Question

Let the curve be represented by $x=2(\cos t+t\sin t)$, $y=2(\sin t-t\cos t)$. Then normal at any point '$t$' of the curve is at a distance of _____ units from the origin.

• A. 1
• B. 0
• C. 2
• D. 4

Hint:

Calculus Tip: This specific parametric curve is the "involute of a circle". A fascinating property of the involute of a circle is that its normal is always tangent to the base circle, meaning the normal's distance from the origin is always exactly equal to the base circle's radius.

Answer 1

The Correct Option is C

Concept: Calculus - Derivatives of Parametric Functions and Application of Derivatives (Tangents and Normals).

Step 1: Find the derivatives of $x$ and $y$ with respect to $t$. Given $x = 2(\cos t + t\sin t)$, apply the product rule to differentiate: $\frac{dx}{dt} = 2(-\sin t + \sin t + t\cos t) = 2t\cos t$. Given $y = 2(\sin t - t\cos t)$, differentiate similarly: $\frac{dy}{dt} = 2(\cos t - (\cos t - t\sin t)) = 2(\cos t - \cos t + t\sin t) = 2t\sin t$.

Step 2: Determine the slope of the normal to the curve. First, find the slope of the tangent, which is $\frac{dy}{dx}$. Using the parametric derivative formula, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t\sin t}{2t\cos t} = \tan t$. The slope of the normal ($m_n$) is the negative reciprocal of the tangent's slope: $m_n = -\frac{1}{dy/dx} = -\frac{1}{\tan t} = -\cot t = -\frac{\cos t}{\sin t}$.

Step 3: Write the equation of the normal line. The point slope form of a line is $Y - y_1 = m(X - x_1)$. Let $(X,Y)$ be general coordinates on the line. Substitute the coordinates $(x,y)$ and the slope of the normal: $Y - 2(\sin t - t\cos t) = -\frac{\cos t}{\sin t} \cdot [X - 2(\cos t + t\sin t)]$.

Step 4: Simplify the equation of the normal into standard form. Cross-multiply by $\sin t$: $Y\sin t - 2\sin^2 t + 2t\sin t\cos t = -X\cos t + 2\cos^2 t + 2t\sin t\cos t$. The $2t\sin t\cos t$ terms cancel out on both sides. Move $X\cos t$ to the left side and $2\sin^2 t$ to the right side: $X\cos t + Y\sin t = 2\sin^2 t + 2\cos^2 t$. Factor out the 2 to get $2(\sin^2 t + \cos^2 t)$. Since $\sin^2 t + \cos^2 t = 1$, the standard equation of the normal line is $X\cos t + Y\sin t - 2 = 0$.

Step 5: Calculate the perpendicular distance from the origin. The perpendicular distance $d$ from the origin $(0,0)$ to the line $AX + BY + C = 0$ is given by $d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$.
Substitute our line's coefficients ($A = \cos t$, $B = \sin t$, $C = -2$): $d = \frac{|(0)\cos t + (0)\sin t - 2|}{\sqrt{\cos^2 t + \sin^2 t}} = \frac{|-2|}{\sqrt{1}} = 2$ units. $$ \therefore \text{The normal is at a constant distance of 2 units from the origin.} $$