What is the smallest distance between the origin and a point on the graph of $y=(x^2-3)/(sqrt{2})
I read that as
y = (x^2 - 3)/√2
dy/dx = (1/√2)(2x) = 2x/√2
let that closest point be P(x,y)
so the slope of the tangent at P = 2x/√2
slope of PO = y/x
at the closest point, PO is perpendicular to the tangent at P
then,
y/x = -√2/(2x)
y = - √2/2
for the x:
(x^2-3)/√2 = -√2/2
x^2 - 3 = -1
x^2 = 2
x = ± √2
looking at my sketch , I will pick
(√2 , -√2/2)
To find the smallest distance between the origin and a point on the graph of $y=\frac{x^2-3}{\sqrt{2}}$, we need to find the point on the graph that is closest to the origin.
The distance between any point $(x, y)$ on the graph and the origin $(0, 0)$ can be found using the distance formula:
$d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$
Now, we can substitute the equation of the graph into the distance formula:
$d = \sqrt{x^2 + \left(\frac{x^2-3}{\sqrt{2}}\right)^2} = \sqrt{x^2 + \frac{x^4 - 6x^2 + 9}{2}}$
To find the smallest distance, we need to minimize this distance function. To find the minimum of a function, we can take the derivative and set it equal to zero:
$\frac{d}{dx} \left(\sqrt{x^2 + \frac{x^4 - 6x^2 + 9}{2}}\right) = 0$
Taking the derivative of the function, we use the chain rule:
$\frac{x}{\sqrt{x^2 + \frac{x^4 - 6x^2 + 9}{2}}} + \frac{2x^3 - 12x}{2\sqrt{x^2 + \frac{x^4 - 6x^2 + 9}{2}}} = 0$
To solve this equation for $x$, we can simplify the equation by multiplying through by the denominator:
$x + x^3 - 6x = 0$
Combining like terms:
$x^3 - 5x = 0$
Factoring out an $x$:
$x(x^2 - 5) = 0$
Setting each factor equal to zero:
$x = 0$ or $x^2 - 5 = 0$
Solving for $x$, we have two possible values:
$x = 0$ or $x = \sqrt{5}$
Now, we substitute these values back into the distance function to find the corresponding $y$ values:
For $x = 0$, we get $y = \frac{(0^2 - 3)}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$
For $x = \sqrt{5}$, we get $y = \frac{(\sqrt{5}^2 - 3)}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
So, we have two candidate points: $(0, -\frac{3}{\sqrt{2}})$ and $(\sqrt{5}, \sqrt{2})$
To determine which point is closest to the origin, we can compare their distances:
For $(0, -\frac{3}{\sqrt2})$, the distance is $d_1 = \sqrt{0^2 + \left(-\frac{3}{\sqrt{2}}\right)^2} = \frac{3}{\sqrt{2}}$
For $(\sqrt{5}, \sqrt{2})$, the distance is $d_2 = \sqrt{\left(\sqrt{5}\right)^2 + \left(\sqrt{2}\right)^2} = \sqrt{5 + 2} = \sqrt{7}$
Comparing $d_1$ and $d_2$, we can see that $d_1 < d_2$.
Therefore, the smallest distance between the origin and a point on the graph of $y=\frac{x^2-3}{\sqrt{2}}$ is $\frac{3}{\sqrt{2}}$.