At any point ‘ $t$ ’ on the curve $x = a (t + \sin t),$ $y = a (1 - \cos t)$ , find the lengths of tangent, normal, subtangent an subnormal.

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Detailed Solution

Given curves $x = a (t + \sin t); y = a (1 - \cos t)$

diff w.r. to ‘ $t$ ’

$\frac{d x}{d t} = a [1 + \cos t] \frac{d y}{d t} = a [0 - (- \sin t)]$

$= a 2 \cos^{2} t / 2 = a \sin t$

$= 2 a \cos^{2} t / 2 = a (2 \sin t / 2 \cos t / 2)$

$m = \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 a \sin t / 2 \cos t / 2}{2 a \cos^{2} t / 2} = \tan t / 2$

Length of tangent $= |\frac{y}{m} \sqrt{1 + m^{2}}|$

$= |\frac{a (1 - \cos t)}{\tan t / 2} \sqrt{1 + \tan^{2} t / 2}|$

$= |a \frac{2 \sin^{2} t / 2}{\frac{\sin t / 2}{\cos t / 2}} \sqrt{\sec^{2} t / 2}|$

$= | 2 a \sin t / 2 \cos t / 2 (\sec t / 2) |$

$= |2 a \sin t / 2 \cos t / 2 \frac{1}{\cos t / 2}| = | 2 a \sin t / 2 | = 2 a \sin t / 2$

Length of normal $= |y \sqrt{1 + m^{2}}|$

$= |a (1 - \cos t) \sqrt{1 + \tan^{2} t / 2}| = |a 2 \sin^{2} t / 2 \sqrt{\sec^{2} t / 2}|$

$= | 2 a \sin t / 2 \sin t / 2 \sec t / 2 |$

$= |2 a \sin t / 2 \frac{\sin t / 2}{\cos t / 2}| = | 2 a \sin t / 2 \tan t / 2 |$

Length of sub tangent $= |\frac{y}{m}|$

$= |\frac{a (1 - \cos t)}{\tan t / 2}| = |\frac{a 2 \sin^{2} t / 2}{\frac{\sin t / 2}{\cos t / 2}}| = | 2 a \sin t / 2 \cos t / 2 |$

$= |a \cdot \sin 2 (\frac{t}{2})| = | a \cdot \sin t |$

Length of sub normal $= | y m |$

$= | a (1 - \cos t) \tan t / 2 | =∣ a 2 \sin^{2} t / 2) \tan t / 2 ∣$

$= |2 a \sin^{2} t / 2 \tan t / 2|$

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