For $f\left(t\right)=(\frac{1}{2t-3},\frac{t}{\sqrt{{t}^{2}+1}})$ what is the distance between f(0) and f(1)?

Alissa Hancock
2022-07-01
Answered

For $f\left(t\right)=(\frac{1}{2t-3},\frac{t}{\sqrt{{t}^{2}+1}})$ what is the distance between f(0) and f(1)?

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Elijah Benjamin

Answered 2022-07-02
Author has **10** answers

Step 1

Find the points, then use the distance formula to calculate the distance between them.

$f\left(t\right)=(x\left(t\right),y\left(t\right))$

The first part the parametric equation gives the x coordinate and the second part gives the y coordinate.

$f\left(0\right)=(\frac{1}{2\left(0\right)-3},\frac{0}{\sqrt{{0}^{2}+1}})=(-\frac{1}{3},0)$

$f\left(1\right)=(\frac{1}{2\left(1\right)-3},\frac{1}{\sqrt{{1}^{2}+1}})=(-1,\frac{1}{\sqrt{2}})$

The distance between $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ is $d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$ so the distance between $f\left(0\right)$ and $f\left(1\right)$ is given by:

$d=\sqrt{{(-1-(-\frac{1}{3}))}^{2}+{(\frac{1}{\sqrt{2}}-0)}^{2}}$

$d=\sqrt{{(-\frac{2}{3})}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}}$

$d=\sqrt{\frac{4}{9}+\frac{1}{2}}=\sqrt{\frac{17}{18}}=\frac{\sqrt{17}}{3\sqrt{2}}$

Find the points, then use the distance formula to calculate the distance between them.

$f\left(t\right)=(x\left(t\right),y\left(t\right))$

The first part the parametric equation gives the x coordinate and the second part gives the y coordinate.

$f\left(0\right)=(\frac{1}{2\left(0\right)-3},\frac{0}{\sqrt{{0}^{2}+1}})=(-\frac{1}{3},0)$

$f\left(1\right)=(\frac{1}{2\left(1\right)-3},\frac{1}{\sqrt{{1}^{2}+1}})=(-1,\frac{1}{\sqrt{2}})$

The distance between $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ is $d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$ so the distance between $f\left(0\right)$ and $f\left(1\right)$ is given by:

$d=\sqrt{{(-1-(-\frac{1}{3}))}^{2}+{(\frac{1}{\sqrt{2}}-0)}^{2}}$

$d=\sqrt{{(-\frac{2}{3})}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}}$

$d=\sqrt{\frac{4}{9}+\frac{1}{2}}=\sqrt{\frac{17}{18}}=\frac{\sqrt{17}}{3\sqrt{2}}$

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