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)

Related questions
See all questions in Introduction to Parametric Equations
Impact of this question
1612 views around the world
You can reuse this answer
Creative Commons License