Amber Quinn

2022-06-25

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

SuefsSeeltHeRn8

Expert

Step 1
Find the points first:
$f\left(2\right)=\left(\frac{1}{1-2},-\frac{1}{1+{2}^{2}}\right)=\left(-1,-\frac{1}{5}\right)$
$f\left(5\right)=\left(\frac{1}{1-5},-\frac{1}{1+{5}^{2}}\right)=\left(-\frac{1}{4},-\frac{1}{26}\right)$
The distance between the points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is calculated through $d=\sqrt{{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({x}_{2}-{x}_{1}\right)}^{2}}$
$d=\sqrt{{\left(-\frac{1}{26}-\left(-\frac{1}{5}\right)\right)}^{2}+{\left(-\frac{1}{4}-\left(-1\right)\right)}^{2}}$
$d=\sqrt{{\left(\frac{21}{130}\right)}^{2}+{\left(\frac{3}{4}\right)}^{2}}$
Step 2
You can go ahead and make this a decimal answer, but you can also do some nice simplification if you feel up to it.
$d=\sqrt{{\left(\frac{3\cdot 7}{2\cdot 5\cdot 13}\right)}^{2}+{\left(\frac{3}{{2}^{2}}\right)}^{2}}$
$d=\sqrt{\frac{{3}^{2}\cdot {7}^{2}}{{2}^{2}\cdot {5}^{2}\cdot {13}^{2}}+\frac{{3}^{2}}{{2}^{4}}}$
$d=\sqrt{\frac{{2}^{2}\cdot {3}^{2}\cdot {7}^{2}}{{2}^{4}\cdot {5}^{2}\cdot {13}^{2}}+\frac{{3}^{2}\cdot {5}^{2}\cdot {13}^{2}}{{2}^{4}\cdot {5}^{2}\cdot {13}^{2}}}$
$d=\sqrt{\frac{{3}^{2}\left({2}^{2}\cdot {7}^{2}+{5}^{2}\cdot {13}^{2}\right)}{{2}^{4}\cdot {5}^{2}\cdot {13}^{2}}}$
$d=\frac{3}{{2}^{2}\cdot 5\cdot 13}\sqrt{{\left(2\cdot 7\right)}^{2}+{\left(5\cdot 13\right)}^{2}}$
$d=\frac{3\sqrt{4421}}{260}\approx 0.767199$

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