What is the smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 18)?

Answer 1

We can use the formula for the distance between a point $(x, y)$ and the origin $(0,0)$:

$$d = \sqrt{x^2 + y^2}$$

To minimize this distance, we can minimize $d^2 = x^2 + y^2$. We know that the point $(x,y)$ lies on the graph of $y = \frac{1}{2}(x^2 - 18)$, so we can substitute this into the expression for $d^2$:

$$d^2 = x^2 + y^2 = x^2 + \left(\frac{1}{2}(x^2 - 18)\right)^2$$

Simplifying this expression gives:

$$d^2 = x^2 + \frac{1}{4}(x^4 - 36x^2 + 324)$$

To find the minimum value of $d^2$, we can take the derivative with respect to $x$ and set it equal to 0:

$$\frac{d}{dx} (x^2 + \frac{1}{4}(x^4 - 36x^2 + 324)) = 2x + \frac{1}{4}(4x^3 - 72x) = 0$$

Simplifying this equation gives:

$$5x^3 - 36x = 0$$

Which factors as:

$$x(5x^2 - 36) = 0$$

The solutions are $x=0$ and $x=\pm\frac{6\sqrt{5}}{5}$. To determine which one gives the minimum distance, we can check the second derivative:

$$\frac{d^2}{dx^2} (x^2 + \frac{1}{4}(x^4 - 36x^2 + 324)) = 2 + 3x^2$$

At $x=0$, this is positive, so $x=0$ gives a minimum. The corresponding $y$ value is:

$$y = \frac{1}{2}(0^2 - 18) = -9$$

So the closest point on the graph to the origin is $\left(0, -9\right)$, and the distance between them is:

$$d = \sqrt{0^2 + (-9)^2} = \boxed{9}$$