Problem 46 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The differential equation of the family of general circles is (A) \(y^{\prime \prime \prime}\left(1+y^{\prime 2}\right)-3 y^{\prime} y^{\prime \prime 2}=0\) (B) \(y^{\prime \prime \prime}\left(1+y^{\prime 2}\right)+3 y^{\prime} y^{\prime \prime 2}=0\) (C) \(y^{\prime \prime \prime}\left(1+y^{\prime 2}\right)-3 y^{\prime \prime} y^{\prime 2}=0\) (D) None of these

Step-by-Step Solution

Answer

Option (A) matches the differential equation for circles.

Step 1: Understand the General Form of a Circle

A general circle in the coordinate plane has the equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Express the Circle in Parametric Form

For simplicity and differentiation, express the circle in a parametric form: \(x = h + r\cos(\theta)\) and \(y = k + r\sin(\theta)\), where \(\theta\) is the parameter.

Step 3: Differentiate to Find First Derivative \(y'\)

Using implicit differentiation, find \(\frac{dy}{dx}\). Start by differentiating implicitly: \(x - h = r\cos(\theta)\) becomes \(dx = -r\sin(\theta) d\theta\). Likewise, \(y - k = r\sin(\theta)\) becomes \(dy = r\cos(\theta) d\theta\). So, \(y' = \frac{dy}{dx} = -\cot(\theta)\).

Step 4: Differentiate Again to Find Second Derivative \(y''\)

Use the derivative \(y' = -\cot(\theta)\) to find \(y''\). Differentiate again with respect to \(x\): \(y'' = \frac{d}{dx}(-\cot(\theta)) = \csc^2(\theta) \cdot \frac{d\theta}{dx}\).

Step 5: Use Chain Rule for Third Derivative \(y'''\)

Continue differentiating to find the third derivative: \(y''' = \frac{d}{dx} (\csc^2(\theta) \cdot \frac{d\theta}{dx}) = -2\csc^2(\theta)\cot(\theta)\cdot \left(\frac{d\theta}{dx}\right)^2\). Use chain rule and trigonometric identities in the differentiation.

Step 6: Analyze Options Given and Verify

Compare the derived differential equation forms with the provided options (A), (B), (C) to determine which matches the circle. Match the derived structure with these options to validate correctness.

Key Concepts

Parametric Equations

Parametric equations offer a way to express the coordinates of points on a curve as functions of a single variable, called the parameter. For example, consider a circle with center \(h, k\) and radius \(r\).
Using parametric equations, we can represent it as \(x = h + r\cos(\theta)\) and \(y = k + r\sin(\theta)\).

• Parameter: The variable \(\theta\) is the parameter, often representing an angle in the context of circles.
• Benefits: Parametric equations simplify the differentiation and analysis of curves, especially those that cannot be described by simple functions.

They are particularly useful because they allow us to describe more complex trajectories and curves by eliminating the need for a single dependent variable form.
This form is especially helpful in cases like differentiating circle equations, which typically result in cumbersome algebra when handled directly.

Implicit Differentiation

Implicit differentiation is a technique used when it is difficult or impossible to solve for one variable in terms of another directly. When given an equation that defines \(y\) in terms of \(x\), but not explicitly as \(y = f(x)\), we use implicit differentiation to find derivatives.
For instance, a circle's equation \( (x - h)^2 + (y - k)^2 = r^2 \) cannot be easily rearranged as \(y = f(x)\).

• Steps: Differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) (\(y = y(x)\)).
• Chain Rule: When differentiating terms involving \(y\), apply the chain rule. For example, the derivative of \(y^2\) with respect to \(x\) is \(2y \cdot \frac{dy}{dx}\).

Implicit differentiation is critical in handling equations where direct differentiation is not feasible and is extensively applied in multiple calculus problems including this exercise.

Higher Order Derivatives

Higher order derivatives assess how functions' rates of change themselves change. In practical terms, while the first derivative \(y'\) gives the slope of a tangent, the second derivative \(y''\) describes the concavity or curvature. The third derivative, \(y'''\), looks into the rate of change of the curvature.

• Application: In the formulation of the exercise involving circles, calculating up to the third derivative, \(y'''\), helps determine how the slope and concavity change with \(\theta\).
• Usage: Higher order derivatives often enter scenarios in physics and engineering where acceleration, jerk, and other rates of change are analyzed.

In this particular exercise, examining the derivatives shows the interplay of trigonometric identities with these rates.
Each differentiation step builds from understanding \(y' = -\cot(\theta)\) and leverages these concepts progressively through \(y''\) and \(y'''\).

Circle Equations

Circle equations allow us to represent circles in a standard mathematical form. The most common version is \( (x - h)^2 + (y - k)^2 = r^2 \), depicting a circle centered at \( (h, k) \) with a radius \( r \).

• Key Components: \( (h, k) \) defines the center of the circle, \( r \) is the radius, and the terms \( (x - h) \) and \( (y - k) \) ensure the distance formula is maintained.
• Transformation: By expressing this equation in parametric form, we observe motion along the circumference described by functions of a parameter, generally an angle \( \theta \).

This expression highlights the fundamental geometry of a circle and provides a toolset for reasoning about loci and constant distances from a central point.
Such equations are not just geometric but also integral in calculus when analyzing properties like circumferential path or iterations into other complex curves.