1. **State the problem:** We want to show that the curvature $\kappa$ of the helix given by the vector function $$\mathbf{r}(t) = \langle a \cos t, a \sin t, b t \rangle$$ is $$\kappa = \frac{a}{a^2 + b^2}.$$\n\n2. **Recall the formula for curvature:** For a space curve $\mathbf{r}(t)$, curvature is given by $$\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}.$$\n\n3. **Compute the first derivative:** $$\mathbf{r}'(t) = \left\langle -a \sin t, a \cos t, b \right\rangle.$$\n\n4. **Compute the second derivative:** $$\mathbf{r}''(t) = \left\langle -a \cos t, -a \sin t, 0 \right\rangle.$$\n\n5. **Compute the cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$:**\n$$\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a \sin t & a \cos t & b \\ -a \cos t & -a \sin t & 0 \end{vmatrix} = \langle a b \sin t, -a b \cos t, a^2 \rangle.$$\n\n6. **Calculate the magnitude of the cross product:**\n$$\| \mathbf{r}'(t) \times \mathbf{r}''(t) \| = \sqrt{(a b \sin t)^2 + (-a b \cos t)^2 + (a^2)^2} = \sqrt{a^2 b^2 (\sin^2 t + \cos^2 t) + a^4} = \sqrt{a^2 b^2 + a^4} = a \sqrt{a^2 + b^2}.$$\n\n7. **Calculate the magnitude of the first derivative:**\n$$\| \mathbf{r}'(t) \| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + b^2} = \sqrt{a^2 (\sin^2 t + \cos^2 t) + b^2} = \sqrt{a^2 + b^2}.$$\n\n8. **Compute the curvature:**\n$$\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} = \frac{a \sqrt{a^2 + b^2}}{(\sqrt{a^2 + b^2})^3} = \frac{a \sqrt{a^2 + b^2}}{(a^2 + b^2)^{3/2}} = \frac{a}{a^2 + b^2}.$$\n\n**Final answer:** $$\boxed{\kappa = \frac{a}{a^2 + b^2}}.$$