For #f(t)= (lnt/e^t, te^t)# what is the distance between #f(1)# and #f(2)#?

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Answer 1 · 2017-04-14T20:28:19

#d~~12.06#

First, plug in #1#:

#f(1)=(ln(1)/e^1,1(e^1))#

#f(1)=(0,e^1)#

Now plug in #2#:

#f(2)=(ln(2)/e^2,2(e^2))#

Now use the distance formula:

#d=sqrt((ln(2)/e^2-0)^2+(2e^2-e^1)^2)#

#d~~12.06#