Calc 3: Easiest way to do this?

[SeeDRankLou]

I have this problem, and I got the right answer, I'm just not sure I did this in the easiest manner that I could have. Here's what I did:

Find the curvature of r(t) in terms of t
r(t) = t cos(t) i + t sin(t) j

curvature (K) = | y'' x' - x'' y' | / ((x' ² + y' ²)^(3/2))

x(t) = t cos(t)
x'(t) = cos(t) - t sin(t)
x''(t) = -sin(t) - ( sin(t) + t cos(t) ) = -2 sin(t) - t cos(t)

y(t) = t sin(t)
y'(t) = sin(t) + t cos(t)
y''(t) = cos(t) + ( cos(t) - t sin(t) ) = 2 cos(t) - t sin(t)

(1) y'' x' = ( 2 cos(t) - t sin(t) ) ( cos(t) - t sin(t) ) = 2 cos²(t) - 3t cos(t) sin(t) + t² sin²(t)
(2) x'' y' = (-2 sin(t) - t cos(t) ) ( sin(t) + t cos(t) ) = -2 sin²(t) - 3t sin(t) cos(t) - t² cos²(t)
(1) - (2) = 2 ( cos²(t) + sin²(t) ) + t² ( sin²(t) + cos²(t) ) = 2 + t²

(3) x' ² = ( cos(t) - t sin(t) )² = cos²(t) - 2t cos(t) sin(t) + t² sin²(t)
(4) y' ² = ( sin(t) + t cos(t) )² = sin²(t) + 2t sin(t) cos(t) + t² cos²(t)
(3) + (4) = ( cos²(t) + sin²(t) ) + t² ( sin²(t) + cos²(t) ) = 1 + t²

K = | 2 + t² | / (( 1 + t² )^(3/2))
2 + t² > 0
K = ( 2 + t² ) / (( 1 + t² )^(3/2))

This seems like the long way of doing this. My question is if there was something simplier I could have done that I am overlooking. Like converting the vector function to an xy-function or something like that.

[Aerith's Knight]

Hmmm, maybe.

r(t) = t (cos(t)[x] + sin(t)[y])

can(?) be written as:

r(t) = t (cos(t) + i * sin(t)) - where i is imaginary, etc etc.

Therefore by the Euler transform:

r(t) = t * e^it

I'm not sure whether such conjecture is allowed, though.

You can find the curvature formula for polar coordinates here.