Question

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: 

• A. \( \frac{4}{\pi} \sqrt{1 + \pi^2} \)
• B. \( 4 \sqrt{1 + \pi^2} \)
• C. \( 4\pi \)
• D. \( \frac{4}{\pi} \)

Hint:

For parametric equations, the length of the normal can be calculated using the reciprocal of the slope, ensuring correct differentiation and evaluation at the given parameter.

Answer 1

The Correct Option is B

Step 1: Compute the Derivatives We are given the parametric equations: \[ x = 2(\cos 2t + t \sin 2t) \] \[ y = 4(\sin 2t + t \cos 2t) \] Differentiate both equations with respect to \( t \): \[ \frac{dx}{dt} = 2\left(-2\sin 2t + t \cos 2t + \sin 2t\right) = 2\left(-2\sin 2t + \sin 2t + t \cos 2t\right) \] \[ \frac{dx}{dt} = 2\left(-\sin 2t + t \cos 2t\right) \] \[ \frac{dy}{dt} = 4\left(2\cos 2t + t(-\sin 2t) + \cos 2t\right) \] \[ \frac{dy}{dt} = 4\left(3\cos 2t - t \sin 2t\right) \]

Step 2: Find the Slope of the Tangent and Normal The slope of the tangent is given by: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4(3\cos 2t - t \sin 2t)}{2(-\sin 2t + t \cos 2t)} \] \[ = \frac{2(3\cos 2t - t \sin 2t)}{-\sin 2t + t \cos 2t} \] The slope of the normal is the negative reciprocal: \[ m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{-\sin 2t + t \cos 2t}{2(3\cos 2t - t \sin 2t)} \]

 Step 3: Compute the Length of the Normal The length of the normal is given by: \[ \text{Normal Length} = \frac{\sqrt{1 + \left(\frac{dy}{dx}\right)^2}}{\left|\frac{dx}{dt}\right|} \] Substituting \( t = \frac{\pi}{4} \), we get: \[ \text{Normal Length} = 4 \sqrt{1 + \pi^2} \]