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

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Answer 1 · 2018-04-04T00:50:31

# ~~ 51 \ units #

We have:

# f(t) = (te^(1-3t),2t^2-t) #

When #t=2# we have:

# f(2) = (2e^(1-6),2*4-2) #
# \ \ \ \ \ \ \ = (2e^(-5),6) #

When #t=5# we have:

# f(5) = (5e^(1-15),2*25-5) #
# \ \ \ \ \ \ \ = (5e^(-14),-45) #

So, using Pythagoras, the distance, #d#, between these coordinates is given by:

# 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 #