The angle made by the tangent of the curve x | vedclass

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Question

(A) Given the parametric equations of the curve: $x = a(t + \sin t \cos t) = a(t + \frac{1}{2} \sin 2t)$ $y = a(1 + \sin t)^2$ First,differentiate $x$

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$\frac{1}{4}(\pi + 2t)$

Vedclass · Answer

(A) Given the parametric equations of the curve: $x = a(t + \sin t \cos t) = a(t + \frac{1}{2} \sin 2t)$ $y = a(1 + \sin t)^2$ First,differentiate $x$ and $y$ with respect to $t$: $\frac{dx}{dt} = a(1 + \cos 2t) = a(2 \cos^2 t) = 2a \cos^2 t$ $\frac{dy}{dt} = 2a(1 + \sin t) \cos t$ The slope of the tangent $m = 	an 	heta = \frac{dy/dt}{dx/dt}$: $	an 	heta = \frac{2a(1 + \sin t) \cos t}{2a \cos^2 t} = \frac{1 + \sin t}{\cos t}$ Using trigonometric identities: $1 + \sin t = (\cos \frac{t}{2} + \sin \frac{t}{2})^2$ $\cos t = \cos^2 \frac{t}{2} - \sin^2 \frac{t}{2} = (\cos \frac{t}{2} - \sin \frac{t}{2})(\cos \frac{t}{2} + \sin \frac{t}{2})$ Thus,$	an 	heta = \frac{\cos \frac{t}{2} + \sin \frac{t}{2}}{\cos \frac{t}{2} - \sin \frac{t}{2}}$ Dividing numerator and denominator by $\cos \frac{t}{2}$: $	an 	heta = \frac{1 + 	an \frac{t}{2}}{1 - 	an \frac{t}{2}} = 	an(\frac{\pi}{4} + \frac{t}{2})$ Therefore,$	heta = \frac{\pi}{4} + \frac{t}{2} = \frac{\pi + 2t}{4}$.

The angle made by the tangent of the curve $x = a(t + \sin t \cos t)$; $y = a(1 + \sin t)^2$ with the $x$-axis at any point on it is

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