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Thank you ahead of time for taking a look at this. My problem is as follows: We have two forms of the equation for the curvature of a space curve $\mathbf{r}(t)$ given by \begin{equation} \kappa(t) = \frac{|\mathbf{T'(t)}|}{|\mathbf{r'(t)}|} \end{equation} and \begin{equation} \kappa(t) = \frac{|\mathbf{r'}(t)\times\mathbf{r''}(t)|}{|\mathbf{r'(t)}|^3} \end{equation} All authors I have been able to find equate the two forms, indicating that both will give the same curvature of a space curve. However, for some space curves, I obtain different forms of the curvature depending on which form of the $\kappa$ equation I use.

Take the example $y = 5e^x$. We parameterize this as \begin{equation*} \mathbf{r}(t)=\ \end{equation*} and using the first form of the curvature formula obtain $|\mathbf{T'}(t)|= 5e^t$ and $|\mathbf{r'}(t)|= \sqrt{1+25e^{2t}}$ which gives \begin{equation*} \kappa(t) = \frac{5e^t}{\sqrt{1+25e^{2t}}} \end{equation*} yet when I use the second curvature formula I obtain \begin{equation*} \kappa(t) = \frac{5e^t}{(1+25e^{2t})^{3/2}} \end{equation*} I have been puzzling at this for a while now and have not been able to ascertain as to why the two forms of the equation should give different answers.

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    where $T$ is the unit tangent ... do you need to normalise it before you differentiate ? – 2017-02-25

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You have to normalise $\mathbf{r}'$ to get $\mathbf{T}$ before you differentiate it: in your case, $$ \mathbf{T}(t) = \frac{(1,5e^t)}{\sqrt{1+25e^{2t}}}. $$ In general, $$ \mathbf{T}' = \left(\frac{\mathbf{r}'}{\lVert \mathbf{r}' \rVert}\right)' = \frac{\mathbf{r}''}{\lVert \mathbf{r}' \rVert} - \frac{\mathbf{r}' (\mathbf{r}' \cdot \mathbf{r}'')}{\lVert \mathbf{r}' \rVert^3}, $$ and taking the cross product eliminates the second term.