Question

dealing with parametric curves
if dy/dx = dy/dt/dx/dt = -sin(t)/-2sin(2t)
now I need to find d^2y/dx^2 and i'm a bit stuck...

Answer 1

\[\frac{d^2y}{dx^2}=\Large{\frac{d}{dt}\left(\frac{dy}{dx}\right)\over\frac{dx}{dt}}\]

Answer 2

\[\frac{ \frac{ d }{ dt }(\frac{ -\sin(t) }{ -2\sin(2t) }) }{ -2\sin(2t) }\]
then I would have\[\frac{ \frac{ \cos (t) }{ 4\cos(2t)} }{ -2\sin(2t) }\]
is that correct?

Answer 3

\[\frac d{dt}(\frac{-\sin t}{-2\sin 2t})\neq\frac{\cos t}{4\cos t}\]remember the quotient rule...

Answer 4

you may do well to simplify first\[\frac{-\sin t}{-2\sin(2t)}=\frac{\sin t}{4\sin t\cos t}=\frac1{4\cos t}=\frac14\sec t\]

Answer 5

that makes more sense