Calculate the curvature of a helix, parametrized by c(t)= < cos(t),sin(t),t >, at time t.

I got root 2 over 2root2 ?

Look at your result and you'll see that it agrees with math fan's:
√2 /(2√2) = ½

Let s = path length. So that
ds² = dx² + dy² + dz²

If you take a position vector parametrized with t,
r_(t) =
and you want to differentiate wrt s, note that
ds² = dx² + dy² + dz² = (x' ² + y' ² + z' ²) dt²

so that
ds/dt = √(x' ² + y' ² + z' ²)

Then note that
d/ds = (dt/ds) d/dt
and that
dt/ds = 1/(ds/dt) = 1/√(x' ² + y' ² + z' ²)

Now the (lineal) curvature of a curve in n-space is the magnitude of the vector, d²r_(t)/ds². So first get the 1st derivative:
dr_(t)/ds = (dr_(t)/dt)/(ds/dt) = /√(x' ² + y' ² + z' ²) = v_ /v

We will need to figure
dv/dt = (d/dt)√(v_•v_) = 1/[2√(v_•v_)] * (v_•a_ + a_•v_) = a_•v_ /v

Then get the 2nd derivative:
d²r_(t)/ds² = (d/ds)(dr_(t)/ds) = (dt/ds) (d/dt)(dr_(t)/ds) = (1/v) (d/dt)(v_ /v)
= (1/v) (a_ /v - (1/v²)(dv/dt)v_) = (1/v) (a_ /v - (v_/v³)(a_•v_))
= (v² a_ - (a_•v_)v_)/v⁴

and the lineal curvature is the magnitude of that:
κ = ||v² a_ - (a_•v_)v_||/v⁴

where
v_(t) = r'_(t)
v(t) = ||v_(t)||
a_(t) = r''_(t)

Now apply that result:
r_(t) =
v_(t) = r'_(t) = <-sin(t), cos(t), 1>
v(t) = √(sin²(t) + cos²(t) + 1) = √2
a_(t) = r''_(t) = <-cos(t), -sin(t), 0>
a_•v_ = sin(t)cos(t) - sin(t)cos(t) + 0 = 0

κ = ||2<-cos(t), -sin(t), 0>||/4 = 2/4 = ½

... and now I need a drink!!

The curvature of a curve is equal to the magnitude of the second derivative of position with respect to arclength. That is to express in an equation, k=Id^2r/ds^2I.
dr/ds=(dr/dt)(dt/ds)=(dr/dt)/(Idr/dtI), which is in this case
(1/2)^(1/2)<-sin(t),cos(t),1>.

So, d^2r/ds^2 is...
d/dt((1/2)^(1/2)<-sin(t),cos(t),1>)
(dt/ds)=
(1/2)<-cos(t),-sin(t),0>, of which the magnitude
(and hence the curvature) is...
1/2.