# For f(t)= (t^2,t^3) what is the distance between f(1) and f(3)?

Nov 18, 2016

$\approx 27.2$ units

#### Explanation:

$f \left(1\right) = \left(1 , 1\right)$
$f \left(3\right) = \left(9 , 27\right)$

By Pythagoras, the distance, $d$, between these points is:

${d}^{2} = {\left(27 - 1\right)}^{2} + {\left(9 - 1\right)}^{2}$
$\therefore {d}^{2} = {26}^{2} + {8}^{2}$
$\therefore {d}^{2} = 676 + 64$
$\therefore {d}^{2} = 740$
$\therefore d = \sqrt{740} \approx 27.2$