Key Concepts and Formulas

• Distance Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Finding Minima/Maxima using Derivatives: If $f'(x) = 0$ and $f''(x) > 0$ at a point $x$, then $f(x)$ has a local minimum at that point.
• Optimization: The process of finding the minimum or maximum value of a function.

Step-by-Step Solution

1. Define the Distance Function

We want to find the minimum distance from the origin $(0, 0)$ to a point $(x, y)$ on the curve $y = x^2 - 4$. The distance $d$ between $(0, 0)$ and $(x, x^2 - 4)$ is given by:

$d = \sqrt{(x - 0)^2 + (x^2 - 4 - 0)^2} = \sqrt{x^2 + (x^2 - 4)^2}$

To simplify the problem, we can minimize the square of the distance, $D = d^2$, instead of $d$, since the minimum of $d^2$ will occur at the same $x$-value as the minimum of $d$. Thus,

$D = x^2 + (x^2 - 4)^2 = x^2 + x^4 - 8x^2 + 16 = x^4 - 7x^2 + 16$

We are minimizing $D(x) = x^4 - 7x^2 + 16$. This simplifies the differentiation process.

2. Find the First Derivative and Critical Points

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

$D'(x) = \frac{dD}{dx} = 4x^3 - 14x$

Setting $D'(x) = 0$:

$4x^3 - 14x = 0$ $2x(2x^2 - 7) = 0$

This gives us three possible critical points: $x = 0$, $x = \sqrt{\frac{7}{2}}$, and $x = -\sqrt{\frac{7}{2}}$.

3. Find the Second Derivative and Check for Minima

To determine whether these critical points correspond to minima, we take the second derivative of $D(x)$:

$D''(x) = \frac{d^2D}{dx^2} = 12x^2 - 14$

Now we evaluate the second derivative at each critical point:

• For $x = 0$: $D''(0) = 12(0)^2 - 14 = -14 < 0$. This indicates a local maximum.

• For $x = \sqrt{\frac{7}{2}}$: $D''\left(\sqrt{\frac{7}{2}}\right) = 12\left(\frac{7}{2}\right) - 14 = 42 - 14 = 28 > 0$. This indicates a local minimum.

• For $x = -\sqrt{\frac{7}{2}}$: $D''\left(-\sqrt{\frac{7}{2}}\right) = 12\left(\frac{7}{2}\right) - 14 = 42 - 14 = 28 > 0$. This indicates a local minimum.

Since both $x = \sqrt{\frac{7}{2}}$ and $x = -\sqrt{\frac{7}{2}}$ yield minima, we can use either value to find the minimum distance.

4. Calculate the Minimum Distance

Let's use $x = \sqrt{\frac{7}{2}}$. Then $y = x^2 - 4 = \frac{7}{2} - 4 = \frac{7}{2} - \frac{8}{2} = -\frac{1}{2}$. The point on the curve is $\left(\sqrt{\frac{7}{2}}, -\frac{1}{2}\right)$.

The square of the distance is: $D = \left(\sqrt{\frac{7}{2}}\right)^4 - 7\left(\sqrt{\frac{7}{2}}\right)^2 + 16 = \left(\frac{7}{2}\right)^2 - 7\left(\frac{7}{2}\right) + 16 = \frac{49}{4} - \frac{49}{2} + 16 = \frac{49 - 98 + 64}{4} = \frac{15}{4}$

Thus, the minimum squared distance is $\frac{15}{4}$. Taking the square root to find the minimum distance $d$:

$d = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$

However, we made an error. Let's recalculate the value of $D$ using $D=x^2 + (x^2 - 4)^2$: If $x^2 = \frac{7}{2}$, then $D = \frac{7}{2} + (\frac{7}{2} - 4)^2 = \frac{7}{2} + (-\frac{1}{2})^2 = \frac{7}{2} + \frac{1}{4} = \frac{14+1}{4} = \frac{15}{4}$. Then $d = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

We are still not getting the correct answer. Let's go back to $D(x) = x^4 - 7x^2 + 16$. We have $x^2 = \frac{7}{2}$. So $x^4 = \frac{49}{4}$. Therefore, $D = \frac{49}{4} - 7(\frac{7}{2}) + 16 = \frac{49}{4} - \frac{98}{4} + \frac{64}{4} = \frac{15}{4}$. Then $d = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

Let's consider $x=0$. Then $y = 0^2 - 4 = -4$. The distance is $\sqrt{0^2 + (-4)^2} = 4$. Since $4 > \frac{\sqrt{15}}{2}$, the minimum distance is not at $x=0$.

There MUST be an error in the problem statement or correct answer. Let us reconsider the derivatives: $D(x) = x^4 - 7x^2 + 16$ $D'(x) = 4x^3 - 14x = 2x(2x^2 - 7) = 0$. So $x = 0, x = \pm \sqrt{\frac{7}{2}}$. $D''(x) = 12x^2 - 14$. $D''(0) = -14 < 0$. So $x=0$ is a local max. $D''(\pm \sqrt{\frac{7}{2}}) = 12(\frac{7}{2}) - 14 = 42 - 14 = 28 > 0$. So $x = \pm \sqrt{\frac{7}{2}}$ are local mins.

At $x = \sqrt{\frac{7}{2}}$, $y = x^2 - 4 = \frac{7}{2} - 4 = -\frac{1}{2}$. $d = \sqrt{x^2 + y^2} = \sqrt{\frac{7}{2} + \frac{1}{4}} = \sqrt{\frac{14+1}{4}} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

However, the given answer is $\frac{\sqrt{19}}{\sqrt{2}} = \sqrt{\frac{19}{2}}$. Let's assume THIS is the correct answer. Then $\frac{19}{2} = x^2 + (x^2-4)^2$. So $2x^4 - 14x^2 + 32 = 19$. Then $2x^4 - 14x^2 + 13 = 0$. $x^2 = \frac{14 \pm \sqrt{196 - 4(2)(13)}}{4} = \frac{14 \pm \sqrt{196 - 104}}{4} = \frac{14 \pm \sqrt{92}}{4} = \frac{14 \pm 2\sqrt{23}}{4} = \frac{7 \pm \sqrt{23}}{2}$.

If the correct answer is actually $\sqrt{\frac{19}{2}}$, then something is wrong with the problem statement. The calculations have been carefully checked and the minimum distance is indeed $\frac{\sqrt{15}}{2}$.

Common Mistakes & Tips

• Squaring the Distance: Remember that minimizing the square of the distance simplifies calculations without changing the location of the minimum.
• Second Derivative Test: Always check the second derivative to confirm whether a critical point is a minimum or maximum.
• Algebra Errors: Double-check all algebraic manipulations to avoid mistakes.

Summary

We found the minimum distance from the curve $y = x^2 - 4$ to the origin by defining a distance function, squaring it for simplicity, finding critical points using the first derivative, and confirming these points corresponded to minima using the second derivative. The minimum distance was calculated to be $\frac{\sqrt{15}}{2}$. While this result does not match the provided correct answer, the calculations have been thoroughly reviewed.

Final Answer

The final answer is $\boxed{\frac{\sqrt{15}}{2}}$, which corresponds to option (C) if the question or provided "correct answer" were actually asking for $\frac{\sqrt{15}}{2}$.