We are given the parameterized curve:

$\mathbf{r}(t) = \left\langle 5 \cos t, \sqrt{7} \sin t, 2 \cos t \right\rangle$

We need to find the curvature using the alternative formula:

$\kappa = \frac{|\mathbf{a} \times \mathbf{v}|}{|\mathbf{v}|^3}$

Here:

• $\mathbf{v}(t)$ is the velocity vector (the first derivative of $\mathbf{r}(t)$),
• $\mathbf{a}(t)$ is the acceleration vector (the second derivative of $\mathbf{r}(t)$).

Step 1: Compute the velocity vector $\mathbf{v}(t)$

Differentiate $\mathbf{r}(t)$ with respect to $t$:

$\mathbf{v}(t) = \frac{d}{dt} \left( 5 \cos t, \sqrt{7} \sin t, 2 \cos t \right) = \left\langle -5 \sin t, \sqrt{7} \cos t, -2 \sin t \right\rangle$

Step 2: Compute the acceleration vector $\mathbf{a}(t)$

Differentiate $\mathbf{v}(t)$ with respect to $t$:

$\mathbf{a}(t) = \frac{d}{dt} \left( -5 \sin t, \sqrt{7} \cos t, -2 \sin t \right) = \left\langle -5 \cos t, -\sqrt{7} \sin t, -2 \cos t \right\rangle$

Step 3: Compute the cross product $\mathbf{a} \times \mathbf{v}$

Now, compute the cross product of $\mathbf{a}(t)$ and $\mathbf{v}(t)$:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 \cos t & -\sqrt{7} \sin t & -2 \cos t \\ -5 \sin t & \sqrt{7} \cos t & -2 \sin t \end{vmatrix}$$ The cross product is given by: $$\mathbf{a}(t) \times \mathbf{v}(t) = \left\langle (2 \sqrt{7} \sin^2 t + 2 \sqrt{7} \cos^2 t), (10 \sin t \cos t - 10 \sin t \cos t), (5 \sqrt{7} \sin^2 t + 5 \sqrt{7} \cos^2 t) \right\rangle$$ Simplify using the Pythagorean identity $$\sin^2 t + \cos^2 t = 1$$: $$\mathbf{a}(t) \times \mathbf{v}(t) = \left\langle 2 \sqrt{7}, 0, 5 \sqrt{7} \right\rangle$$ ### Step 4: Compute the magnitude of $$\mathbf{a} \times \mathbf{v}$$ $$|\mathbf{a} \times \mathbf{v}| = \sqrt{(2 \sqrt{7})^2 + 0^2 + (5 \sqrt{7})^2} = \sqrt{28 + 175} = \sqrt{203}$$ ### Step 5: Compute the magnitude of $$\mathbf{v}(t)$$ $$|\mathbf{v}(t)| = \sqrt{(-5 \sin t)^2 + (\sqrt{7} \cos t)^2 + (-2 \sin t)^2}$$ $$= \sqrt{25 \sin^2 t + 7 \cos^2 t + 4 \sin^2 t}$$ $$= \sqrt{(25 + 4) \sin^2 t + 7 \cos^2 t} = \sqrt{29 \sin^2 t + 7 \cos^2 t}$$ At this stage, to get a more specific curvature, you'd need the value of $$t$$, but you can proceed with this general form or substitute a specific $$t$$ value. Would you like further steps in the process or a specific $$t$$ for evaluation? Here are some follow-up questions for further exploration: 1. What is the interpretation of curvature in 3D space? 2. Can curvature ever be negative, and why or why not? 3. How does this method compare with other curvature formulas? 4. How does the cross product help determine curvature? 5. What are practical applications of curvature in physics or engineering? **Tip:** The cross product of vectors in curvature problems helps determine the "twist" or "bend" in the curve's motion.