## Help me find curvature of r(t) = <(\sqrt 15)t, e^t, e^-t> at point (0,1,1)

So I got this problem on the test and couldn't do it:
Find curvature of r(t) = <(\sqrt 15)t, e^t, e^-t> at point (0,1,1)

I found 1st and 2nd derivatives:
1st = <(\sqrt 15), e^t, -e^-t>
2nd = < 0, e^t, e^-t>

Magnitude of tangent vector: \sqrt ( 15 + e^2 + e^-2) ---- not sure because I plugged in the point into r'(t) and then found magnitude of that. My book only provides example of sin t and cos t when talking about magnitude of tangent vector, so I don't know how to plug in the values.

k(t) = |r'(t) x r"(t)| / |r'(t)|^3

r'(t) x r"(t) = 0, (\sqrt 15)/e^t, (\sqrt 15)*e^t

Magnitude of that I plugged in (0,1,1) again and in the end I get this:

k(t) = e^2 * (/sqrt(15e^4+15)) / ((\sqrt(e^4 + 15e^2 + 1))^3)

When I plug in the values on a calculator I get like 0.0516. I don't know how to simplify the above

The answer choices on the test were: 1/17, /sqrt 2 / 17, /sqrt 2 / 19, 17*/sqrt 2 and only 1/17 is somewhat close (its 0.05882)