# For <mstyle displaystyle="true"> f ( t ) </mrow>

For $f\left(t\right)=\left(\frac{{\left(\mathrm{ln}t\right)}^{2}}{t},{t}^{3}\right)$ what is the distance between f(2) and f(4)?
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Step 1
We have a parametric function giving us cartesian coordinates for the specified parameter:
$f\left(t\right)=\left(\frac{{\left(\mathrm{ln}t\right)}^{2}}{t},{t}^{3}\right)$
This with $t=2$ we get:
$f\left(2\right)=\left(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2},8\right)$
And with $t=4$ we get:
$f\left(4\right)=\left(\frac{{\left(\mathrm{ln}4\right)}^{2}}{4},64\right)$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left(\frac{{\left({\mathrm{ln}2}^{2}\right)}^{2}}{4},64\right)$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left(\frac{{\left(2\mathrm{ln}2\right)}^{2}}{4},64\right)$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left({\left(\mathrm{ln}2\right)}^{2},64\right)$
Step 2
The distance. d, between these two cartesian coordinates is given by pythagoras:
${d}^{2}={\left({\left(\mathrm{ln}2\right)}^{2}-\frac{{\left(\mathrm{ln}2\right)}^{2}}{2}\right)}^{2}+{\left(64-8\right)}^{2}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={\left(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2}\right)}^{2}+{\left(56\right)}^{2}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136$
Giving: $d=\sqrt{\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136}$