To find the point on the curve $y = \sqrt{x - 2} + 1$ that is closest to the point $(4, 1)$, we need to minimize the distance between any point $(x, y)$ on the curve and the given point $(4, 1)$.

Step 1: Distance Formula

The distance $D$ between a point $(x, y)$ and $(4, 1)$ is given by:

$D = \sqrt{(x - 4)^2 + (y - 1)^2}$

Since $y = \sqrt{x - 2} + 1$, we substitute this into the distance formula:

$D(x) = \sqrt{(x - 4)^2 + (\sqrt{x - 2} + 1 - 1)^2}$

Simplify the expression:

$D(x) = \sqrt{(x - 4)^2 + (\sqrt{x - 2})^2}$

$D(x) = \sqrt{(x - 4)^2 + (x - 2)}$

$D(x) = \sqrt{x^2 - 8x + 16 + x - 2}$

$D(x) = \sqrt{x^2 - 7x + 14}$

Step 2: Minimizing the Distance

To find the minimum distance, we need to minimize $D(x)$. However, since the square root is a monotonically increasing function, we can instead minimize the squared distance $D^2(x) = x^2 - 7x + 14$.

Let:

$f(x) = x^2 - 7x + 14$

Step 3: Find the Critical Points

To find the critical points, we take the derivative of $f(x)$ with respect to $x$ and set it to zero:

$f'(x) = 2x - 7$

Set $f'(x) = 0$:

$2x - 7 = 0$

$x = \frac{7}{2} = 3.5$

Step 4: Verify if $x = 3.5$ is within the domain

The curve $y = \sqrt{x - 2} + 1$ is defined for $x - 2 \geq 0$ or $x \geq 2$. Since $3.5 \geq 2$, it is within the domain.

Step 5: Find the Corresponding $y$ and the Minimum Distance

Substitute $x = 3.5$ into the curve equation to find $y$:

$y = \sqrt{3.5 - 2} + 1 = \sqrt{1.5} + 1$

Thus, the point on the curve closest to $(4, 1)$ is $\left(3.5, \sqrt{1.5} + 1\right)$.

Now, calculate the minimum distance:

$D_{\text{min}} = \sqrt{(3.5 - 4)^2 + (\sqrt{1.5} + 1 - 1)^2}$

$D_{\text{min}} = \sqrt{(-0.5)^2 + (\sqrt{1.5})^2}$

$D_{\text{min}} = \sqrt{0.25 + 1.5}$

$D_{\text{min}} = \sqrt{1.75}$

$D_{\text{min}} \approx 1.32$

Conclusion

The point on the curve $y = \sqrt{x - 2} + 1$ closest to $(4, 1)$ is approximately $\left(3.5, \sqrt{1.5} + 1\right)$. The minimum distance between the point $(4, 1)$ and the curve is approximately 1.32 units.

Would you like a detailed breakdown of any part, or have any questions?

Related Questions

1. How do you find the distance between two points in a coordinate plane?
2. What is the domain of the function $y = \sqrt{x - 2} + 1$?
3. How can the derivative test determine the minimum or maximum of a function?
4. How do you calculate the square root of a non-perfect square manually?
5. What are other methods for minimizing functions in calculus?

Tip:

Always check the domain of the functions involved in problems to avoid undefined values and incorrect results.