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

Step 1: Distance Formula

The distance $d$ between the points $(x, y)$ and $(4, -1)$ is given by: $d = \sqrt{(x - 4)^2 + (y + 1)^2}$ Since $y = x^2$, we substitute $y = x^2$ into the distance formula: $d = \sqrt{(x - 4)^2 + (x^2 + 1)^2}$

Step 2: Minimize the Distance

To find the minimum distance, we need to minimize $d^2$ (since minimizing the square of the distance will also minimize the distance itself): $d^2 = (x - 4)^2 + (x^2 + 1)^2$ Expand and simplify this expression: $d^2 = (x^2 - 8x + 16) + (x^4 + 2x^2 + 1)$ $d^2 = x^4 + 3x^2 - 8x + 17$

Step 3: Find the Critical Points

To minimize $d^2$, take the derivative of $d^2$ with respect to $x$ and set it equal to zero: $\frac{d}{dx}(x^4 + 3x^2 - 8x + 17) = 4x^3 + 6x - 8$ Set the derivative equal to zero: $4x^3 + 6x - 8 = 0$

Step 4: Solve the Cubic Equation

This equation is cubic and can be solved numerically (or graphically). Using numerical methods or graphing, the solution closest to zero is found to be approximately $x \approx 1.29$.

Step 5: Find the Closest Point

Once we have the $x$-coordinate, plug it back into $y = x^2$ to find the corresponding $y$-coordinate: $y = (1.29)^2 \approx 1.66$ So the closest point on the curve to $(4, -1)$ is approximately $(1.29, 1.66)$.

Final Answer:

The coordinates of the closest point are approximately $(1.29, 1.66)$.

Would you like more details, or do you have any other questions?

Here are 5 related questions:

1. How would the process change if the curve was $y = x^3$ instead of $y = x^2$?
2. What happens if the external point is on the curve itself?
3. Can you find the distance between the closest point on the curve and the given external point?
4. What is the significance of the derivative when finding the minimum distance?
5. How can you use Lagrange multipliers to solve this problem?

Tip: When solving problems involving the minimum distance, always consider minimizing the square of the distance function to avoid dealing with square roots.