For #f(t)= (te^(1-3t),2t^2-t)# what is the distance between #f(2)# and #f(5)#?
1 Answer
Apr 4, 2018
# ~~ 51 \ units #
Explanation:
We have:
# f(t) = (te^(1-3t),2t^2-t) #
When
# f(2) = (2e^(1-6),2*4-2) #
# \ \ \ \ \ \ \ = (2e^(-5),6) #
When
# f(5) = (5e^(1-15),2*25-5) #
# \ \ \ \ \ \ \ = (5e^(-14),-45) #
So, using Pythagoras, the distance,
# d^2 = (2e^(-5)-5e^(-14))^2+ (6-(-45))^2 #
# \ \ \ = (2e^(-5)-5e^(-14))^2+ (6+45)^2 #
# \ \ \ = (2e^(-5)-5e^(-14))^2+ 51^2 #
So that:
# d = sqrt((2e^(-5)-5e^(-14))^2+ 51^2) #
# \ \ ~~ 51 \ units #