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

For $f\left(t\right)=\left(\frac{1}{2t-3},t{e}^{t}\right)$ what is the distance between $f\left(0\right)$ and $f\left(1\right)$ ?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Feriheashrz
Step 1
We have $f\left(t\right)=\left(\frac{1}{2t-3},t{e}^{t}\right)$ , so:
$f\left(0\right)=\left(-\frac{1}{3},0\right)$
$f\left(1\right)=\left(-1,e\right)$
So then by Pythagoras, the distance, d, between $f\left(0\right)$ and $f\left(1\right)$ is given by:
${d}^{2}={\left(-1-\left(-\frac{1}{3}\right)\right)}^{2}+{\left(e-0\right)}^{2}$

And so:
$d=\sqrt{\frac{4}{9}+{e}^{2}}$
###### Did you like this example?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee