skynugurq7

2022-07-03

For $f\left(t\right)=(\frac{{e}^{t}}{t},4t+\frac{1}{t})$ what is the distance between f(2) and f(5)?

gutinyalk

Beginner2022-07-04Added 11 answers

Step 1

We will substitute first 2 and then 5:

$f\left(2\right)=(\frac{{e}^{2}}{2},4\cdot 2+\frac{1}{2})=(\frac{{e}^{2}}{2},\frac{17}{2})$

$f\left(5\right)=(\frac{{e}^{5}}{5},4\cdot 5+\frac{1}{5})=(\frac{{e}^{5}}{5},\frac{101}{5})$

Step 2

Then we will apply the distance formula:

$d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

$d=\sqrt{{(\frac{{e}^{5}}{5}-\frac{{e}^{2}}{2})}^{2}+{(\frac{101}{5}-\frac{17}{2})}^{2}}$

$=\sqrt{{\left(\frac{2{e}^{5}-5{e}^{2}}{10}\right)}^{2}+{\left(\frac{101\cdot 2-17\cdot 5}{10}\right)}^{2}}$

$=\sqrt{\frac{4{e}^{10}-20{e}^{7}+25{e}^{4}}{100}+\frac{202-85}{100}}$

$=\frac{\sqrt{4{e}^{10}-20{e}^{7}+25{e}^{4}+117}}{10}$

We will substitute first 2 and then 5:

$f\left(2\right)=(\frac{{e}^{2}}{2},4\cdot 2+\frac{1}{2})=(\frac{{e}^{2}}{2},\frac{17}{2})$

$f\left(5\right)=(\frac{{e}^{5}}{5},4\cdot 5+\frac{1}{5})=(\frac{{e}^{5}}{5},\frac{101}{5})$

Step 2

Then we will apply the distance formula:

$d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

$d=\sqrt{{(\frac{{e}^{5}}{5}-\frac{{e}^{2}}{2})}^{2}+{(\frac{101}{5}-\frac{17}{2})}^{2}}$

$=\sqrt{{\left(\frac{2{e}^{5}-5{e}^{2}}{10}\right)}^{2}+{\left(\frac{101\cdot 2-17\cdot 5}{10}\right)}^{2}}$

$=\sqrt{\frac{4{e}^{10}-20{e}^{7}+25{e}^{4}}{100}+\frac{202-85}{100}}$

$=\frac{\sqrt{4{e}^{10}-20{e}^{7}+25{e}^{4}+117}}{10}$

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