To find the coordinates of the point on the curve $y = \sqrt{x}$ that is closest to the point $(3, 0)$, we will use optimization techniques. Here is a step-by-step solution:

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Step 1: Define the distance formula

The distance $D$ between the point $(x, y)$ on the curve and $(3, 0)$ is given by: $D = \sqrt{(x - 3)^2 + y^2}.$ Since $y = \sqrt{x}$, substitute $y$: $D = \sqrt{(x - 3)^2 + (\sqrt{x})^2}.$ Simplify: $D = \sqrt{(x - 3)^2 + x}.$

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Step 2: Minimize the square of the distance

To simplify calculations, we minimize $D^2$ instead of $D$. Let: $f(x) = (x - 3)^2 + x.$ The value of $x$ that minimizes $f(x)$ will also minimize $D$.

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Step 3: Differentiate $f(x)$

Expand $f(x)$: $f(x) = (x - 3)^2 + x = x^2 - 6x + 9 + x = x^2 - 5x + 9.$ Now, compute the derivative: $f'(x) = 2x - 5.$

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Step 4: Solve $f'(x) = 0$ (First Derivative Test)

Set $f'(x) = 0$: $2x - 5 = 0 \implies x = \frac{5}{2}.$

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Step 5: Verify the critical point

To confirm that $x = \frac{5}{2}$ is a minimum, examine the second derivative: $f''(x) = 2.$ Since $f''(x) > 0$, the function $f(x)$ is concave up at $x = \frac{5}{2}$, confirming a local minimum.

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Step 6: Find the corresponding $y$-coordinate

The corresponding $y$-coordinate is: $y = \sqrt{x} = \sqrt{\frac{5}{2}}.$

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Final Answer

The coordinates of the point on the curve closest to $(3, 0)$ are: $\left( \frac{5}{2}, \sqrt{\frac{5}{2}} \right).$

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Would you like further explanation or assistance with plotting? 😊

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5 Follow-Up Questions:

1. How would the method change if the point $(3, 0)$ were on the negative $y$-axis?
2. Can you verify the solution by finding the exact value of the minimum distance?
3. What happens if the curve is changed to $y = x^2$?
4. How does the choice of optimizing $D^2$ instead of $D$ simplify the process?
5. Can we solve this problem using Lagrange multipliers? If so, how?

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One Tip:

Always check the concavity using the second derivative or verify endpoints for a closed interval optimization problem.