For #f(t)= (t^3-t+1,t^2-t)# what is the distance between #f(2)# and #f(5)#?

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Answer 1 · 2017-06-20T06:59:37

The distance between #f(2)# and #f(5)# is #115.4#

As #f(t)=(t^3-t+1,t^2-t)#, we have

#f(2)=(2^3-2+1,2^2-2)# or #(7,2)# and

#f(5)=(5^3-5+1,5^2-5)# or #(121,20)#

Hence distance between the two is

#sqrt((121-7)^2+(20-2)^2)#

= #sqrt(114^2+18^2)#

= #sqrt(12996+324)#

= #sqrt13320#

= #6sqrt370=115.4# (upto #1dp#)