Find the minimum distance between their centers.

The equation for the square of distance I came up with in the previous post was:
dist^2 = (225 - 480t + 256t^2)/(256 - 512t + 256t^2) + t^2

I used symbolab.com to do the polynomial multiplication

(dist^2)' = ((225 - 480t + 256t^2)'(256 - 512t + 256t^2) - (225 - 480t + 256t^2)(256 - 512t + 256t^2)') /(256 - 512t + 256t^2)^2 + 2t = 0

(dist^2)' = ((-480 + 512t)(256 - 512t + 256t^2) - (225 - 480t + 256t^2)(-512 + 512t)) / (256 - 512t + 256t^2)^2 + 2t = 0

=( (131072t^3-385024t^2+376832t-122880) - (-115200+360960t-376832t^2+131072t^3) )/ (256 - 512t + 256t^2)^2 + 2t = 0

=(-8192t^2+15872t-7680)/ (256 - 512t + 256t^2)^2 + 2t = 0
=((-8192t^2+15872t-7680)  + 2t(256 - 512t + 256t^2)^2  )  / (256 - 512t + 256t^2)^2   = 0
=((-8192t^2+15872t-7680)  + (131072t^5-524288t^4+786432t^3-524288t^2+131072t) )  / (256 - 512t + 256t^2)^2   = 0

=( 131072t^5-524288t^4+786432t^3-532480t^2+146944t-7680  )  / (256 - 512t + 256t^2)^2   = 0

131072t^5-524288t^4+786432t^3-532480t^2+146944t-7680 = 0

Symbolab finds 5 roots:
~0.06698, ~0.74999, ~0.93301, ~0.99999, 5/4

The first of these corresponds to the previous answer.

Or, ( 131072t^5-524288t^4+786432t^3-532480t^2+146944t-7680  )  / (256 - 512t + 256t^2)^2
simplifies to (4t-3)(4t-5)(16t^2 - 16t + 1)/(128(t-1)^3)  
The numerator of which has zeros at:
3/4, 5/4, (2 ± √3)/4
and (2 - √3)/4 corresponds to the previous answer ~0.06698

Posted by Larry on 2023-05-14 13:17:31