For #f(t)= (t/sqrt(t+1),t^2-t)# what is the distance between #f(0)# and #f(2)#?

1 Answer
Praveer Sharan
Nov 29, 2017

#(4sqrt(3))/3#

Explanation:

#f(0) = (0/(sqrt(0+1)), 0^2-0)#
#= (0, 0)#
#f(2) = (2/(sqrt(2+1)), 2^2-2)#
#= (2/sqrt(3), 2)#
Let's use the distance formula. In case you don't know what it is, here it is:
The distance #d# between #(a, b)# and #(c, d)# is:
#d = sqrt((d-b)^2 + (c-a)^2)#.
We can apply this to our problem. Using the distance formula,
#d = sqrt((2-0)^2 + (2/sqrt(3)-0)^2)#
#= sqrt((2^2) + (2/sqrt(3))^2)#
#=sqrt(4 + 4/3)#
#=sqrt(12/3 + 4/3)#
#=sqrt(16/3)#
#=4/sqrt(3)#
#=(4sqrt(3))/3# (if you want to rationalize the denominator)

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