Correct Answer - Option 1 : 0
Concept:
Let x = f(t) and y = f(t) then
By Chain Rule, we have
\(\rm \dfrac{d y}{dx} = \dfrac{\dfrac {dy}{dt}}{\dfrac {dx}{dt}}\)
sin (2t) = 2 sin t cos t
Calculations:
Given, x = cos (2t)
⇒\(\rm \dfrac {dx}{dt} = - 2\;sin \;(2t)\)
and y = sin2 t
⇒ \(\rm \dfrac {dy}{dt} = 2\;sin \;tcos\;t\)
⇒\(\rm \dfrac {dy}{dt} = sin \;2t\)
By Chain Rule, we have
\(\rm \dfrac{d y}{dx} = \dfrac{\dfrac {dy}{dt}}{\dfrac {dx}{dt}}\)
\(\rm \dfrac{d y}{dx} =\dfrac {sin \; 2t}{-2sin \;2t}\)
\(\rm \dfrac{d y}{dx} =\dfrac {-1}{2}\)
Differentiating with respect to x, we get
\(\rm \dfrac{d^2 y}{dx^2} = 0\)
Hence, if x = cos (2t) and y = sin2 t, then \(\rm \dfrac{d^2 y}{dx^2}\) equal to 0
