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 th

Q: Why do we differentiate x and y separately? Can't we just find y in terms of x?

My dear student, that is a classic trap! Trying to eliminate $\theta$ to get $y = f(x)$ would lead to a messy, non-algebraic nightmare. In parametric calculus, we treat $\theta$ as the master variable. By differentiating $x$ and $y$ separately with respect to $\theta$, we use the chain rule $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ to find the slope without ever needing to explicitly define the curve in Cartesian coordinates. It is the elegant path, not the hard one!

Q: Why does the normal slope become $-\cot \theta$?

Remember the fundamental rule of coordinate geometry: if two lines are perpendicular, the product of their slopes is $-1$. We found the tangent slope $m_t = \tan \theta$. Therefore, the normal slope $m_n$ must be $-\frac{1}{m_t}$. Since $\frac{1}{\tan \theta} = \cot \theta$, the normal slope is simply $-\cot \theta$. It is a beautiful, direct relationship!

Q: How do I know the distance from the origin is constant?

Once you derive the equation of the normal as $x \cos \theta + y \sin \theta = a$, you are looking at the normal form of a line: $x \cos \alpha + y \sin \alpha = p$. Here, $p$ represents the perpendicular distance from the origin. Since our equation is $x \cos \theta + y \sin \theta = a$, the distance $p$ is clearly $a$, which is a constant independent of $\theta$.

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 makes a constant angle with the $x$-axis

(B) it passes through the origin

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

(D) none of these

Sigma Percentile · Accepted Answer

### The Geometry of the Involute: A Journey into the Normal

Welcome, my aspiring engineers. Today, we are not just solving a calculus problem; we are uncovering the hidden geometry of the Involute of a Circle.

Imagine you are holding a string wrapped tightly around a circular spool. As you unwind the string, keeping it taut, the end of the string traces a path. That path is defined by the parametric equations: 
$$x = a (\cos 	heta + 	heta \sin 	heta)$$
$$y = a (\sin 	heta - 	heta \cos 	heta)$$

Our goal is to find a fundamental property of the normal line at any point $P$ on this path. Let us embark on this journey.

### Phase 1: The Tangent's Secret

To understand the normal, we must first understand the tangent. In the world of parametric equations, we do not have a direct $y = f(x)$ relationship. Instead, we have two variables dancing to the tune of a parameter, $	heta$.

To find the slope of the tangent, $m_t = \frac{dy}{dx}$, we use the chain rule:
$$m_t = \frac{dy/d	heta}{dx/d	heta}$$

Let us differentiate $x$ with respect to $	heta$. Applying the product rule to $	heta \sin 	heta$, we get:
$$\frac{dx}{d	heta} = a [-\sin 	heta + (\sin 	heta + 	heta \cos 	heta)] = a 	heta \cos 	heta$$

Similarly, for $y$, we differentiate $y = a (\sin 	heta - 	heta \cos 	heta)$. Applying the product rule again:
$$\frac{dy}{d	heta} = a [\cos 	heta - (\cos 	heta - 	heta \sin 	heta)] = a 	heta \sin 	heta$$

When we divide these, the $a$ and $	heta$ terms vanish, leaving us with the elegant result:
$$m_t = \frac{a 	heta \sin 	heta}{a 	heta \cos 	heta} = 	an 	heta$$

### Phase 2: The Normal's Perpendicularity

Now that we know the tangent's slope is $	an 	heta$, the normal's slope, $m_n$, is simply the negative reciprocal:
$$m_n = -\frac{1}{	an 	heta} = -\cot 	heta$$

This is the key that unlocks the door. We now have a point $P(x_1, y_1)$ and a slope $m_n$. We use the point-slope form: $y - y_1 = m_n(x - x_1)$.

Substituting our values, we get:
$$y - a(\sin 	heta - 	heta \cos 	heta) = -\frac{\cos 	heta}{\sin 	heta} [x - a(\cos 	heta + 	heta \sin 	heta)]$$

### Phase 3: The Algebraic Collapse

I know this equation looks intimidating, but do not fear the complexity. Mathematics often hides its greatest beauty behind a wall of symbols.

Let us cross-multiply by $\sin 	heta$ to clear the fraction:
$$\sin 	heta [y - a(\sin 	heta - 	heta \cos 	heta)] = -\cos 	heta [x - a(\cos 	heta + 	heta \sin 	heta)]$$

Expanding both sides, we see terms like $a 	heta \sin 	heta \cos 	heta$ appearing on both sides. They cancel out perfectly! We are left with:
$$y \sin 	heta - a \sin^2 	heta = -x \cos 	heta + a \cos^2 	heta$$

Rearranging this, 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 stunningly simple:
$$x \cos 	heta + y \sin 	heta = a$$

### Phase 4: The Geometric Revelation

We have arrived at the final destination. The equation $x \cos 	heta + y \sin 	heta = a$ is the equation of the normal.

Using the perpendicular distance formula $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we substitute the origin $(0,0)$ into our equation $x \cos 	heta + y \sin 	heta - a = 0$. The result is:
$$p = \frac{|-a|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = a$$

The distance is $a$, a constant! No matter where you are on the curve, the normal is always at a distance $a$ from the origin. This is the hallmark of the **involute of a circle**. You have successfully navigated the calculus and uncovered the geometric truth.

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

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Visualized Solution

Parametric Equations of the Curve

Roadmap to the Normal

Differentiating $x$ with respect to $θ$

Differentiating $y$ with respect to $θ$

Slope of the Tangent ( $m_{t}$ )

Slope of the Normal ( $m_{n}$ )

Setting up the Equation of the Normal

Expanding the Equation

Final Equation of the Normal

Analyzing the Normal's Property

Distance from the Origin

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of the Involute

Phase 1

Phase 2

Phase 3

Phase 4

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