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

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Answer 1 · 2016-01-02T13:30:42

The Euclidean distance can be used. (A calculator will be needed)

#d(x,y,z,...)=sqrt(Δx^2+Δy^2+Δz^2+...)#

The distance is 0.9618565

First, we need to find the exact points:

#f(1)=(ln1/e^1,e^1/1)#

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

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

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

The Euclidean distance can generally be calculated through this formula:

#d(x,y,z,...)=sqrt(Δx^2+Δy^2+Δz^2+...)#

Where Δx, Δy, Δz are the differences in each space (axis). Therefore:

#d(1,2)=sqrt((0-ln2/e^2)^2+(e-e^2/2)^2)#

#d(1,2)=sqrt(0.0087998+0.953056684)#

#d(1,2)=0.9618565#