For the curve x = a(theta + sin theta) and y | vedclass

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(A) Given $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{

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$2a \sin \frac{	heta}{2}, a \sin 	heta$

Vedclass · Answer

(A) Given $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$.First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \sin 	heta}{a(1 + \cos 	heta)} = \frac{2 \sin(	heta/2) \cos(	heta/2)}{2 \cos^2(	heta/2)} = 	an(	heta/2)$.Length of the tangent is given by $|y \sqrt{1 + (dx/dy)^2}|$ or $|y \csc(	heta/2)|$. Specifically,the length of the tangent is $|y \frac{\sqrt{1 + (dy/dx)^2}}{dy/dx}| = |\frac{y \sec(	heta/2)}{	an(	heta/2)}| = |\frac{a(1 - \cos 	heta)}{\sin(	heta/2)}| = |\frac{2a \sin^2(	heta/2)}{\sin(	heta/2)}| = 2a \sin(	heta/2)$.Length of the sub-tangent is given by $|y / (dy/dx)| = |\frac{a(1 - \cos 	heta)}{	an(	heta/2)}| = |\frac{2a \sin^2(	heta/2)}{\sin(	heta/2) / \cos(	heta/2)}| = |2a \sin(	heta/2) \cos(	heta/2)| = a \sin 	heta$.

For the curve $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$,the lengths of the tangent and the sub-tangent at point $\theta$ are respectively:

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