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

For $f\left(t\right)=\left(\frac{1}{2t-3},\frac{t}{\sqrt{{t}^{2}+1}}\right)$ what is the distance between f(0) and f(1)?
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Step 1
Find the points, then use the distance formula to calculate the distance between them.
$f\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$
The first part the parametric equation gives the x coordinate and the second part gives the y coordinate.
$f\left(0\right)=\left(\frac{1}{2\left(0\right)-3},\frac{0}{\sqrt{{0}^{2}+1}}\right)=\left(-\frac{1}{3},0\right)$
$f\left(1\right)=\left(\frac{1}{2\left(1\right)-3},\frac{1}{\sqrt{{1}^{2}+1}}\right)=\left(-1,\frac{1}{\sqrt{2}}\right)$
The distance between $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is $d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$ so the distance between $f\left(0\right)$ and $f\left(1\right)$ is given by:
$d=\sqrt{{\left(-1-\left(-\frac{1}{3}\right)\right)}^{2}+{\left(\frac{1}{\sqrt{2}}-0\right)}^{2}}$
$d=\sqrt{{\left(-\frac{2}{3}\right)}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}}$
$d=\sqrt{\frac{4}{9}+\frac{1}{2}}=\sqrt{\frac{17}{18}}=\frac{\sqrt{17}}{3\sqrt{2}}$