Problem 164 - A Complete Resource Book in Mathematics for JEE Main

Question

The normal to the curve \(x=a(\cos \theta+\theta \sin \theta), y=\) \(a(\sin \theta-\theta \cos \theta)\) at any point \(\theta\) is such that \(\quad\) [2005] (A) It passes through the origin (B) It makes angle \(\frac{\pi}{2}+\theta\) with the \(x\)-axis (C) It passes through \(\left(a \frac{\pi}{2},-a\right)\) (D) It is at a constant distance from the origin

Step-by-Step Solution

Answer

The correct answer is (B).

Step 1: Understand the problem

We need to determine a property of the normal to the curve defined parametrically by \( x = a(\cos \theta + \theta \sin \theta) \) and \( y = a(\sin \theta - \theta \cos \theta) \). Specifically, we need to validate if the normal is characterized by any of the given options (A, B, C, or D).

Step 2: Find the slope of the tangent

First, calculate derivatives of \( x \) and \( y \) with respect to \( \theta \):\[ \frac{dx}{d\theta} = a(-\sin \theta + \theta \cos \theta + \sin \theta) = a\theta \cos \theta \] \[ \frac{dy}{d\theta} = a(\cos \theta + \theta \sin \theta - \cos \theta) = a\theta \sin \theta \]The slope of the tangent is given by:\[ \frac{dy}{dx} = \frac{\theta \sin \theta}{\theta \cos \theta} = \tan \theta \]

Step 3: Determine the slope of the normal

The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore, it is:\[ m_\text{normal} = -\cot \theta \]

Step 4: Find equation of the normal at (x,y) in terms of theta

The equation of the normal can be expressed as:\[ y - a(\sin \theta - \theta \cos \theta) = -\cot \theta (x - a(\cos \theta + \theta \sin \theta)) \]Simplify this equation: \[ y = -\cot \theta x + a\csc \theta \] Here, a represents a constant, and \( \csc \theta \) is the cosecant function.

Step 5: Analyze properties of normal line

To determine the characteristics presented in each option, observe the equation derived in Step 4. If we evaluate this for known points, either the origin is satisfied (for option A), or constant properties validate options (B) or (C). Evaluate if y-intercept indicates unique distance for option (D).

Step 6: Evaluate option B

From \( y = -\cot \theta x + a\csc \theta \), the angle \(\phi\) the normal makes with the x-axis is given by \(\cot^{-1}(-\cot \theta) = \pi/2 + \theta \), confirming option B.

Key Concepts

Tangent Lines

A tangent line is a straight line that touches a curve at a single point without crossing it. This single point is often referred to as the point of tangency. The slope of the tangent line to a curve is essential in understanding the behavior and direction of the curve at a specific point. In the context of parametric curves, these are defined by separate equations for each coordinate in terms of a third parameter, often denoted as \( \theta \).

To find the slope of the tangent line for a parametric curve, we first need to calculate the derivatives of both the \(x\) and \(y\) coordinates with respect to the parameter \(\theta\):

• \( \frac{dx}{d\theta} \) - represents how the \(x\) coordinate changes as \(\theta\) changes.
• \( \frac{dy}{d\theta} \) - represents how the \(y\) coordinate changes as \(\theta\) changes.

The slope of the tangent line is given by the ratio \( \frac{dy}{dx} \). Specifically, in this exercise, computing these derivatives helps us find \( \tan \theta \), which informs the angle at which the tangent line is inclined relative to the horizontal axis.

Normal Lines

Normal lines are perpendicular to tangent lines at the point of tangency on a curve. They play a crucial role in geometry and physics, where understanding perpendicularity helps in calculating forces and angles. For any given curve, once we determine the tangent's slope, the normal line’s slope is simply the negative reciprocal of the tangent slope.

In this exercise, the slope of the tangent line is \( \tan \theta \). Thus, the normal line's slope is \(-\cot \theta\). This nature of a normal line makes it particularly fascinating because it reveals the orientation of a curve with respect to a coordinate system. This perpendicular relationship defines how we can draw normals at various points, potentially crossing specific key positions like the origin or employing specific angles from the x-axis, as examined in the chosen option \(B\).

Calculus Derivatives

Derivatives are a fundamental concept in calculus, representing how a function changes at any given point. They are instrumental in finding rates of change, slopes of curves, and instantaneous behavior of dynamical systems. In parametric equations, derivatives involve differentiating each component of the curve with respect to a common parameter like \(\theta\).

In this exercise, derivatives helped find the slopes for both the tangent and normal lines of the curve. Specifically, by determining \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), we obtained a comprehensive understanding of how the curve behaves as \(\theta\) varies.

• Calculating \( \frac{dx}{d\theta} = a\theta \cos \theta \) and \( \frac{dy}{d\theta} = a\theta \sin \theta \).
• The slope of the tangent line was established through \( \frac{dy}{dx} = \tan \theta \).
• This led to identifying the slope of the normal line as \(-\cot \theta\).

By mastering derivatives, students can delve into more complex motion and geometry concepts, enabling them to decisively solve problems similar to the one in this exercise.