Frank Day

2022-07-15

For $f\left(t\right)=(\frac{{\left(\mathrm{ln}t\right)}^{2}}{t},{t}^{3})$ what is the distance between f(2) and f(4)?

Freddy Doyle

Beginner2022-07-16Added 20 answers

Step 1

We have a parametric function giving us cartesian coordinates for the specified parameter:

$f\left(t\right)=(\frac{{\left(\mathrm{ln}t\right)}^{2}}{t},{t}^{3})$

This with $t=2$ we get:

$f\left(2\right)=(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2},8)$

And with $t=4$ we get:

$f\left(4\right)=(\frac{{\left(\mathrm{ln}4\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=(\frac{{\left({\mathrm{ln}2}^{2}\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=(\frac{{\left(2\mathrm{ln}2\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=({\left(\mathrm{ln}2\right)}^{2},64)$

Step 2

The distance. d, between these two cartesian coordinates is given by pythagoras:

$d}^{2}={({\left(\mathrm{ln}2\right)}^{2}-\frac{{\left(\mathrm{ln}2\right)}^{2}}{2})}^{2}+{(64-8)}^{2$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={\left(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2}\right)}^{2}+{\left(56\right)}^{2}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136$

Giving: $d=\sqrt{\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136}$

$\approx 56.000515...$

We have a parametric function giving us cartesian coordinates for the specified parameter:

$f\left(t\right)=(\frac{{\left(\mathrm{ln}t\right)}^{2}}{t},{t}^{3})$

This with $t=2$ we get:

$f\left(2\right)=(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2},8)$

And with $t=4$ we get:

$f\left(4\right)=(\frac{{\left(\mathrm{ln}4\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=(\frac{{\left({\mathrm{ln}2}^{2}\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=(\frac{{\left(2\mathrm{ln}2\right)}^{2}}{4},64)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=({\left(\mathrm{ln}2\right)}^{2},64)$

Step 2

The distance. d, between these two cartesian coordinates is given by pythagoras:

$d}^{2}={({\left(\mathrm{ln}2\right)}^{2}-\frac{{\left(\mathrm{ln}2\right)}^{2}}{2})}^{2}+{(64-8)}^{2$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={\left(\frac{{\left(\mathrm{ln}2\right)}^{2}}{2}\right)}^{2}+{\left(56\right)}^{2}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136$

Giving: $d=\sqrt{\frac{{\left(\mathrm{ln}2\right)}^{4}}{4}+3136}$

$\approx 56.000515...$

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