To compute the curvature $\kappa$ of the curve defined by $r(t) = \left( e^t \sqrt{2}, t \sqrt{2} \right)$ at $t = 0$, we follow these steps:

1. Define the position vector $r(t)$: $r(t) = \left( x(t), y(t) \right) = \left( e^t \sqrt{2}, t \sqrt{2} \right)$

2. Compute the first derivatives: $x'(t) = \frac{d}{dt}(e^t \sqrt{2}) = e^t \sqrt{2}$ $y'(t) = \frac{d}{dt}(t \sqrt{2}) = \sqrt{2}$

3. Compute the second derivatives: $x''(t) = \frac{d}{dt}(e^t \sqrt{2}) = e^t \sqrt{2}$ $y''(t) = \frac{d}{dt}(\sqrt{2}) = 0$

4. Curvature formula: The curvature $\kappa$ for a parametrized curve $r(t) = (x(t), y(t))$ is given by: $\kappa = \frac{|x' y'' - y' x''|}{\left( (x')^2 + (y')^2 \right)^{3/2}}$

5. Substitute values at $t = 0$: At $t = 0$:

   • $x'(0) = e^0 \sqrt{2} = \sqrt{2}$
   • $y'(0) = \sqrt{2}$
   • $x''(0) = e^0 \sqrt{2} = \sqrt{2}$
   • $y''(0) = 0$

   Plugging these into the formula:

   = \frac{|0 - 2|}{\left( 2 + 2 \right)^{3/2}} = \frac{2}{(4)^{3/2}}$$

6. Simplify: $\kappa = \frac{2}{8} = \frac{1}{4} = 0.25$

Answer: The curvature $\kappa$ at $t = 0$ is $0.25$.

Would you like more details or further explanation on any steps?

Here are some related questions that might interest you:

1. How would the curvature change if the function were scaled by a constant factor?
2. Can curvature be negative, and under what circumstances?
3. What is the physical interpretation of curvature for this curve?
4. How does the formula for curvature differ in three-dimensional space?
5. What is the curvature at points other than $t = 0$ for this curve?

Tip: The curvature of a curve provides insight into how sharply it bends at a particular point, with larger values indicating tighter curves.