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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 - 18)

 Jul 12, 2022
 #1
avatar+13793 
0

What is the smallest distance?


Hello Guest!

 

The smallest distance is r (determined graphically with 4.179).

\(f(x) = \frac{1}{\sqrt{2}} (x^2 - 18)\\ g(x)= (r^2-x^2)^{0.5}\\ f'(x)=\frac{2x}{\sqrt 2}\\ g'(x)=0.5(r^2-x^2)^{-0,5}\cdot (-2x)\\ g'(x)=-\frac{x}{\sqrt{r^2-x^2}}\)

\(f'(x)=g'(x)\\ \sqrt 2\cdot x=-x(r^2-x^2)^{-0.5}\\ \sqrt{2}\cdot (r^2-x^2)^{0.5}=-1\)

\(-\sqrt{2}=\sqrt{r^2-x^2}\)

To be continued.

laugh  !

 Jul 12, 2022
edited by asinus  Jul 12, 2022
 #3
avatar+13793 
0

I was totally on the wrong track there! Thanks CPhill!

laugh  !

asinus  Jul 12, 2022
 #2
avatar+124525 
+2

Let the point  we are looking for  be  ( x  ,  (1/sqrt (2)) (x^2 -18) )

 

Using the square of the distance formula

 

D^2  = x^2  + [ (1/sqrt (2))(x^2 -18)]^2   

 

D^2 =  x^2  + (1/2) ( x^2 - 18)^2        take the derivative of  this and  set to 0

 

D^2'  =   2x  + 2x ( x^2 - 18)  = 0

 

2x ( x^2 - 18 + 1)  =  0

 

2x ( x^2 - 17)  = 0

 

The second factor set = 0  is what we want

 

x^2  - 17   =  0

 

x^2  = 17

 

x = sqrt (17)

 

y = -1/sqrt (2)

 

The shortest distance is

 

sqrt [  (sqrt (17))^2  + ( -1/sqrt (2))^2 ] =  sqrt [ 17 + 1/2 ]  =  sqrt (17.5) ≈   4.183 units

 

 

cool cool cool

 Jul 12, 2022

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