For #f(t)= (lnt/t, t^2/e^t)# what is the distance between #f(2)# and #f(4)#?

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Answer 1 · 2017-04-24T01:35:27

#(4(e^2-4))/e^4 ##

We have:

# f(t)= (lnt/t, t^2/e^t)#

Put #t=2#:

# f(2)= (ln2/2, 4/e^2)#

Put #t=4#:

# f(4)= (ln4/4, 16/e^4)#
# \ \ \ \ \ \ \= (ln2^2/4, 16/e^4)#
# \ \ \ \ \ \ \= (2ln2/4, 16/e^4)#
# \ \ \ \ \ \ \= (ln2/2, 16/e^4)#

By Pythagoras:

# d^2 = Delta x^2 + Delta y^2 #
# \ \ \ = (0)^2 + (4/e^2-16/e^4)^2 #

# :. d = 4/e^2-16/e^4 #
# \ \ \ \ \ \ \ = (4(e^2-4))/e^4 #