The normal to the curve x = a(cos θ + θ sin θ), y = a(sin θ - θ cos θ) at any point 'θ' is such that

Q: Wait, why do we use parametric differentiation instead of just finding $y$ in terms of $x$?

Great question! In many JEE problems, the relationship between $x$ and $y$ is implicit or transcendental, making it nearly impossible to isolate $y$ as a function of $x$. Parametric differentiation is our most powerful tool here—it allows us to treat $x$ and $y$ as independent functions of a single parameter $\theta$, bypassing the need for an explicit algebraic relationship. It is the elegant way to handle motion along a curve.

Q: Is the cancellation of the $a\theta \sin \theta \cos \theta$ term just a lucky coincidence?

Not at all! In mathematics, especially in geometry, such 'lucky' cancellations are usually a sign that you have uncovered a fundamental property of the curve. The involute of a circle is defined by this specific geometry, and the cancellation is the mathematical proof that the normal line is always tangent to the base circle. It is the universe's way of confirming your derivation is correct.

Q: How do I interpret the final equation $x \cos \theta + y \sin \theta = a$?

This is the 'normal form' of a straight line, written as $x \cos \alpha + y \sin \alpha = p$. In this form, $p$ represents the perpendicular distance from the origin to the line. Since our equation is $x \cos \theta + y \sin \theta = a$, we can immediately see that the distance from the origin is $a$. Because $a$ is a constant, the normal line remains at a fixed distance from the origin for every point on the curve.

Question

The normal to the curve $x = a(\cos 	heta + 	heta \sin 	heta)$, $y = a(\sin 	heta - 	heta \cos 	heta)$ at any point '$	heta$' is such that

Options:

(A) it passes through the origin

(B) it makes an angle $\frac{\pi}{2} + 	heta$ with the $x$-axis

(C) it passes through $(a\frac{\pi}{2}, -a)$

(D) it is at a constant distance from the origin

Sigma Percentile · Accepted Answer

### The Involute's Secret: A Journey into Geometry

Imagine you are standing in a workshop, holding a circular spool of thread. You tie a knot at the end of the thread and, keeping it taut, you begin to unwind it.

As the thread peels away from the circle, the tip of the thread traces a path in the air. This path is what mathematicians call the **involute of a circle**. It is a curve of profound elegance, and today, we are going to peel back its layers to uncover a hidden geometric truth.

### The Calculus of Motion

We are given the parametric equations for this curve:

$$x = a(\cos 	heta + 	heta \sin 	heta)$$

$$y = a(\sin 	heta - 	heta \cos 	heta)$$

To understand the behavior of the normal line at any point $P(	heta)$, we must first understand the tangent. In the world of parametric equations, we use the chain rule by differentiating $x$ and $y$ with respect to the parameter $	heta$ independently.

Let's look at $x$ first. Applying the product rule to the term $	heta \sin 	heta$, we get:

$$\frac{dx}{d	heta} = a [-\sin 	heta + (\sin 	heta + 	heta \cos 	heta)] = a	heta \cos 	heta$$

Notice how the $-\sin 	heta$ and $+\sin 	heta$ vanish? It is as if the math is clearing the path for us. Now, let's do the same for $y$:

$$\frac{dy}{d	heta} = a [\cos 	heta - (\cos 	heta - 	heta \sin 	heta)] = a	heta \sin 	heta$$

Again, the $\cos 	heta$ terms cancel out, leaving us with a beautifully simple derivative. This is the beauty of the involute—the derivatives are clean, symmetric, and ready for the next step.

### The Slope of the Normal

With our derivatives in hand, finding the slope of the tangent $m_t$ is straightforward:

$$m_t = \frac{dy/d	heta}{dx/d	heta} = \frac{a	heta \sin 	heta}{a	heta \cos 	heta} = 	an 	heta$$

This tells us that the tangent line at any point $	heta$ has a slope of $	an 	heta$. But we are interested in the **normal** line—the line perpendicular to the tangent.

The slope of the normal, $m_n$, is the negative reciprocal of the tangent's slope:

$$m_n = -\frac{1}{	an 	heta} = -\cot 	heta = -\frac{\cos 	heta}{\sin 	heta}$$

### The Elegant Cancellation

Now, we construct the equation of the normal line using the point-slope form: $y - y_1 = m_n(x - x_1)$. Substituting our point $P(x_1, y_1)$ and our slope $m_n$, we get:

$$y - a(\sin 	heta - 	heta \cos 	heta) = -\frac{\cos 	heta}{\sin 	heta} [x - a(\cos 	heta + 	heta \sin 	heta)]$$

This looks intimidating, but let's take a breath and multiply both sides by $\sin 	heta$ to clear the fraction:

$$y \sin 	heta - a \sin^2 	heta + a	heta \sin 	heta \cos 	heta = -x \cos 	heta + a \cos^2 	heta + a	heta \sin 	heta \cos 	heta$$

Look closely at the term $a	heta \sin 	heta \cos 	heta$ on both sides. They are identical! When we subtract them, they vanish completely.

We are left with:

$$y \sin 	heta - a \sin^2 	heta = -x \cos 	heta + a \cos^2 	heta$$

Rearranging the terms to bring $x$ and $y$ to one side, we get:

$$x \cos 	heta + y \sin 	heta = a(\cos^2 	heta + \sin^2 	heta)$$

Since $\cos^2 	heta + \sin^2 	heta = 1$, the equation simplifies to the incredibly elegant form:

$$x \cos 	heta + y \sin 	heta = a$$

### The Geometric Revelation

We have arrived at the normal form of a line, $x \cos 	heta + y \sin 	heta = a$. In coordinate geometry, this form tells us that the perpendicular distance from the origin $(0,0)$ to this line is exactly $a$.

Think about what this means. No matter where you are on the curve—no matter what the value of $	heta$ is—the normal line is always at a constant distance $a$ from the origin.

This confirms that the normal line is always tangent to a fixed circle of radius $a$ centered at the origin. We started with a complex parametric curve, navigated through the calculus of derivatives, and arrived at a simple, beautiful geometric truth. You have mastered the involute!

Animated Solution for Mathematics - Differentiation: The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

Select Answer:

Visualized Solution

Visualizing the Curve $P (θ)$

Differentiating $x$ with respect to $θ$

Differentiating $y$ with respect to $θ$

Finding the Slope of the Tangent

Finding the Slope of the Normal

Setting up the Equation of the Normal

Simplifying the Equation

Arriving at the Standard Form

Finding the Distance from the Origin

Conclusion & Geometric Insight

The Sigma Insight: Tangents, Normals and Rate Measure

The Involute's Secret

The Calculus of Motion

The Slope of the Normal

The Elegant Cancellation

The Geometric Revelation

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