the parametric equation of a curve are x=cos2Øand y=1+sin2Ø find dy/dx and d^2y/dx^2 at Ø=pai/6.find the relationship between x and y

Question

sir steve help

see my work
dx/dØ=-2sin2Ø dy/dØ=2cos2Ø

dy/dx=dy/dØ*dØ/dx

dy/dx=2cos2Ø*1/-2sin2Ø

dy/dx=-cot2Ø

d^2y/dx^2=-2cosec^2(2Ø)

plz round it up for me am practising

Answer 1

I'll just use t for ease of typing. Those big Øs just distract me from the math.

And that's "pi" not "pai" or "pie" or a dozen other ways it gets misspelled!

x = cos(2t)
y = 1+sin(2t)

dy/dt = 2cos(2t
dx/dt = -2sin(2t)
dy/dx = dy/dt / dx/dt = -cot(2t)

d^2y/dx^2 = d/dx (dy/dx)
= d/dt (dy/dx) / dx/dt
= 2csc^2(2t)/-2sin(2t) = csc^3(2t)

or, directly,
y = 1+√(1-x^2)
y' = -x/√(1-x^2)
y" = -1/(1-x^2)^(3/2) = -csc^3(2t)

getting d^2y/dx^2 is not just so simple as you tried to make it.