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(C) Given the parametric equations of the curve are $x = 12(t + \sin t \cos t)$ and $y = 12(1 + \sin t)^2$.First,we find the derivatives with respect

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$27$

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(C) Given the parametric equations of the curve are $x = 12(t + \sin t \cos t)$ and $y = 12(1 + \sin t)^2$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 12(1 + \cos^2 t - \sin^2 t) = 12(1 + \cos 2t) = 12(2 \cos^2 t) = 24 \cos^2 t$.$\frac{dy}{dt} = 12 	imes 2(1 + \sin t) \cos t = 24(1 + \sin t) \cos t$.The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24(1 + \sin t) \cos t}{24 \cos^2 t} = \frac{1 + \sin t}{\cos t}$.Since the angle made with the positive $x$-axis is $\frac{\pi}{3}$,the slope is $	an(\frac{\pi}{3}) = \sqrt{3}$.So,$\frac{1 + \sin t}{\cos t} = \sqrt{3} \Rightarrow 1 + \sin t = \sqrt{3} \cos t$.Dividing by $2$,we get $\frac{1}{2} \sin t + \frac{\sqrt{3}}{2} \cos t = \frac{1}{2} \Rightarrow \sin(t + \frac{\pi}{3}) = \frac{1}{2}$.Since $0 Thus,$t + \frac{\pi}{3} = \frac{5\pi}{6} \Rightarrow t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.Now,substitute $t = \frac{\pi}{6}$ into the expression for $y$:$y_{0} = 12(1 + \sin(\frac{\pi}{6}))^2 = 12(1 + \frac{1}{2})^2 = 12(\frac{3}{2})^2 = 12 	imes \frac{9}{4} = 27$.

If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x=12(t+\sin t \cos t)$,$y =12(1+\sin t )^{2}$,$0 < t < \frac{\pi}{2}$,with the positive $x$-axis is $\frac{\pi}{3}$,then $y_{0}$ is equal to

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