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

myntfalskj4
2022-07-16
Answered

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

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Feriheashrz

Answered 2022-07-17
Author has **8** answers

Step 1

We have $f\left(t\right)=(\frac{1}{2t-3},t{e}^{t})$ , so:

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

$f\left(1\right)=(-1,e)$

So then by Pythagoras, the distance, d, between $f\left(0\right)$ and $f\left(1\right)$ is given by:

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

$={(-\frac{2}{3})}^{2}+{\left(e\right)}^{2}$

$=\frac{4}{9}+{e}^{2}$

And so:

$d=\sqrt{\frac{4}{9}+{e}^{2}}$

We have $f\left(t\right)=(\frac{1}{2t-3},t{e}^{t})$ , so:

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

$f\left(1\right)=(-1,e)$

So then by Pythagoras, the distance, d, between $f\left(0\right)$ and $f\left(1\right)$ is given by:

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

$={(-\frac{2}{3})}^{2}+{\left(e\right)}^{2}$

$=\frac{4}{9}+{e}^{2}$

And so:

$d=\sqrt{\frac{4}{9}+{e}^{2}}$

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