For $f (t) = (\frac{e^{t}}{t}, 4 t + \frac{1}{t})$ $f(t)= (e^t/t,4t+1/t)$ what is the distance between $f (2)$ $f(2)$ and $f (5)$ $f(5)$ ?

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Answer 1 · 2017-11-02T05:42:42

The distance is $\frac{\sqrt{4 e^{10} - 20 e^{7} + 25 e^{4} + 117}}{10}$ $sqrt(4e^10-20e^7+25e^4+117)/10$

we will substitute first 2 and then 5:

$f (2) = (\frac{e^{2}}{2}, 4 \cdot 2 + \frac{1}{2}) = (\frac{e^{2}}{2}, \frac{17}{2})$ $f(2)=(e^2/2,4*2+1/2)=(e^2/2,17/2)$

$f (5) = (\frac{e^{5}}{5}, 4 \cdot 5 + \frac{1}{5}) = (\frac{e^{5}}{5}, \frac{101}{5})$ $f(5)=(e^5/5,4*5+1/5)=(e^5/5,101/5)$

Then we will apply the distance formula:

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$d = \sqrt{{(\frac{e^{5}}{5} - \frac{e^{2}}{2})}^{2} + {(\frac{101}{5} - \frac{17}{2})}^{2}}$ $d=sqrt((e^5/5-e^2/2)^2+(101/5-17/2)^2)$

$= \sqrt{{(\frac{2 e^{5} - 5 e^{2}}{10})}^{2} + {(\frac{101 \cdot 2 - 17 \cdot 5}{10})}^{2}}$ $=sqrt(((2e^5-5e^2)/10)^2+((101*2-17*5)/10)^2)$

$= \sqrt{\frac{4 e^{10} - 20 e^{7} + 25 e^{4}}{100} + \frac{202 - 85}{100}}$ $=sqrt((4e^10-20e^7+25e^4)/100+(202-85)/100)$

$= \frac{\sqrt{4 e^{10} - 20 e^{7} + 25 e^{4} + 117}}{10}$ $=sqrt(4e^10-20e^7+25e^4+117)/10$

For f(t)=(ett,4t+1t)f(t)= (e^t/t,4t+1/t) what is the distance between f(2)f(2) and f(5)f(5)?