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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.

Find an equation for the circle of curvature of the curve r(t) = t i + sin(t) j at the point (π/2, 1). The curve parameterizes the graph of y = sin(x) in the xy-plane.

Explore how to find the circle of curvature for the parametric curve r(t) = t i + sin(t) j at the point (π/2, 1). This problem involves calculating the curvature and using it to determine the radius of curvature, with detailed steps in velocity and acceleration vectors.

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