For #f(t)= (sint-cost,t^2)# what is the distance between #f(2)# and #f(5)#?

1 Answer
Oct 20, 2016

# sqrt(2-2cos7) #

Explanation:

# f(t) = (sint-cost,t^2) #

# f(2)=(sin2-cos2,4) #
# f(5)=(sin5-cos5,25) #

By Pythagoras, the difference between these points is;
distance = # sqrt((sin5-sin2)^2+(-cos5--cos2)^2) #
= # sqrt((sin5-sin2)^2+(-cos5+cos2)^2) #
= # sqrt(sin^2 5-2sin5sin2+sin^2 2+cos^2 5-2cos5cos2+cos^2 2) #
= # sqrt(sin^2 5+cos^2 5-2sin5sin2-2cos5cos2+cos^2 2+sin^2 2) #
= # sqrt(1-2sin5sin2-2cos5cos2+1) #
= # sqrt(2-2sin5sin2-2cos5cos2) #
= # sqrt(2-2(sin5sin2+cos5cos2)) #
= # sqrt(2-2cos(5+2)) #
= # sqrt(2-2cos(7)) #