سوال ۳۱ کنکور ارشد علوم کامپیوتر ۱۴۰۱

Setup. For a curve given in polar form $r = r(\theta)$, the curvature is

$\kappa = \frac{\left| r^{2} + 2\,(r')^{2} - r\,r'' \right|}{\left( r^{2} + (r')^{2} \right)^{3/2}}.$

Derivatives. Let $u = \theta - \tfrac{\pi}{6}$, so $r = 2\cos u$. Then

$r' = -2\sin u, \qquad r'' = -2\cos u.$

Evaluate at $\theta = \tfrac{\pi}{6}$ (so $u = 0$):

$r = 2\cos 0 = 2, \qquad r' = -2\sin 0 = 0, \qquad r'' = -2\cos 0 = -2.$

Plug in.

$\text{numerator} = \left| r^{2} + 2(r')^{2} - r\,r'' \right| = \left| 4 + 0 - (2)(-2) \right| = |4 + 4| = 8,$

$\text{denominator} = \left( r^{2} + (r')^{2} \right)^{3/2} = (4 + 0)^{3/2} = 4^{3/2} = 8.$

$\kappa = \frac{8}{8} = 1.$

Geometric check. An equation of the form $r = 2a\cos(\theta - \alpha)$ is a circle of radius $a$ (passing through the origin), here with $2a = 2$, i.e. $a = 1$. A circle of radius $1$ has constant curvature $\kappa = 1/a = 1$ everywhere, matching the formula.

The curvature is $1$, which is option (2).

Exam trick: Recognize $r = 2a\cos(\theta - \alpha)$ as a circle of radius $a$ — its curvature is the constant $1/a$ at every point, so no derivative grinding is needed.