intersection point is \(-1\). We found the derivatives are negative reciprocals of each other, confirming this condition.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in describing various physical phenomena and can model a wide range of problems. These equations can be used to find unknown functions when certain conditions are met.
First-order differential equations, like those in our discussion, involve the first derivatives of the function. In the problem at hand, both differential equations express the rate of change of the function \( y \) with respect to \( x \).
Solving differential equations involves finding all possible functions that satisfy the equation. In this scenario, the process involves understanding not just individual solutions but also the relationship between families of solutions. The provided solutions in the exercise aid in analyzing how different families of curves relate to each other through their derivatives.

Orthogonality

Orthogonality is a concept often used in mathematics to describe perpendicularity. When two things are orthogonal, they form a right angle and have a special geometric relationship. This concept extends beyond the physical shape to more abstract mathematical entities, like functions and vectors.
In the context of curves, orthogonality means that the curves intersect at right angles, akin to perpendicular lines. This is specifically determined by examining the slopes of the curves at their points of intersection. The condition for two curves to be orthogonal is that the product of their slopes equals \(-1\).
In our exercise, we see that the given differential equations define families of curves with one derivative being the negative reciprocal of the other. This relationship directly fulfills the orthogonality condition, confirming that the curves intersect at right angles.

Curves Intersection

The intersection of curves refers to the points at which two or more curves meet or cross. This is where their underlying equations have simultaneous solutions. Understanding intersections is important because they reveal points of interaction between different mathematical objects.
In the case of our differential equations, finding intersections involves analyzing where their solutions overlap. Based on the problem's conditions, the derivative comparison indicates the nature of these intersections.
The exercise confirms that the families of curves not just intersect, but do so at right angles, as their derivative products equate to \(-1\). This particular intersection characteristic highlights the orthogonal nature, which is a deeper geometric relationship than mere touching points.