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

Question

Christopher Allers · Accepted Answer

The answer is #=sqrt2/2=0.707#

#cos(pi/4)=sqrt2/2#

#sin(pi/4)=sqrt2/2#

#sinpi=0#

#cospi=-1#

Let's calculate #f(pi/4)# and #f(pi)#

#f(t)=(sin^2t, cos^2t)#

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

#=((sqrt2/2)^2,(sqrt2/2)^2)#

#=(1/2,1/2)#

#f(pi)=(sin^2(pi), cos^2(pi))#

#=(0,1)#

The distance between 2 points #(x_1,y_1)# and #(x_2,y_2)# is

#d=sqrt((x_2-x_1)^2+y_2-y_1)^2)#

The distance is

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

#=sqrt(1/4+1/4)#

#=sqrt2/2=0.707#

Luke Alvarez · Answer

undefined To find the distance between the points $ f(\frac{\pi}{4}) $ and $ f(\pi) $ for the function $ f(t) = (\sin^2 t, \cos^2 t) $, we can use the distance formula in two dimensions:

$$ \text{Distance} = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } $$

Given $ f(t) = (\sin^2 t, \cos^2 t) $, we have:

$$ f\left(\frac{\pi}{4}\right) = \left(\sin^2 \left(\frac{\pi}{4}\right), \cos^2 \left(\frac{\pi}{4}\right)\right) = \left(\frac{1}{2}, \frac{1}{2}\right) $$

and

$$ f(\pi) = \left(\sin^2 \pi, \cos^2 \pi\right) = (0, 1) $$

Now, plug the values into the distance formula:

$$ \text{Distance} = \sqrt{ \left(0 - \frac{1}{2}\right)^2 + \left(1 - \frac{1}{2}\right)^2 } $$

$$ = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} $$

$$ = \sqrt{\frac{1}{4} + \frac{1}{4}} $$

$$ = \sqrt{\frac{1}{2}} $$

$$ = \frac{\sqrt{2}}{2} $$

So, the distance between $ f\left(\frac{\pi}{4}\right) $ and $ f(\pi) $ is $ \frac{\sqrt{2}}{2} $.