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

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

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Answer 1 · 2016-04-19T11:11:08

The distance between $f (0)$ $f(0)$ and $f (1)$ $f(1)$ is $0.866$ $0.866$ .

Explanation:

The coordinates of point for $f (0)$ $f(0)$ is $(\frac{0}{0 - 3}, \frac{0}{\sqrt{0^{2} + 1}})$ $(0/(0-3),0/sqrt(0^2+1))$ or $(0, 0)$ $(0,0)$

and for $f (1)$ $f(1)$ we have $(\frac{1}{1 - 3}, \frac{1}{\sqrt{1^{2} + 1}})$ $(1/(1-3),1/sqrt(1^2+1))$ or $(- \frac{1}{2}, \frac{1}{\sqrt{2}})$ $(-1/2,1/sqrt2)$

The distance between the two points is

$\sqrt{{(0 - (- \frac{1}{2}))}^{2} + {(0 - \frac{1}{\sqrt{2}})}^{2}} = \sqrt{\frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = \frac{1.732}{2} = 0.866$ $sqrt((0-(-1/2))^2+(0-1/sqrt2)^2)=sqrt(1/4+1/2)=sqrt(3/4)=sqrt3/2=1.732/2=0.866$

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

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