For #f(t)= ((t+3)^2,t^3-t)# what is the distance between #f(2)# and #f(4)#?

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Answer 1 · 2018-05-05T13:47:32

The distance
#D= sqrt{(49-25)^2 + (60-6)^2}=sqrt[576+2916]=sqrt3492=59.093#

i am not sure , but any way

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

#f(2)= ((2+3)^2,(2)^3-2)=(25,6)#

#f(4)= ((4+3)^2,(4)^3-4)=(49,60)#

The distance between two points given by

point1=#(25,6)#

point2=#(49,60)#

#D= sqrt{(x_2 -x_1)^2 + (y_2- y_1)^2}#

#D= sqrt{(49-25)^2 + (60-6)^2}=sqrt[576+2916]=sqrt3492=59.093#