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 \( \mathbf{r}''(t) \):

\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\cos(t) & -\sin(t) & 2 \\
-\sin(t) & -\cos(t) & 0
\end{vmatrix}
\]

\[
\mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{i} \left( (-\sin(t))(0) - (-\cos(t))(2) \right) - \mathbf{j} \left( \cos(t)(0) - (-\sin(t))(2) \right) + \mathbf{k} \left( \cos(t)(-\cos(t)) - (-\sin(t))(-\sin(t)) \right)
\]

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

find the curvature of (sin(t), cos(t), 2t)

Discover how to calculate the curvature of the parametric curve r(t) = (sin(t), cos(t), 2t) with step-by-step solutions. Learn key calculus concepts, including cross products, vector differentiation, and curvature. Suitable for undergraduate students studying multivariable calculus.

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