We are tasked with finding the coordinates of points on the graph of the parabola $y = x^2$ that are closest to the point $(10, \frac{1}{2})$. Let us solve this step by step.

Step 1: Distance Formula

The distance $d$ between any point $(x, x^2)$ on the parabola and $(10, \frac{1}{2})$ is given by: $d = \sqrt{(x - 10)^2 + \left(x^2 - \frac{1}{2}\right)^2}.$

To minimize $d$, we can minimize $d^2$ (to avoid dealing with the square root). Thus, we minimize the function: $f(x) = (x - 10)^2 + \left(x^2 - \frac{1}{2}\right)^2.$

Step 2: Expanding $f(x)$

Expand $f(x)$: $f(x) = (x - 10)^2 + \left(x^2 - \frac{1}{2}\right)^2.$ $f(x) = (x^2 - 20x + 100) + \left(x^4 - x^2 + \frac{1}{4}\right).$ $f(x) = x^4 + x^2 - x^2 - 20x + 100 + \frac{1}{4}.$ $f(x) = x^4 - 20x + 100.25.$

Step 3: Finding Critical Points

To minimize $f(x)$, find its derivative and set it to 0: $f'(x) = 4x^3 - 20.$ Set $f'(x) = 0$: $4x^3 - 20 = 0 \implies x^3 = 5 \implies x = \sqrt[3]{5}.$

Step 4: Verify the Second Derivative

The second derivative is: $f''(x) = 12x^2.$ Since $f''(x) > 0$ for all $x$, the critical point $x = \sqrt[3]{5}$ is a minimum.

Step 5: Compute Corresponding $y$-Value

The point on the parabola is $(x, x^2)$. For $x = \sqrt[3]{5}$: $y = (\sqrt[3]{5})^2.$

Thus, the closest point is: [ \left(\sqrt[3]{5}, (\sqrt[3]{5})^2\right). ]

If you'd like, I can provide numeric approximations or additional steps. Let me know! Here are some follow-up questions:

1. How is the derivative used to minimize the distance?
2. Would you like the exact numeric coordinates of the closest point?
3. How do second derivatives confirm a minimum?
4. What other methods could solve this problem (e.g., Lagrange multipliers)?
5. Would you like me to generalize this approach for other parabolas?

Tip: Always check the boundary points of the domain, if applicable, when solving minimization problems.