To find the equation of the circle of curvature of the given curve, we can break this problem down into a few steps.

The curve is given by the parameterization:

$r(t) = t \hat{i} + \sin(t) \hat{j}$

and we're tasked to find the circle of curvature at the point $\left( \frac{\pi}{2}, 1 \right)$. This curve corresponds to the graph of $y = \sin(x)$ in the $xy$-plane, parameterized by $t$.

The circle of curvature is characterized by its center (called the center of curvature) and its radius (the radius of curvature), which are related to the curvature $\kappa$ of the curve at a given point.

Steps to Solve:

1. Find the velocity vector $v(t) = \frac{dr}{dt}$: $v(t) = \frac{d}{dt}\left( t \hat{i} + \sin(t) \hat{j} \right) = \hat{i} + \cos(t) \hat{j}$

2. Find the acceleration vector $a(t) = \frac{d^2 r}{dt^2}$: $a(t) = \frac{d}{dt}(v(t)) = -\sin(t) \hat{j}$

3. Find the curvature $\kappa(t)$ using the formula: $\kappa(t) = \frac{|v(t) \times a(t)|}{|v(t)|^3}$ First, compute $|v(t)|$ and $|v(t) \times a(t)|$:

   • $|v(t)| = \sqrt{1^2 + \cos^2(t)} = \sqrt{1 + \cos^2(t)}$
   • $v(t) \times a(t)$ is a scalar value (since we are in 2D), and it results in: $v(t) \times a(t) = \det \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \cos(t) & 0 \\ 0 & -\sin(t) & 0 \end{vmatrix} = 1 \cdot 0 - \cos(t) \cdot (-\sin(t)) = \cos(t) \sin(t)$ Therefore, $|v(t) \times a(t)| = |\cos(t) \sin(t)|$.

   Thus, the curvature is: $\kappa(t) = \frac{|\cos(t) \sin(t)|}{(1 + \cos^2(t))^{3/2}}$

4. Evaluate the curvature at $t = \frac{\pi}{2}$: At $t = \frac{\pi}{2}$, we have: $\cos\left(\frac{\pi}{2}\right) = 0, \quad \sin\left(\frac{\pi}{2}\right) = 1$ Thus: $\kappa\left(\frac{\pi}{2}\right) = \frac{0 \cdot 1}{(1 + 0)^{3/2}} = 0$

   Therefore, at $t = \frac{\pi}{2}$, the curvature is $0$, which means the radius of curvature is infinite. This corresponds to a point of inflection on the curve, indicating no well-defined circle of curvature at that specific point.

This suggests that while the circle of curvature typically gives a local approximation to the curve, at the point $\left( \frac{\pi}{2}, 1 \right)$, the curve flattens out, so no finite circle can be constructed to match the curvature.

Would you like further details or assistance in exploring other aspects of curvature?

Here are 5 related questions to explore:

1. How is the center of curvature determined geometrically for a given curve?
2. What happens to the curvature and circle of curvature at inflection points of a curve?
3. Can curvature be interpreted differently in 3D space as opposed to 2D?
4. How does the radius of curvature relate to the second derivative of a function in general?
5. How can the Frenet-Serret formulas be used to study the curvature and torsion of space curves?

Tip: Curvature is a measure of how sharply a curve bends at a point. A higher curvature indicates a tighter bend. At points where curvature is zero, the curve is flat.