Question

For `f(t)= (sint,cos^2t/t)` what is the distance between `f(pi/4)` and `f(pi)`?

Answer 1

`sqrt(1/2+1/(pi^2)`

Explanation:

Here,
`f(t)=(sint,cos^2t/t)`
`=>f(pi/4)=(sin(pi/4),cos^2(pi/4)/(pi/4))=(1/sqrt(2),(1/2)/(pi/4))=(1/sqrt(2),2/pi)`
similarly,
`f(pi)=(sinpi,cos^2pi/pi)=(0,(-1)^2/pi)=(0,1/pi)`.
Distance Formula:
If `A(x_1,y_1) and B(x_2.y_2),` then
`d(A,B)=sqrt((x_1-x_2)^2+(x_2-y_2)^2)`
`=>d(f(pi/4),f(pi))=sqrt((1/sqrt(2)-0)^2+(2/pi-1/pi)^2)=sqrt(1/2+1/(pi^2)`