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

1 Answer
Alan P.
Apr 19, 2016

#2/3sqrt(10)#

Explanation:

If #f(t)=(sqrt(t)/(t+1),t^2-t)#
then
#color(white)("XXX")f(0)=(sqrt(0)/(0+1),0^2-0) =(0,0)#

#color(white)("XXX")f(2)=(sqrt(2)/(2+1),2^2-2)=(sqrt(2)/3,2)#

The distance between #f(0)# and #f(2)# can be calculated using the Pythagorean Theorem
#color(white)("XXX")d=sqrt((sqrt(2)/3-0)^2+(2-0)^2)#

#color(white)("XXX")=sqrt(4/9+4)#

#color(white)("XXX")=sqrt(40/9)#

#color(white)("XXX")=2/3sqrt(10)#

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