$$  $$Find the minimum distance from the curve y = x to the point (1, 0). Test your critical value in the second derivative to show it is a minimum.

To find the minimum distance from the curve $ y = x $ to the point $ (1, 0) $, we need to minimize the distance function between any point $(x, y)$ on the curve and the point $(1, 0)$. Let's proceed with the solution step by step.

### 1. Distance Formula
The distance $ D $ between a point $(x, y)$ on the curve and the point $(1, 0)$ is given by:
\[
D = \sqrt{(x - 1)^2 + (y - 0)^2}
\]
Since the point $(x, y)$ lies on the curve $ y = x $, we can substitute $ y = x $ into the distance formula:
\[
D(x) = \sqrt{(x - 1)^2 + (x)^2} = \sqrt{(x - 1)^2 + x^2}
\]
### 2. Simplify the Distance Function
Simplify the expression under the square root:
\[
D(x) = \sqrt{(x - 1)^2 + x^2} = \sqrt{(x^2 - 2x + 1) + x^2} = \sqrt{2x^2 - 2x + 1}
\]
### 3. Minimize the Distance Function
To minimize $ D(x) $, it is easier to minimize the square of the distance function $ D(x)^2 $, as it avoids dealing with the square root:
\[
D(x)^2 = 2x^2 - 2x + 1
\]
We need to find the critical points by taking the derivative of $ D(x)^2 $ with respect to $ x $ and setting it to zero.

### 4. First Derivative
Find the first derivative of $ D(x)^2 $:
\[
\frac{d}{dx}[D(x)^2] = \frac{d}{dx}[2x^2 - 2x + 1] = 4x - 2
\]
Set the derivative equal to zero to find the critical points:
\[
4x - 2 = 0 \quad \Rightarrow \quad x = \frac{1}{2}
\]

### 5. Second Derivative Test
To confirm that this critical point is a minimum, we take the second derivative of $ D(x)^2 $ and check its sign at $ x = \frac{1}{2} $:
\[
\frac{d^2}{dx^2}[D(x)^2] = \frac{d}{dx}[4x - 2] = 4
\]
Since the second derivative is positive ($ 4 > 0 $), the function $ D(x)^2 $ has a local minimum at $ x = \frac{1}{2} $.

### 6. Conclusion
The minimum distance occurs when $ x = \frac{1}{2} $. The corresponding point on the curve is $ \left(\frac{1}{2}, \frac{1}{2}\right) $. To find the minimum distance, substitute $ x = \frac{1}{2} $ into the original distance formula:
\[
D\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2} - 1\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707
\]
Thus, the minimum distance from the curve $ y = x $ to the point $ (1, 0) $ is $ \frac{1}{\sqrt{2}} $.

Would you like more details on any step?

### Relative Questions:
1. What is the geometric interpretation of the result?
2. How would you find the minimum distance if the curve was $ y = x^2 $?
3. Can you generalize the method to find the minimum distance from any point to any line?
4. How does the second derivative test confirm a minimum?
5. What would the distance function look like for a 3D problem?
6. What happens if you consider the distance from the point to the curve in a different metric, like Manhattan distance?
7. How can you verify this result using a different method, like Lagrange multipliers?
8. How would the problem change if the point was outside the first quadrant?

### Tip:
Always check the second derivative when minimizing or maximizing a function to confirm the nature of the critical points.

Learn how to minimize the distance from the curve y = x to the point (1, 0) using calculus. Detailed step-by-step solution includes finding critical points, using the second derivative test, and calculating the minimum distance. Suitable for advanced high school students and calculus enthusiasts.

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