2.10 Q-11

Question Statement

Find the point on the curve $y = x^2 - 1$ that is closest to the point $(3, -1)$.

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Background and Explanation

This problem asks us to find the closest point on the curve $y = x^2 - 1$ to a given point $(3, -1)$. To solve this, we need to minimize the distance between the point on the curve and the given point using the distance formula. The task involves using calculus to find the value of $x$ that minimizes the distance, then finding the corresponding $y$-coordinate on the curve.

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Solution

1. Define the distance function: Let $l$ be the distance between a point $(x, y)$ on the curve $y = x^2 - 1$ and the point $(3, -1)$. Using the distance formula:

$$l = \sqrt{(x - 3)^2 + (y - (-1))^2}$$

Substituting $y = x^2 - 1$ into the equation:

$$l = \sqrt{(x - 3)^2 + (x^2 - x + 1)^2}$$

2. Simplify the distance function: Simplifying the terms inside the square root:

$$l = \sqrt{(x - 3)^2 + (x^2 - x + 1)^2}$$ $$l = \sqrt{(x - 3)^2 + x^4 - 2x^3 + 3x^2 - 2x + 1}$$

3. Differentiate the distance function: To find the value of $x$ that minimizes the distance, we differentiate $l$ with respect to $x$. Since $l$ involves a square root, we use the chain rule:

$$\frac{dl}{dx} = \frac{1}{2\sqrt{(x - 3)^2 + (x^2 - x + 1)^2}} \times \left( 2(x - 3) + 4x^3 - 2x^2 - 2x + 2 \right)$$

Simplifying:

$$\frac{dl}{dx} = \frac{1}{\sqrt{(x - 3)^2 + (x^2 - x + 1)^2}} \times \left( 2x^3 - 2x^2 + 2x - 6 \right)$$

4. Solve for the critical points: Set $\frac{dl}{dx} = 0$ to find the critical points:

$$2x^3 - 2x^2 + 2x - 6 = 0$$

Factor the equation:

$$(x - 1)(2x^2 + 2x + 3) = 0$$

This gives two possibilities:

• $x = 1$
• The quadratic $2x^2 + 2x + 3 = 0$ gives imaginary roots, so we discard this solution.

5. Determine the minimum: To confirm that $x = 1$ corresponds to a minimum, we check the sign of $\frac{dl}{dx}$ around $x = 1$:

   • For $x = 1 - \epsilon$ (where $\epsilon$ is a small positive number), $x - 1 = -\epsilon$, so $\frac{dl}{dx}$ is negative.
   • For $x = 1 + \epsilon$, $x - 1 = \epsilon$, so $\frac{dl}{dx}$ is positive.

   Since the derivative changes from negative to positive at $x = 1$, this is a local minimum.

6. Find the corresponding $y$-coordinate: Substituting $x = 1$ into the equation of the curve $y = x^2 - 1$:

$$y = (1)^2 - 1 = 0$$

Therefore, the point on the curve closest to $(3, -1)$ is $(1, 0)$.

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Key Formulas or Methods Used

• Distance formula: $l = \sqrt{(x - x_1)^2 + (y - y_1)^2}$
• First derivative: To find critical points and identify the minimum distance.
• Second derivative test: To confirm that $x = 1$ corresponds to a minimum.

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Summary of Steps

1. Define the distance function using the distance formula.
2. Simplify the distance function by substituting $y = x^2 - 1$.
3. Differentiate the distance function to find the critical points.
4. Solve for $x$ by setting the derivative equal to zero.
5. Check the second derivative or sign of the derivative to confirm a minimum at $x = 1$.
6. Calculate the corresponding $y$-coordinate on the curve: $y = 0$.
7. The closest point on the curve is $(1, 0)$.

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