What is the arclength of $(ln (t + 3), \frac{ln t}{t})$ $(ln(t+3),lnt/t)$ on $t \in [1, 2]$ $t in [1,2]$ ?

Question

1398 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-09T09:46:49

$\approx 0.4287742274$ $approx 0.4287742274$

We have

$x (t) = ln (t + 3)$ $x(t)=ln(t+3)$ then

$x'(t)=1/(t+3)$

$y(t)=ln(t)/t$

then

$y'(t)=(1/t*t-ln(t))/t^2$

$y'(t)=(1-ln(t))/t^2$
and we have to integrate
$int_1^2sqrt((1/(t+3))^2+((1-ln(t))/t^2)^2)dt$
by a numerical method we get

$approx 0.4287742274$

What is the arclength of (ln(t+3),lnt/t)(ln(t+3),lntt)(ln(t+3),lnt/t) on t in [1,2]t∈[1,2]t in [1,2]?