What is the smallest distance between the origin and a point on the graph of y = \frac{1}{\sqrt{2}} (x^2 - 18)
+14915 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.
!
+128475 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
