For #f(t)= (t+21,-3t^2-2t)# what is the distance between #f(2)# and #f(5)#?

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Answer 1 · 2017-03-12T17:25:31

The distance is #=69.1#

#f(t)=(t+21,-3t^2-2t)#

We start by calculating

#f(2)=(23,-12-4)=(23,-16)#

#f(5)=(26,-75-10)=(26,-85)#

The distance between 2 points #(x_1,y_1)# and #(x_2,y_2)# is

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Therefore,

#d=sqrt((26-23)^2+(-85+16)^2)#

#=sqrt(3^2+69^2)#

#=sqrt(9+4761)#

#=sqrt4770#

#=69.1#