To find the curvature $\kappa$ of a curve parameterized by $\mathbf{r}(t) = (x(t), y(t), z(t))$, we use the following formula:

$\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}$

Given the curve $\mathbf{r}(t) = (\sin(t), \cos(t), 2t)$, let's calculate the curvature step by step.

Step 1: Compute the first derivative $\mathbf{r}'(t)$

We differentiate each component of $\mathbf{r}(t)$:

$\mathbf{r}'(t) = \left( \frac{d}{dt} \sin(t), \frac{d}{dt} \cos(t), \frac{d}{dt} 2t \right)$ $\mathbf{r}'(t) = (\cos(t), -\sin(t), 2)$

Step 2: Compute the second derivative $\mathbf{r}''(t)$

Now, differentiate $\mathbf{r}'(t)$:

$\mathbf{r}''(t) = \left( \frac{d}{dt} \cos(t), \frac{d}{dt} (-\sin(t)), \frac{d}{dt} 2 \right)$ $\mathbf{r}''(t) = (-\sin(t), -\cos(t), 0)$

Step 3: Compute the cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$

Next, we calculate the cross product of $\mathbf{r}'(t)$ and $\mathbf{r}''(t)$:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos(t) & -\sin(t) & 2 \\ -\sin(t) & -\cos(t) & 0 \end{vmatrix}$$ $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{i} \left( (-\sin(t))(0) - (-\cos(t))(2) \right) - \mathbf{j} \left( \cos(t)(0) - (-\sin(t))(2) \right) + \mathbf{k} \left( \cos(t)(-\cos(t)) - (-\sin(t))(-\sin(t)) \right)$$ Simplifying: $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{i}(2\cos(t)) - \mathbf{j}(2\sin(t)) + \mathbf{k}(-\cos^2(t) - \sin^2(t))$$ Using the identity $$\cos^2(t) + \sin^2(t) = 1$$, we get: $$\mathbf{r}'(t) \times \mathbf{r}''(t) = (2\cos(t), -2\sin(t), -1)$$ ### Step 4: Compute the magnitude of $$\mathbf{r}'(t) \times \mathbf{r}''(t)$$ The magnitude of the cross product is: $$|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(2\cos(t))^2 + (-2\sin(t))^2 + (-1)^2}$$ $$|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{4\cos^2(t) + 4\sin^2(t) + 1}$$ $$|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{4(\cos^2(t) + \sin^2(t)) + 1} = \sqrt{4 + 1} = \sqrt{5}$$ ### Step 5: Compute the magnitude of $$\mathbf{r}'(t)$$ The magnitude of $$\mathbf{r}'(t)$$ is: $$|\mathbf{r}'(t)| = \sqrt{(\cos(t))^2 + (-\sin(t))^2 + (2)^2}$$ $$|\mathbf{r}'(t)| = \sqrt{\cos^2(t) + \sin^2(t) + 4} = \sqrt{1 + 4} = \sqrt{5}$$ ### Step