Question

Given that: \[ x = a \sin(2t) (1 + \cos(2t)), \quad y = a \cos(2t) (1 - \cos(2t)) \] Find \(\frac{dy}{dx}\).

• A. \( \frac{a \tan(t)}{b} \)
• B. \( \frac{a \tan(t)}{b} \)
• C. \( \frac{b \tan(t)}{a} \)
• D. \( \frac{b}{a \tan(t)} \)

Hint:

Remember: When applying the chain rule, we need to differentiate both the numerator and the denominator to find the derivative \( \frac{dy}{dx} \).

Answer 1

The Correct Option is D

Step 1: Apply the chain rule for differentiation. The chain rule states: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] We must compute \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). Step 2: Differentiate \( y = a \cos(2t) (1 - \cos(2t)) \) with respect to \( t \). Using the product rule: \[ \frac{dy}{dt} = a \left[ \frac{d}{dt} \left( \cos(2t) \right) (1 - \cos(2t)) + \cos(2t) \frac{d}{dt} \left( (1 - \cos(2t)) \right) \right] \] The individual derivatives are: \[ \frac{d}{dt} \left( \cos(2t) \right) = -2 \sin(2t), \quad \frac{d}{dt} \left( (1 - \cos(2t)) \right) = 2 \sin(2t) \] Substituting back: \[ \frac{dy}{dt} = a \left[ -2 \sin(2t) (1 - \cos(2t)) + \cos(2t) \cdot 2 \sin(2t) \right] \] Simplifying the expression: \[ \frac{dy}{dt} = 2a \sin(2t) \left[ \cos(2t) - (1 - \cos(2t)) \right] \] \[ \frac{dy}{dt} = 2a \sin(2t) \left[ 2 \cos(2t) - 1 \right] \] Step 3: Differentiate \( x = a \sin(2t) (1 + \cos(2t)) \) with respect to \( t \). Applying the product rule: \[ \frac{dx}{dt} = a \left[ \frac{d}{dt} \left( \sin(2t) \right) (1 + \cos(2t)) + \sin(2t) \frac{d}{dt} \left( (1 + \cos(2t)) \right) \right] \] The derivatives are: \[ \frac{d}{dt} \left( \sin(2t) \right) = 2 \cos(2t), \quad \frac{d}{dt} \left( (1 + \cos(2t)) \right) = -2 \sin(2t) \] Substituting these values: \[ \frac{dx}{dt} = a \left[ 2 \cos(2t) (1 + \cos(2t)) + \sin(2t) (-2 \sin(2t)) \right] \] Simplifying the expression: \[ \frac{dx}{dt} = 2a \cos(2t) (1 + \cos(2t)) - 2a \sin^2(2t) \] Step 4: Compute \( \frac{dy}{dx} \). Using the chain rule formula: \[ \frac{dy}{dx} = \frac{2a \sin(2t) \left( 2 \cos(2t) - 1 \right)}{2a \cos(2t) \left( 1 + \cos(2t) \right) - 2a \sin^2(2t)} \] Simplifying the fraction: \[ \frac{dy}{dx} = \frac{\sin(2t) \left( 2 \cos(2t) - 1 \right)}{\cos(2t) \left( 1 + \cos(2t) \right) - \sin^2(2t)} \] Answer: The final answer is option (4): \( \frac{b}{a \tan(t)} \).