What is the arc length of $f(t)=(lnt,5-lnt)$ over $t in [3,4]$ ?

Question

1492 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-29T14:32:51

Arc length $l= sqrt(2)*(ln 4-ln 3)$

$l=0.406844$

$l=int sqrt((dy/dt)^2+(dx/dt)^2) dt$

Solve for $dx/dt$ given $x=ln t$

$dx/dt=(1/t)$

Solve for $dy/dt$ given $y=5-ln t$

$dy/dt=-1*1/t$

Solve for the length $l$

$l=int_a^b sqrt((dy/dt)^2+(dx/dt)^2) dt$

$a=3$ and $b=4$

$l=int_3^4 sqrt((dy/dt)^2+(dx/dt)^2) dt$

$l=int_3^4 sqrt((-1/t)^2+(1/t)^2) dt$

$l=int_3^4 sqrt(2/t^2) dt$

$l=int_3^4 sqrt(2)/t dt$

$l= sqrt(2)*(ln 4-ln 3)$

$l=0.406844$

God bless....I hope the explanation is useful.

What is the arc length of f(t)=(lnt,5-lnt) f(t)=(lnt,5-lnt) over t in [3,4] t in [3,4] ?