Reparametrize by arc length and find the signed curvature Let 
$$g(t)=(e^t \cos(t),e^t \sin(t))$$
Reparametrize $g(t)$ by arc length starting at $0$, then find the signed curvature of the unit-speed reparametization.
I found that the $e^t=\frac{s+\sqrt2}{\sqrt2}$ and $t=ln(\frac{s+\sqrt2}{\sqrt2)})$, which I substituted into $g$ to get
$$g(t(s))=\frac{s+\sqrt2}{\sqrt2}\cos(\ln(\frac{s+\sqrt2}{\sqrt2})),\frac{s+\sqrt2}{\sqrt2}(\sin(\ln(\frac{s+\sqrt2}{\sqrt2}))$$
This is where I get lost: The signed curvature $k_s=\frac{d\theta}{ds}$, where 
$$\theta=\ln\Bigl(\frac{s+\sqrt2}{\sqrt2}\Bigr)$$
Can someone help me out here and tell me what I'm doing wrong? Any help would be appreciated.
 A: Everything you have done seems correct. We have $g^\prime(t)=(e^t\cos t-e^t\sin t, e^t\sin t+e^t\cos t)$ and so $$\left|g^\prime(t)\right|^2=(e^t\cos t-e^t\sin t)^2+(e^t\sin t+e^t\cos t)^2=2e^{2t}.$$ Hence $\left|g^\prime(t)\right|=\sqrt 2 e^t$. Using $s=\int_0^t\left|g^\prime(t)\right|dt$ you get $t=\ln\left(\frac{s+\sqrt 2}{\sqrt 2}\right)$.
From here, we can now reparamtrize with respect to arc length. That is, we form a new curve $\widetilde g=(g\circ t)(s)$, where $t(s)$ is the inverse of arc-length $s(t)$. You have $\widetilde g$ is unit speed. Indeed by the chain rule $$\left|\frac{d\widetilde g}{ds}(s)\right|=\left|\frac{dg}{dt}(t(s))\right|\left|\frac{dt}{ds}(s)\right|=\left|\frac{dg}{dt}(t(s))\right|\frac{1}{\left|\frac{ds}{dt}(t(s))\right|}$$where the last equality follows from the inverse function theorem. However, we have $\frac{ds}{dt}(t(s))=\left|g^\prime(t(s))\right|$ by the fundamental theorem of calculus.
You can verify explicitly this works in your example. 
Now that you have this unit speed reparametrization calculating the curvature is just one more derivative. It's the norm of the acceleration $\left|\widetilde g^{\prime\prime}(s)\right|$.
