What is the smallest distance between the origin and a point on the graph of y = \frac{1}{\sqrt{2}} (x^2 - 13)

Guest Oct 17, 2022

#1**0 **

What is the smallest distance between the origin and a point on the graph?

**Hello Guest!**

\(y = \frac{1}{\sqrt{2}} (x^2 - 13)\\ y=\frac{x^2}{\sqrt{2}}-\frac{13}{\sqrt{2}}\\ y'=x\cdot\sqrt{2}\ |\ the\ perpendicular\ through\ the\ origin\ is:\\ y=-\frac{1}{\sqrt{2}}x\\ \frac{x^2}{\sqrt{2}}-\frac{13}{\sqrt{2}}=-\frac{1}{\sqrt{2}}x\ |\ \cdot \sqrt{2}\\ x^2+x-13=0\)

\(x\in \{-\frac{1}{2}-\frac{\sqrt{53}}{2},\color{blue}\frac{\sqrt{53}}{2}-\frac{1}{2}\}\)

The smallest distance between the origin and a point on the graph

\(y = \frac{1}{\sqrt{2}} (x^2 - 13)\)** is 3.14005. **Not correct! See answer 7#.

asinus Oct 17, 2022

#2**+2 **

Here are two possible methods.

1. Algebraic.

If (x, y) is a point on he curve, then its distance from the origin d, will be such that

\(d^{2}=x^{ 2}+y^{2}=x^{2}+(1/2)(x^{2}-13)^{2} \\ =x^{2}+(1/2)(x^{4}-26x^{2}+169)=(1/2)(x^{4}-24x^{2}+169) \\ =(1/2)(x^{4}-24x^{2}+144-144+169) \\ =(1/2)\{(x^{2}-12)^{2}+25\}.\)

The minimum value for d^2 (and so the minimum for d), occurs when x^2 = 12.

So,

\(d^{2}_{\text{min}}=25/2, \\ d_{\text{min}}=5\sqrt{2}/2 \approx 3.5355.\)

2 Calculus

\(d^{2}=x^{2}+y^{2}, \\ 2d\frac{d}{dx}(d)=2x+2y\frac{dy}{dx} \\ d\frac{dd}{dx} =x+\frac{1}{\sqrt{2}}(x^{2}-13)x\sqrt{2}=x(x^{2}-12).\)

The minimum for d will be when dd/dx = 0, i.e. when x^2 = 12.

\(x=\sqrt{12}\:\quad y=(1/\sqrt{2})(12-13)=-1/\sqrt{2}, \\ d^{2}=12+(1/2)=25/2 \)

as earlier, etc.

Guest Oct 18, 2022

#3**+1 **

It can be seen from this graph that guest answer is most likely correct.

asinus, I am sorry but your answer is not correct. (maybe it is just a careless error, I have not checked)

Melody Oct 19, 2022

#6**+1 **

The parabola and the circle should just touch.

The blue circle is your solution. As you can see your circl does not touch the parabola.

So the parabola is always more than 3.14 units from the origin.

Your answer is not correct.

The green circle is guests solution. As you can see, his/her cirlce does appear to be tangential to the parabola.

So the green answer appears to be correct.

Melody
Oct 20, 2022

#7**+2 **

I made an embarrassing mistake and I apologize.

Here is my correct solution:

\(y = \frac{1}{\sqrt{2}} (x^2 - 13)\\ r^2=x^2+y^2\\ r^2=x^2+\frac{1}{2}(x^2-13)^2\\ \frac{d(r^2)}{dx}=2x+\frac{1}{2}\cdot 2(x^2-13)\cdot 2x=0\\ 2x+2x^3-26x=0\\ 2x^3-24x=0\\ x(x^2-12)=0\)

\(x\in \{0,-\sqrt{12},{\color{blue}\sqrt{12}}\}\\ r^2=x^2+\frac{1}{2}(x^2-13)^2\\ r^2=12+\frac{1}{2}(12-13)^2\\ r^2=12.5\\ \color{blue}r=\sqrt{12.5}=3.5355 \)

**This is Guest 2#'s solution.**

**Sorry again!**

!

asinus
Oct 21, 2022