For #f(t)= (te^(t-1),t^2-t+1)# what is the distance between #f(0)# and #f(2)#?
1 Answer
Jul 5, 2016
Find the x,y coordinates of the two points at t=0 and t=2, then find the distance between them (
Explanation:
#x=te^(t-1)#
#y=t^2-t+1# - At t=0
#x=0*e^-1=0#
#y=0-0+1=1#
#:.f(0)=(0,1)# - At t=2
#x=2*e^(2-1)=5.44#
#y=4-2+1=3#
#:.f(2)=(5.44,3)# - The distance between
#f(0)# and#f(2)# is
#sqrt((5.44-0)^2+(3-1)^2)=sqrt(33.59)#
#=5.80#
