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What is the smallest distance between the origin and a point on the graph of $y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)?$

 

 

Edit:

This was not displaying properly for me.. SO

I added dollar signs and now it displays properly for me. I wonder if it displays properly for other people??  Any comments?

Guest Jan 12, 2018
edited by Melody  Jan 17, 2018
 #1
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\(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\)

 

The distance  is  given by :

 

√ [ x^2 +  (x^4 - 6x^2 + 9) / 2]   =

 

√  [ 2x^2  + x^4 - 6x^2  + 9] * √[ 1/2] =

 

√ [ x^4 - 4x^2  + 9] *  √[1/2 ]  =

 

[ x^4/2 - 2x^2  + 9/2 ] ^(1/2)

 

Taking the derivative of this and setting to 0 we have

 

[ 2x^3 - 4x] /  [  2 [ x^4/2 - 2x^2  + 9/2 ] ^(1/2)]  = 0

 

Multiply through by the denominator and we have that

 

[ 2x^3  - 4x]   =  0      factor

 

2x [ x^2 - 2]  =  0

 

Solving this for x  gives x = 0,  x  = √2  and x  = -√ 2

 

We could take the second derivative and determine which of these values gives a minimum, but it's messy.....a better approach  is to graph the given function  and  the circle x^2 + y^2  = 9/2  ....the reason for this is that when x  = 0 , the associated function point lies on the circle 

 

See the graph here : https://www.desmos.com/calculator/upxjjsafyd

 

But....when  x  = ± √ 2.....the associated points on the function  (±√ 2, - 1/ √ 2)  will fall inside this circle

 

So....using either point,  the minimum distance is 

 

√ [  2  +  1/2  ]  =

 

√ [  2.5 ] units   ≈   1.58 units

 

 

cool cool cool

CPhill  Jan 12, 2018

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