Deriving the equation for centripetal acceleration

I'm going to be completely honest about my last post in which I looked at how fast you would have to spin a spaceship to make artificial gravity.  I didn't remember how to calculate centripetal force, and I didn't have my phone or computer or physics books to look up the equation.  I was able to do it on one bar napkin though.

Assuming an object orbits a central point and has its position defined by the functions:

x(t) = r * cos(ωt)

y(t) = r * sin(ωt)

Take the first derivative to get velocity and you have:

vx(t) = x'(t) = -ω * r * sin(ωt)

vy(t) = y'(t) = ω * r * cos(ωt)

Take another derivative for acceleration:

ax(t) = vx'(t) = x''(t) = -ω² * r * cos(ωt)

ay(t) = vy'(t) = y''(t) = -ω² * r * sin(ωt)

So now we can just use the Pythagorean Theorem (x² + y² = r²) to find the magnitude of centripetal acceleration.  Rearranging:

a = √(ax² + ay²) 

It looks like it's going to get hairy with all the squaring and square rooting, but a quick substitution will simplify it. Substitute and multiply out:

a = √( (-ω² * r² * cos(ωt) )² + (-ω² * r² * sin(ωt) )² ) 

a = √(   ω⁴ * r² * cos²(ωt)   +   ω⁴ * r² * sin²(ωt)   )

Divide out ω⁴ * r²:

a = √(  ω⁴ * r² * ( cos²(ωt) + sin²(ωt)  )

a = ω² * r * √( cos²(ωt) + sin²(ωt) )

Remember the property of sin²ϑ + cos²ϑ = 1?  If we just apply that, we are left with the equation of a = ω² * r.  Derivation complete!!