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

Question

Isabella Ackerman · Accepted Answer

#sqrt13/4#

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#

Luke Alvarez · Answer

undefined To find the distance between $ f(\frac{\pi}{4}) $ and $ f(\pi) $, we can use the distance formula in two dimensions, which is the distance between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ given by $ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $.

For $ f(\frac{\pi}{4}) $, $ t = \frac{\pi}{4} $, so $ f(\frac{\pi}{4}) = (\sin^2(\frac{\pi}{4}), \frac{\frac{\pi}{4}}{\pi} - 2) $.

For $ f(\pi) $, $ t = \pi $, so $ f(\pi) = (\sin^2(\pi), \frac{\pi}{\pi} - 2) $.

$ \sin^2(\frac{\pi}{4}) = \sin^2(\frac{\pi}{4}) = (\frac{\sqrt{2}}{2})^2 = \frac{1}{2} $.

Now, calculate the distance using the formula:

$ d = \sqrt{(\sin^2(\pi) - \sin^2(\frac{\pi}{4}))^2 + (\frac{\pi}{\pi} - 2 - (\frac{\frac{\pi}{4}}{\pi} - 2))^2} $.

Substitute the values:

$ d = \sqrt{(\frac{0}{1} - \frac{1}{2})^2 + (1 - 2 - (\frac{1}{4} - 2))^2} $.

$ d = \sqrt{(\frac{-1}{2})^2 + (1 - 2 - (\frac{1}{4} - 2))^2} $.

$ d = \sqrt{(\frac{1}{4}) + (-1 - (\frac{-7}{4}))^2} $.

$ d = \sqrt{(\frac{1}{4}) + (-1 + \frac{7}{4})^2} $.

$ d = \sqrt{(\frac{1}{4}) + (\frac{3}{4})^2} $.

$ d = \sqrt{(\frac{1}{4}) + (\frac{9}{16})} $.

$ d = \sqrt{\frac{4}{16} + \frac{9}{16}} $.

$ d = \sqrt{\frac{13}{16}} $.

$ d = \frac{\sqrt{13}}{4} $.

So, the distance between $ f(\frac{\pi}{4}) $ and $ f(\pi) $ is $ \frac{\sqrt{13}}{4} $.