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

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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.