2.10 Q-12

Question Statement

Find the point on the curve $y = x^2 + 1$ that is closest to the point $(18, 1)$.

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Background and Explanation

This problem asks us to determine the point on the curve $y = x^2 + 1$ that is closest to a given point $(18, 1)$. To solve this, we will use the distance formula to express the distance between any point on the curve and the point $(18, 1)$. By minimizing this distance, we can find the point on the curve that is closest to $(18, 1)$. We will need to use calculus to minimize the distance function.

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Solution

1. Define the distance function: The distance $l$ between a point $(x, y)$ on the curve $y = x^2 + 1$ and the point $(18, 1)$ is given by the distance formula:

$$l = \sqrt{(x - 18)^2 + (y - 1)^2}$$

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

$$l = \sqrt{(x - 18)^2 + (x^2 + 1 - 1)^2}$$

Simplifying:

$$l = \sqrt{(x - 18)^2 + x^4}$$

2. Simplify the distance function: Now, we need to differentiate this function to minimize the distance. First, let’s rewrite the expression:

$$l = \sqrt{(x - 18)^2 + x^4}$$

3. Differentiate the distance function: To find the minimum, we need to differentiate $l$ with respect to $x$. First, we square both sides to avoid the square root:

$$l^2 = (x - 18)^2 + x^4$$

Differentiate both sides:

$$\frac{d}{dx}(l^2) = \frac{d}{dx}\left((x - 18)^2 + x^4\right)$$

This gives:

$$2l \cdot \frac{dl}{dx} = 2(x - 18) + 4x^3$$

To find the critical points, we solve $\frac{dl}{dx} = 0$, which simplifies to:

$$2(x - 18) + 4x^3 = 0$$

Simplifying further:

$$4x^3 + 2x - 36 = 0$$

4. Solve for $x$: We now solve the cubic equation $4x^3 + 2x - 36 = 0$. First, check if $x = 2$ is a root by substituting into the equation:

$$4(2)^3 + 2(2) - 36 = 0$$

This simplifies to:

$$32 + 4 - 36 = 0$$

So, $x = 2$ is a root.

5. Factor the cubic equation: Since $x = 2$ is a root, we can factor $4x^3 + 2x - 36$ as $(x - 2)(4x^2 + 8x + 18) = 0$. Now, solve for $x$:

$$x - 2 = 0 \quad \Rightarrow \quad x = 2$$

The quadratic $4x^2 + 8x + 18 = 0$ has no real solutions because its discriminant is negative. So, we discard this factor.

6. Confirm that $x = 2$ is a minimum: To confirm that $x = 2$ corresponds to a minimum, we check the second derivative of $l^2$:

$$\frac{d^2}{dx^2} = 12x^2 + 2$$

Substituting $x = 2$ into this gives:

$$\frac{d^2}{dx^2} = 12(2)^2 + 2 = 48 + 2 = 50$$

Since the second derivative is positive, $x = 2$ corresponds to a minimum.

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

$$y = (2)^2 + 1 = 4 + 1 = 5$$

Therefore, the point on the curve closest to $(18, 1)$ is $(2, 5)$.

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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 minimize the distance.
• Second derivative test: To confirm that $x = 2$ corresponds to a minimum.

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Summary of Steps

1. Define the distance function using the distance formula.
2. Substitute $y = x^2 + 1$ into the distance formula.
3. Differentiate the distance function and set the derivative equal to zero to find critical points.
4. Solve the cubic equation and find $x = 2$ as the root.
5. Confirm that $x = 2$ is a minimum by checking the second derivative.
6. Substitute $x = 2$ into the equation of the curve to find $y = 5$.
7. The closest point on the curve is $(2, 5)$.

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