For #f(t)= (1/t,-1/t^2)# what is the distance between #f(2)# and #f(5)#?

Question

Emma Aldridge · Accepted Answer

#(3sqrt149)/100approx0.3662#

Find the two points described by #f(2)# and #f(5)#.

#f(2)=(1/2,-1/(2^2))=(1/2,-1/4)#

#f(5)=(1/5,-1/5^2)=(1/5,-1/25)#

Use the distance formula to find the distance between the points.

#d=sqrt((1/2-1/5)^2+(-1/4-(-1/25))^2)#

#=sqrt((5/10-2/10)^2+(-25/100+4/100)^2)#

#=sqrt((3/10)^2+(-21/100)^2)#

#=sqrt(9/100+441/10000)#

#=sqrt((900+441)/10000#

#=sqrt1341/100=(3sqrt149)/100approx0.3662#

Scarlett Allison · Answer

undefined To find the distance between $ f(2) $ and $ f(5) $ for the function $ f(t) = \left(\frac{1}{t}, -\frac{1}{t^2}\right) $, we first need to find the values of $ f(2) $ and $ f(5) $, and then calculate the distance between these two points.

When $ t = 2 $, $ f(2) = \left(\frac{1}{2}, -\frac{1}{2^2}\right) = \left(\frac{1}{2}, -\frac{1}{4}\right) $.

When $ t = 5 $, $ f(5) = \left(\frac{1}{5}, -\frac{1}{5^2}\right) = \left(\frac{1}{5}, -\frac{1}{25}\right) $.

Now, we use the distance formula to find the distance between these two points:

$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
$$ \text{Distance} = \sqrt{\left(\frac{1}{5} - \frac{1}{2}\right)^2 + \left(-\frac{1}{25} + \frac{1}{4}\right)^2} $$
$$ \text{Distance} = \sqrt{\left(\frac{2}{10} - \frac{5}{10}\right)^2 + \left(-\frac{1}{25} + \frac{5}{25}\right)^2} $$
$$ \text{Distance} = \sqrt{\left(-\frac{3}{10}\right)^2 + \left(\frac{4}{25}\right)^2} $$
$$ \text{Distance} = \sqrt{\frac{9}{100} + \frac{16}{625}} $$
$$ \text{Distance} = \sqrt{\frac{5625 + 1600}{62500}} $$
$$ \text{Distance} = \sqrt{\frac{7225}{62500}} $$
$$ \text{Distance} = \frac{85}{250} $$
$$ \text{Distance} = 0.34 $$ (rounded to two decimal places)

So, the distance between $ f(2) $ and $ f(5) $ is approximately 0.34 units.