Question

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

Answer 1

`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))`