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.