The length of the normal drawn at t=frac{pi}{ | vedclass

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(B) Given the curve $x = 2(\cos 2t + t \sin 2t)$ and $y = 4(\sin 2t - t \cos 2t)$.First, find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$\fr

Vedclass · Accepted Answer

$4 \sqrt{1+\pi^2}$

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(B) Given the curve $x = 2(\cos 2t + t \sin 2t)$ and $y = 4(\sin 2t - t \cos 2t)$.First, find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$\frac{dx}{dt} = 2(-2 \sin 2t + \sin 2t + 2t \cos 2t) = 2(2t \cos 2t - \sin 2t)$.$\frac{dy}{dt} = 4(2 \cos 2t - \cos 2t + 2t \sin 2t) = 4(\cos 2t + 2t \sin 2t)$.At $t = \frac{\pi}{4}$:$\frac{dx}{dt} = 2(2(\frac{\pi}{4}) \cos \frac{\pi}{2} - \sin \frac{\pi}{2}) = 2(0 - 1) = -2$.$\frac{dy}{dt} = 4(\cos \frac{\pi}{2} + 2(\frac{\pi}{4}) \sin \frac{\pi}{2}) = 4(0 + \frac{\pi}{2}) = 2\pi$.Therefore, the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\pi}{-2} = -\pi$.The slope of the normal is $-\frac{1}{dy/dx} = \frac{1}{\pi}$.At $t = \frac{\pi}{4}$, $y = 4(\sin \frac{\pi}{2} - \frac{\pi}{4} \cos \frac{\pi}{2}) = 4(1 - 0) = 4$.The length of the normal is given by $|y| \sqrt{1 + (\frac{dy}{dx})^2} = |4| \sqrt{1 + (-\pi)^2} = 4 \sqrt{1 + \pi^2}$.

The length of the normal drawn at $t=\frac{\pi}{4}$ on the curve $x=2(\cos 2t + t \sin 2t)$, $y=4(\sin 2t - t \cos 2t)$ is

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