For #f(t)= (1/(2t-3), t/sqrt(t^2+1) )# what is the distance between #f(0)# and #f(1)#?

1 Answer
mason m
Mar 1, 2017

#sqrt17/(3sqrt2)#

Explanation:

Find the points, then use the distance formula to calculate the distance between them.

#f(t)=(x(t),y(t))#

The first part the parametric equation gives the #x# coordinate and the second part gives the #y# coordinate.

#f(0)=(1/(2(0)-3),0/sqrt(0^2+1))=(-1/3,0)#

#f(1)=(1/(2(1)-3),1/sqrt(1^2+1))=(-1,1/sqrt2)#

The distance between #(x_1,y_1)# and #(x_2,y_2)# is #d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)# so the distance between #f(0)# and #f(1)# is given by:

#d=sqrt((-1-(-1/3))^2+(1/sqrt2-0)^2)#

#d=sqrt((-2/3)^2+(1/sqrt2)^2)#

#d=sqrt(4/9+1/2)=sqrt(17/18)=sqrt17/(3sqrt2)#

Related questions
See all questions in Introduction to Parametric Equations
Impact of this question
1083 views around the world
You can reuse this answer
Creative Commons License