For #f(t)= (sin^2t,t/pi-2)# what is the distance between #f(pi/4)# and #f(pi)#?

1 Answer
mason m
Oct 25, 2016

#sqrt13/4#

Explanation:

Find the two points by plugging in #pi/4# and #pi# for #t# in the point #f(t)#:

#f(pi/4)=(sin^2(pi/4),(pi/4)/pi-2)#

#color(white)(f(pi/4))=((sqrt2/2)^2,1/4-2)#

#color(white)(f(pi/4))=(2/4,1/4-8/4)#

#color(white)(f(pi/4))=(1/2,-7/4)#

And

#f(pi)=(sin^2(pi),pi/pi-2)#

#color(white)(f(pi))=(0^2,1-2)#

#color(white)(f(pi))=(0,-1)#

So, we need to find the distance between the points #(1/2,-7/4)# and #(0,-1)# using the distance formula, which states that 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)#

Thus the distance we want is

#d=sqrt((0-1/2)^2+(-1-(-7/4))^2)#

#color(white)d=sqrt((-1/2)^2+(-1+7/4)^2)#

#color(white)d=sqrt((-1/2)^2+(3/4)^2)#

#color(white)d=sqrt(1/4+9/16)#

#color(white)d=sqrt(13/16)#

#color(white)d=sqrt13/4#

Related questions
See all questions in Introduction to Parametric Equations
Impact of this question
1401 views around the world
You can reuse this answer
Creative Commons License