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

 Oct 17, 2022
 #1
avatar+14087 
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#.

laugh

 Oct 17, 2022
edited by asinus  Oct 20, 2022
edited by asinus  Oct 21, 2022
 #2
avatar
+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.

 Oct 18, 2022
 #3
avatar+118132 
+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)

 

 Oct 19, 2022
 #4
avatar+14087 
0

Melody, I think my answer is correct.

The second line in your post proves that.

\(x^2+y^2=3.14^2\)

The exact value is: \(3.14005494464\)

You switched colors. Not correct! See answer 7#.

laugh  !

asinus  Oct 20, 2022
edited by asinus  Oct 20, 2022
edited by asinus  Oct 21, 2022
 #6
avatar+118132 
+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
avatar+14087 
+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!

laugh  !

asinus  Oct 21, 2022
 #8
avatar+118132 
0

Nice work asinus.  And guest too of course. :)

Melody  Oct 21, 2022

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